Class 11 Engineering Maths - Linear Inequalities

Solve the following inequalities graphically in two-dimensional plane:
$x >- 3$

Yellow line is

$x=-3$

$x>-3$

$x\in(-3,\infty)$

Solve 4x - 12 $\displaystyle \geq$ 0 graphically in two dimensional plane

Graph of 4x - 12

$\displaystyle \geq$ 0 is given in the fig We select a point say (0, 0) and substituting it in given inequality we see that
4(0) - 12

$\displaystyle \geq$ 0 or -12

$\displaystyle \geq$ 0 which is false Thus the solution region is the shaded region on the right hand side of the line x = 3

1063943_311108_ans_c16db2b861e14d36adbcfe752dab4f32.bmp

1063943_311108_ans_c16db2b861e14d36adbcfe752dab4f32.bmp

Solve the given inequalities graphically:
$x {\geq} 3, y {\geq} 2$

$x {\geq} 3, y {\geq} 2$

$\Rightarrow x\in [3,\infty), y\in [2,\infty)$
Blue line is

$x=3$ and yellow line is

$y=2$
Now we have to look over the region where

$x\geq3\;\; and\;\; y\geq2$

Solve the inequalities and represent the solution graphically on number line.
$5x+1 > -24, 5x-1 < 24$

We have,

$5x+1>-24\Rightarrow 5x>-24-1=-25\Rightarrow x>-5$
and

$5x-1<24\Rightarrow 5x<24+1=25\Rightarrow x<5$
Thus combining above we get

$-5< x< 5$
Solution is shown in above graph.

Solve $3x+2y>6$ graphically.

1323238_1063156_ans_246a21cbaf4345a79965681460f928f7.jpg

State which of the following are equation (with variable). Given reasons for your an mention which variable is used from the equation with a variable.
$9n-9>5$

$9n-9>5$

No, It is not an equation with variable there is no equal sign in the expression

Solve: $|2x-5|< 1$ .

$|2x−5| \lt 1$

$2x−5 \lt ^+_- 1$

On solving: $2x -5 \lt 1$

We get: $2x \lt 6$

$x \lt 3$

On Solving: $2x−5 \gt – 1$

We get : $2x \gt 4$

$x \gt 2$

Hence, $x \in (2,3)$

Hence the value of x lies in $x \in (2,3)$

Find the solution of the inequality $3\left( 1-x \right) <2\left( x+4 \right)$ and draw on the number line.

The number line is shown in the figure above.

The dark lint shows the values of

$x$

The solution is given below:

$3(1-x)<2(x+4) \Rightarrow 3-3x < 2x+8 \Rightarrow x>-1$

Solve: $-(x-2)+4 > - 3x+10$ .

$-(x-2)+4>-3x+10$

$\implies -x+2+4>-3x+10$

On rearranging, we have

$-x+3x>-6+10$

$\implies 2x>4$

$\implies x>2$

This is the required solution.

Fill in the blank with $>,\ <$ or $=$ sign
$(-25)-(-42)$ ________ $(-42)-(-25)$

According to question,

L.H.S=

$-25+42$

$17$

R.H.S=

$-42+25$

$-17$

L.H.S is greater than R.H.S

So, sign

$>$ is used to fill the blank

Fill in the blank with $>,\ <$ or $=$ sign
$45-(-11)$ ________ $57+(-4)$

According to question,

L.H.S=

$45+11$

$56$

R.H.S=

$57-4$

$53$

L.H.S is greater than R.H.S

So, sign

$>$ is used to fill the blank

Solve $\dfrac{3}{x-2}<1$ , when $x\in R$

$\dfrac{3}{x-2}-1<0\\ \Rightarrow \dfrac{3-x+2}{x-2}<0\\ \Rightarrow \dfrac{5-x}{x-2}<0$

$\therefore (5-x < 0\ and\ x-2 > 0)$ or

$(5-x>0\ and\ x-2 <0)$

$\Rightarrow (x>5\ and\ x>2)$ or

$(x<5\ and x<2)$

$\Rightarrow (x>5\ or\ x<2)\Rightarrow x\in(-\infty,2)\cup (5,\infty)$

Solve :
$3x-2y=1$
$2x+y=3$

$3x-2y=1\\2x+y=3\\y=3-2x\\3x-2(3-2x)=1\\$

$3x-6+4x=1\\7x=7\\x=1\\3(1)-2y=1\\$

$1-2y=1\\2y=2\\y=1$

State True or False and write the correct statement.
In the cartesian plane the horizontal line is called Y-axis.
In the cartesian plane the vertical line is called Y-axis.
The point which lies both the axes is called origin.
The point $(2,-3)$ lies in the third quadrant.
$(-5,-8)$ lies in the fourth quadrant.
The point (-x,-y) lies in the first quadrant where $x<0, y,0.$

1. False. In cartesian plane, the horizontal line is called X-axis.

2. True.

3. True.

4. False. The point (2, -3) lies in the fourth quadrant where x coordinate is positive and y coordinate is negative.

5. False. The point (-5, -8) lies in the third quadrant where both x & y coordinates are negative.

6. True.

Solve the inequalities for real $x$ -
$3\left( 2-x \right) \ge 2\left( 1-x \right)$

Consider the given inequality.

$3\left( 2-x \right)\ge 2\left( 1-x \right)$

$6-3x\ge 2-2x$

$6-2\ge 3x-2x$

$4\ge x$

$x\le 4$

Hence, this is the answer.

Find the pairs of consecutive even positive integers, both of which are larger than $5$ such that their sum is less than $23$ .

Let one number is

$x$ and another number is

$x+2$ that are consecutive even numbers greater than

$5$ then according to the given condition-

$x+x+2<23$

$2x+2<23$

$x=10.5$

Hence, the pair of numbers are

$(6,8),(8,10),(10,12).$

Solve the following inequation and represent the solution set on a number line
$-8\cfrac{1}{2}< -\cfrac{1}{2}-4x\le 7\cfrac{1}{2},x\in 1$

$-8\dfrac{1}{2} < \dfrac{-1}{2} -4x \leq 7\dfrac{1}{2}$

$\dfrac{-17}{2}+\dfrac{1}{2} < -4x \leq \dfrac{15}{2}+\dfrac{1}{2}$

$-8 < -4x \leq 8$

$8 > 4x \geq -8$

$2 > x \geq -2$

1076392_1098485_ans_833eb9d873b24354b6f904f0fa9a871a.png

Represent the following inequalities on real number line:
$-5<x \leq-1$

$-5<x \leq-1$

Solution on the number line is shown in figure.

1792025_1827948_ans_a550e83a3dad411683173b1a4ed243b6.png

Enter 1 if it is true else enter 0.
If $x\, \epsilon\, W$ , then the solution set of $x<2$ is $x = \{0,1\}$

1887777_1803211_ans_eb3020f3f5bf45e4bd177b117f088217.jpeg

Fill in the blanks with correct inequality sign $(>, <, \ge, \le )$ .
$p-q=-3\Rightarrow p.....q$

$p-q=-3\Rightarrow p<q$ .

Solve $2(x – 3) < 1, x ∈ {1, 2, 3, …. 10}$

Given inequation,

$2(x - 3) < 1$

$\Rightarrow 2x - 6 < 1$

$\Rightarrow 2x < 7$

$\Rightarrow x < \dfrac72$

But, $x \in \{1, 2, 3, …. 10\}$

Hence, the solution set is $\{1, 2, 3\}$

Solve the equation
$7 \leq \dfrac{3 x+11}{2} \leq 11$

$Given$

$7 \leq \dfrac{3 x+11}{2} \leq 11$

$\Rightarrow 14 \leq 3 x+11 \leq 22$

$\Rightarrow 14-11 \leq 3 x \leq 22-11$

$\Rightarrow \dfrac{3}{3} \leq x \leq \dfrac{11}{3}$

$\Rightarrow 1 \leq x \leq \dfrac{11}{3}$

$\therefore x \in\left[1, \dfrac{11}{3}\right]$

If $x \in I$ , find the smallest value of $x$ which satisfies the inequation $2x+\dfrac{5}{2}>\dfrac{5x}{3}+2$

Given inequation,

$\ \ 2x+\dfrac{5}{2}>\dfrac{5x}{3}+2$

$\Rightarrow \dfrac{4x + 5}2 > \dfrac{5x + 6}3 \qquad\text{[ Taking L.C.M.]}$

$\Rightarrow 3 (4x + 5) > 2 (5x + 6) \qquad\text{[ Multiplying both sides by 6 ]}$

$\Rightarrow 12x + 15 > 10x + 12$

$\Rightarrow 12x - 10x > 12 - 15$

$\Rightarrow 2x > -3$

$\Rightarrow x > -\dfrac32$

Hence, for $x \in I$ the smallest value of $x$ is $-1$ .

Solve the inequation $- 3 \le 3 - 2x < 9, x \in R.$ Represent your solution on a number line.

Given inequation, $- 3 ≤ 3 - 2x < 9$

$\Rightarrow - 3 - 3 ≤ - 2x < 9 - 3$

$\Rightarrow -6 ≤ -2x < 6$

$\Rightarrow -3 ≤ -x < 3$

$\Rightarrow -3 < x ≤ 3$

As $x \in R$

The solution set is $\{x: x \in R, -3 < x ≤ 3\}$

Representing the solution on a number line:

1663878_1782935_ans_4f416fe36acd4ed58743724dcecc5f29.png

$Solve:$ $4x + 3 < 5x + 7$

$Given:$

$4x + 3 < 5x + 7$

$\Rightarrow -7 + 3 < 5x - 4x$

$\Rightarrow -4 < x$

$\therefore$

$Solution\space Set\space = (-4, \infty).$

Solve the inequality and show the graph of the solution on number line:
$3 (1 -x) < 2 (x + 4)$

Given,

$3(1-x)< 2(x+4)$

${\Rightarrow 3-3x < 2x +8}$

${\Rightarrow 3-8 <2x +3x}$

${\Rightarrow -5 < 5x}$

$\Rightarrow x>-1 \text{or} \;\; x\in (-1, \infty)$
Blue line is

$y=3-3x$ and red line is

$y=2x+8$
Both line intersect at

$x=-1$ and It is clearly observed from graph that
For

$x>-1\;\; \Rightarrow\;\; 2x+8>3-3x$

Solve & graph the solution set of $3x + 6 \geq 9$ and $-5x > -15$ , $x \in R$ .

$\displaystyle 3x+6\geq 9\: \: and\: \: -5x>-15$

$\displaystyle \Rightarrow 3x\geq 3$ and

$-x>-3$

$\displaystyle \Rightarrow x\geq 1$ and

$x< 3$
combining the solution So the solution is

$\displaystyle x\in [1,3)$

Solve & graph the solution set of $\displaystyle 3x+6\geq 9$ and -5x > -15 $\displaystyle x\: \epsilon \: R$

$\displaystyle 3x+6\geq 9\: and\: -5x>-15$

$\displaystyle \Rightarrow 3x\geq 3\Rightarrow -x>-3$

$\displaystyle \Rightarrow x\geq 1\: \: \: \: \Rightarrow \: \: x<3$
Combining the solution
So the solutions is

$\displaystyle x\in [1,3)$
1071552_329126_ans_2ef274bc78f94373bb343e7b51fb79bb.bmp

Solve $24 x < 100$ , when $x$ is a natural number.

$24x < 100 \Rightarrow \cfrac{24x}{24}< \cfrac{100}{24}\Rightarrow x<\cfrac{25}{6}$
When

$x$ is a natural number, the following values of

$x$ make the statement true:

$1, 2, 3, 4$

Solve the inequality and show the graph of the solution on number line:
$3x- 2 < 2x + 1$

Given,

$3x -2 < 2x +1$

${\Rightarrow 3x -2x <1 +2}$

${\Rightarrow x<3} \;\;\text{or} \;\;x\in (-\infty, 3)$
The lines

$y=3x-2$ and

$y=2x+1$ both will intersect at

$x=3$

Clearly, the dark line shows the solution of

$3x-2<2x+1$ .

Solve the inequality and show the graph of the solution on number line:
$\displaystyle {\frac{x}{2}\geq \frac{(5x-2)}{3}-\frac{(7x-3)}{5}}$

Given,

$\displaystyle{\frac{x}{2}\geq \frac{(5x-2)}{3}-\frac{(7x-3)}{5}}$

$\displaystyle{\Rightarrow \frac{x}{2}\geq \frac{5(5x-2)-3(7x-3)}{15}}$

${\Rightarrow\displaystyle \frac{x}{2}\geq \frac{25x-10-21x +9}{15}}$

${\Rightarrow \displaystyle\frac{x}{2}\geq \frac{4x-1}{15}}$

${\Rightarrow\displaystyle 15 x \geq 2(4x-1)}$

${\Rightarrow\displaystyle 15 x \geq 8x-2}$

${\Rightarrow \displaystyle 7x \geq -2}$

${\Rightarrow \displaystyle x\geq -\frac{2}{7}}$

Solve the inequality and show the graph of the solution on number line:
$5x -3 \geq 3x- 5$

Given,

${ 5x -3 \geq 3x -5}$

${\Rightarrow 5x -3x \geq -5 +3}$

${\Rightarrow 2x \geq -2}$

${\Rightarrow x\geq -1} \;\; \text{or} \;\; x\in [-1, \infty)$

Solve the following system of inequalities graphically
$\displaystyle x+2y\leq 8$ ......(1)
$\displaystyle 2x+y\leq 8$ ......(2)
$\displaystyle x\geq 0$ ......(3)
$\displaystyle y\geq 0$ ......(4)

We draw the graphs of the lines x + 2y = 8 and 2x + y = 8 The inequality (1) and (2) represent the region below the two lines including the point on the respective lines
Since

$\displaystyle x\geq 0, y\geq 0$ every point in the shaded region in the first quadrant represent a solution of the given system of inequalities
1063946_311159_ans_891d3934530e496e8780cd540ecc9c70.bmp

1063946_311159_ans_891d3934530e496e8780cd540ecc9c70.bmp

Find x satisfying | x - 5 | $\displaystyle \leq$ 3

$\displaystyle |x-a|\leq r\Leftrightarrow a-r\leq x\leq a+r$ i.e.

$\displaystyle x\in [a-r,a+r]$

$\displaystyle \therefore |x-5|\leq 3\Leftrightarrow 5-3\leq x\leq 5+3$ i.e.

$\displaystyle 2\leq x\leq 8$ i.e.

$\displaystyle x\in [2,8]$
1071584_329165_ans_4d9b3e6a72924a4eb0af14da85422f07.bmp

$2x -3y > 6$

$2x-3y>6$

$3y<2x-6$

$y<\dfrac{2x-6}{3}$
So we have the region below the line.away from origin.

Solve the following inequalities graphically in two-dimensional plane:
$y < -2$

$y\lt-2$ is the shaded region shown in the graph.

If we put

$O(0,0)$ in the LHS,we get LHS

$=0$

Now

$LHS>-2$

Means

$O(0,0)$ lies on opposite side of "less than " side of the line.

Hence required region is "non-origin" side of the line.

Solve the following inequalities graphically in two-dimensional plane:
3x + 4y ${ \leq}$ 12

Put

$O(0,0)$ in the LHS we get

$LHS=0$

$0<12$

So origin is on "less than " side of the line.

$x + y < 5$

$x+y\lt5$ is the shaded region shown in the graph.

If we put

$O(0,0)$ in the LHS,we get LHS

$=0$

Now

$LHS<5$

Means

$O(0,0)$ lies on "less than " side of the line.

Hence required region is "origin" side of the line.

Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23

Let

$x$ be the smaller of the two consecutive even positive integers .

Then the other integer is

$x+2.$
Since both the integers are larger than

$5, x > 5$ ....(1)
Also the sum of the two integers is less than

$23$ .

$x+(x+2)<23$

${\Rightarrow 2x +2< 23}$

${\Rightarrow 2x <23 -2}$

${\Rightarrow 2x <21}$

${\Rightarrow x< \dfrac{21}{2}}$

${\Rightarrow x< 10.5}$ ....(2)
From (1) and (2) we obtain

$5 < x< 10.5.$
Since

$x$ is an even number,

$x$ can take the values

$6,8$ and

$10.$
Thus the required possible pairs are

$(6,8), (8,10)$ and

$(10,12)$ .

Solve the following inequalities graphically in two-dimensional plane:
$-3x + 2y {\geq} -6$

$-3x + 2y \ge -6$

Lets first draw graph of

$x$

$0$

$2$

$y$

$-3$

$0$

$-3x + 2y = -6$

Putting

$x = 0$ in (1)

$-3(0) + 2y = -6$

$0 + 2y = -6$

$2y = -6$

$y = \dfrac{-6}{2}$

$y = -3$

Putting

$y = 0$ in (1)

$-3x + 2(0) = -6$

$-3x + 0 = -6$

$-3x = -6$

$x = \dfrac{-6}{-3}$

$x = 2$

Drawing graph

$x$

$0$

$2$

$y$

$-3$

$0$

Point to be plotted are

$(0, -3), (2, 0)$

Checking for

$(0,0)$

Putting

$x = 0, y = 0$

$-3x + 2y \ge -6$

$-3(0) + 2(0) \ge -6$

$0 \ge -6$

Which is true

Hence origin lies in the plane

$-3x + 2y \ge -6$ .

So, we shade left side of line.

1521818_419403_ans_e4b23168005442468e7abd47ce608143.png

$x -y {\leq} 2$

$x-y\le2$ is the shaded region shown in the graph.

If we put

$O(0,0)$ in the LHS,we get LHS

$=0$

Now

$LHS<0$

Means

$O(0,0)$ lies on "less than " side of the line.

Hence required region is "origin" side of the line.

The longest side of a triangle is $3$ times the shortest side and the third side is $2$ cm shorter than the longest side. If the perimeter of the triangle is at least $61$ cm, find the minimum length in cm. of the shortest side.

Let the length of the shortest side of the triangle be

$x$ cm.
Then length of the longest side

$= 3x$ cm.
Thus the length of the third side

$=(3x-2)$ cm.
Since the perimeter of the triangle is at least

$61$ cm,

$x+3x+(3x-2)$

${\geq}$

$61$

${\Rightarrow 7x -2 \geq 61}$

${\Rightarrow 7x \geq 61 +2}$

${\Rightarrow 7x \geq 63}$

${\Rightarrow x\geq 9}$
Thus the minimum length of the shortest side is

$9$ cm.

Solve the following inequalities graphically in two-dimensional plane: $2x + y {\geq} 6$

The graphical representation of

$2x+y\geq 6$ is shown in the graph. Shaded region represents the given condition.

$y + 8 {\geq}2x$

$y\geq2x-8$
So all the points lies on line or lying origin side will be your answer

Solve the given inequalities graphically:
$x + y$ ${\leq}$ $6, x + y$ ${\geq}$ $4$

Blue line is

$x+y=6$ and red line is

$x+y=4$
Now we have to look the region which is below or on the

$x+y=6$ line and above or on the

$x+y=4$ line.
So required region is in between the two lines.

Solve the given inequalities graphically:
$x + y$ ${\geq}$ $4$ and $2x -y > 0$

Given equations are

$x+y\ge 4, 2x-y>0$

Let

$f(x)=2x$ and

$g(x)=4-x$

The graph of these equations in shown in figure.

Solve the following inequations graphically:
$2x- y >1, x -2y < -1$

Solve the system of inequalities graphically:
$2x + y$ ${\geq}$ $4, x + y$ ${\leq}$ $3, 2x- 3y$ ${\leq}$ $6$

$2x+y>4$ for all set of points which lies above the line

$2x+y=4$

$x+y<3$ for all set of points which lies below the line

$x+y=3$

solution is shaded part.

Solve the system of inequalities graphically:
${x- 2y \leq 3, 3x + 4y \geq 12, x \geq 0 , y \geq 1}$

First we solve

$x-2y \le 3$

Lets first draw graph of

$x-2y=3$

Putting

$x=0$ in (1)

$0-2y=3$

$-2y=3$

$y=\dfrac{3}{-2}$

$y=-1.5$

Putting

$y=0$ in (1)

$x-2(0)=3$

$x-0=3$

$x=3$

Refer image 1.

$\begin{matrix} x & 0 & 3 \\ y & -1.5 & 0 \end{matrix}$

Points to be plotted are

$(0,6),(3,0)$

Checking for

$(0,0)$

Putting

$x=0$ ,

$y=0$

$x-2y\le 3$

$0-2(0)\le 3$

$0\le 3$

Which is true Hence origin lies in plane

$x-2y\le 3$

So, we shade left of line

Now we solve

$3x+4y\ge 12$

Lets first draw graph of

$3x+4y=12$

Putting

$x=0$ in (1)

$3(0)+4y=12$

$0+4y=12$

$4y=12$

$y=\dfrac{12}{4}$

$y=3$

Putting

$y=0$ in (1)

$3x+4(0)=12$

$3x+0=12$

$3x=12$

$x=\dfrac{12}{3}$

$x=4$

Refer image 2.

$\begin{matrix} x & 0 & 4 \\ y & 3 & 0 \end{matrix}$

Points to be plotted are

$(0,3),(4,0)$

checking for

$(0,0)$

Putting

$x=0,y=0$

$3+4y\ge 12$

$3(0)+2(0)\ge 12$

$0\ge 12$

which is false

Hence origin does not lie in plane

$3x+2y > 6$

So, we shade rights side of line.

Also

$y\ge 1$

So, for all values

$x$

Refer image 3.

$x\,\,\,\,\,\,0\,-1\,\,4$

$y\,\,\,\,\,\,1\,\,\,\,\,\,\,1\,\,\,\,1$

Points to be plotted are

$(0,1),(-1,1),(4,1)$

Also

$x\ge 0$

So, the shaded region will lie on the right side of y-axis

Hence the shaded region represents the given inequality.

1522859_419737_ans_9d08062e74e84a0bbea83c1b74484375.png

Solve the given inequalities graphically:
$2x + y$ ${\geq}$ $8, x + 2y$ ${\geq}$ $10$

Blue line is

$2x+y=8$ and red line is

$x+2y=10$
Now according to question we have to look over the region which

$2x+y\geq8\;\; and\;\;x+2y\geq10$ they intersect at point

$(2,4)$
So for

$x>2$ region is above or on the line

$x+2y=10$
and for

$x<2$ region is above or on the line

$2x+y=8$

Solve the given inequalities graphically:
$2x + y$ ${\geq}$ $6$ and $3x + 4y$ ${\leq}$ $12$

Given equations are

$2x+y\ge6\Rightarrow y=6-2x$

and

$3x+4y\le 12\Rightarrow y=\dfrac {12-3x}{4}$

Let

$f(x)=6-2x$

and

$g(x)=\dfrac {12-3x}{4}$

Intersection of this two lines is shown in figure.

Region is shown in the graph.

Solve the given inequalities graphically:
$3x + 2y$ ${\leq}$ $12, x$ ${\geq}$ $1, y$ ${\geq}$ $2$

Draw the line

$3x+2y=12, x = 1$ and

$y =2$ . Now check for the position of origin w.r.t. the given lines to determine the solution of the inequalities.

For

$3x+2y-12$ at

$(0,0)$ ,

$3(0)+2(0) -12 < 0$ . Hence,

$(0,0)$ lies in the inequality

$3x+2y \leq 12$

For

$x -1$ at

$(0,0)$ ,

$0 - 1 < 0$ . Hence,

$(0,0)$ doesn't lie in the inequality

$x \geq 1$

For

$y-2$ at

$(0,0)$ ,

$0-2<0$ . Hence,

$(0,0)$ doesn't lie in the inequality

$y \geq 2$ .

The shaded part in the above graph represents the solution of the given inequlities.

Solve the system of inequalities graphically:
$5x + 4y$ ${\leq}$ $20, x$ ${\geq}$ $1, y$ ${\geq}$ $2$

First we solve

$5x + 4y \ge 20$

Let first draw graph of

$x$

$0$

$4$

$y$

$5$

$0$

$5x + 4y = 20$ ___(1)

Putting

$x = 0$ in (1)

$5(0) +4y =20$

$0 + 4y = 20$

$4y = 20$

$y = \dfrac{20}{4}$

$y = 5$

Putting

$y = 0$ in (1)

$5x + 4(0) = 20$

$5x + 0 = 20$

$5x = 20$

$x = \dfrac{20}{5}$

$x = 4$

Drawing graph

$x$

$0$

$4$

$y$

$5$

$0$

Points to be plotted are

$(0, 5), (4, 0)$

Checking for

$(0,0)$

$5x + 4y \le 20$

Putting

$x = 0 , y = 0$

$5(0) + 4(0) \le 20$

$0 \le 20$

Which is true

Hence origin lies in plane

$5x + 4y \le 20$

So, we shade left side of line

REF. IMAGE 1

Also,

$y \ge 2$

So, for all values of

$x, where\ y = 2$

$x$

$0$

$-1$

$4$

$y$

$2$

Points to be plotted are

$(0,2), (-1, 2), (4, 2)$

Also,

$x \ge 1$

So, for all values of

$y, where \ x = 1$

$x$

$1$

$y$

$0$

$-1$

$3$

Point to be plotted are

$(1, 0), (1, -1), (1, 3)$

Hence the shaded region represents the given in equality

REF. IMAGE 2

Shaded region shows the intersection of given inequality.

1521810_419734_ans_8fb14707ec3243939f1a1039a4a81faa.png

Solve the given inequalities graphically:
$x + y {\leq} 9, y > x, x {\geq} 0$

Observe the lines

$x+y=9$ and

$y=x$
Now we have to look for

$x\geq0$ region
Region is area of triangle of formed by two lines and y axis as shown in graph

Solve, and express the answer graphically.
$\displaystyle\frac{4x+4}{x-4} \le 0$

Solve with an

$=$

$\displaystyle\frac{4x+4}{x-4} = 0$

$4x + 4 = 0$

$x = -1$
Critical values are

$x = -1$ and

$x = 4$ (denominator

$= 0$ )

$ck\left( -2 \right) :\displaystyle\frac { 4\left( -2 \right) +4 }{ \left( -2 \right) -4 } =\displaystyle\frac { -4 }{ -6 } =\displaystyle\frac { 2 }{ 3 }$ ; Not

$\le 0$

$ck\left( 0 \right) :\displaystyle\frac { 4\left( 0 \right) +4 }{ \left( 0 \right) -4 } =\displaystyle\frac { +4 }{ -4 } =-1$ ; yes

$\le 0$

$ck\left( 5 \right) :\displaystyle\frac { 4\left( 5 \right) +4 }{ \left( 5 \right) -4 } =\displaystyle\frac { +24 }{ 1 } =24$ ; Not

$\le 0$
Solution:

$-1 \le x < 4$
Graph:
1036081_427044_ans_b701069c6a814d67b75f0facd5fa92a9.png

Solve the inequalities and represent the solution graphically on number line.
$3x-7 >2(x-6), 6-x> 11-2x$

Red line is

$y=3x-7$
Green line is

$y=2x-12$
black line

$y=11-2x$
blue line is

$y=6-x$
Since

$3x-7>2x-12$ for all

$x>-5$
and

$6-x>11-2x$ for all

$x>5$

Solve the system of inequalities graphically
${3x + 2y \leq 150, x + 4y \leq 80, x \leq 15, y \geq 0, x \geq 0}$

solution is shaded part.

Solve the given inequalities graphically:
${x + 2y \leq 10, x + y \geq 1, x- y \leq 0, x \geq 0, y \geq 0}$

First we solve

$x+2y\le 10$

Lets first draw graph of

$x+2y=10$ ..........(1)

Putting

$x=0$ in (1)

$0+2y=10$

$2y=10$

$y=\dfrac{10}{2}$

$y=5$

$y=0$ in (1)

$x+2(0)=10$

$x+0=10$

$x=10$

Drawing graph

Refer fig.1

$x\,\,\,\,\,0\,\,10$

$y\,\,\,\,\,5\,\,\,\,0$

Points to be plotted are

$(0,5),(10,0)$

Checking for

$(0,0)$

Putting

$x=0,y=0$

$x+2y\le 10$

$0\le 10$

Which is true

So, we shade left of line.

Now we solve

$x+y\ge 1$

Lets first draw graph of

$x+y=1$ ..........(2)

Putting

$y=0$ in (1)

$x+0=1$

$x=1$

Putting

$x=0$ in (1)

$0+y=1$

$y=1$

Drawing graph

Refer fig .2

$x\,\,\,\,0\,\,1$

$y\,\,\,\,1\,\,0$

Points to be plotted are

$(0,1),(0,1)$

Checking for

$(0,0)$

Putting

$x=0,y=0$

$x+y\ge 1$

$0+0\ge 1$

$0\ge 1$

Which is false

Hence origin does not lie in plane

$x+y\ge 1$

So, we shade right upper side of line

Now we solve

$x-y\le 0$

Lets first draw graph of

$x-y=0$ ..........(3)

Putting

$x=0$ in (3)

$0-y=0$

$-y=0$

$y=0$

Putting

$y=2$ in (2)

$x-2=0$

$x=0+2$

$x=2$

Drawing graph

Refer fig. 3

$x\,\,\,\,\,\,0\,\,2$

$y\,\,\,\,\,\,0\,\,2$

Points to be plotted are

$(0,0),(2,2)$

Checking for

$(10,0)$

Putting

$x=10,y=0$

$x-y\le 0$

$10-0\le 0$

$10\le 0$

Which is false

Hence

$(10,0)$ does not lie in plane

$x>y$

So, we shade left side of line.

Also, given

Refer fig. 4

$x\ge 0,y\ge 0$

So shaded region will lie in first quadrant.

Hence the shaded region represents the given inequality.

1522872_419742_ans_88f3a0d8e0334608bff170aa364c7bc2.png

Solve the following inequality
$\displaystyle \frac{-2x+5}{x+6}>-2$

$\dfrac{-2x+5}{x+6}>-2$

$\dfrac{-2x+5}{x+6}+2>0$

$\dfrac{-2x+5+2x+12}{x+6}>0$

$\therefore \dfrac{17}{x+6}>0$

The above equation represents

that

$\dfrac{17}{x+6}>0$ must be positive i.e

$>0$

for

$\dfrac{17}{x+6}>0, x+6$ must be greater

$x+6>0$

$\therefore x>-6...(1)$

Also

$\dfrac{-2x+5}{x+6}>-2$

$\dfrac{+2x+6}{-2x+5}<\dfrac{-1}{2}$ [Taking reciprocal on both side changes the sign of inequality]

$\dfrac{x+6}{-2+5}+\dfrac{1}{2}<0$

$\dfrac{2\times 12+(-2\times +5)}{-2x +5}<0$

$\dfrac{17}{-2x +5}<0$

For the above equation to be true

$-2x+5<0$

$x>\dfrac{5}{2} ....(2)$

After combining both the solutions

the required solution we get is

$\boxed {x>\dfrac{5}{2}}$

Solve the inequality and represent the solution graphically on number line.
$2(x-1)< x+5, 3(x+2)> 2-x$

Red line is

$y=2x-2$
Green line is

$y=x+5$
black line

$y=2-x$
blue line is

$y=3x+6$
Since

$2x-1<x+5 \;\; for$ all

$x<7$
and

$3(x+2)>2-x\;\;for\;\;all\;\;x>-1$
So intersection of these

$x\in(-1,7)$

Solve the inequality and represent the solution graphically on number line.
$5(2x-7)-3(2x+3)\le 0, 2x+19\le 6x+47$

We have,

$5(2x-7)-3(2x+3)\le 0$

$\Rightarrow 10x-35-6x-9\le 0$

$\Rightarrow 4x-44\le 0\Rightarrow x\le 11$
And

$2x+19\le 6x+47$

$\Rightarrow 2x-6x\le 47-19$

$\Rightarrow -4x\le 28\Rightarrow x\ge -7$
Thus using above two we have

$-7\le x\le 11$
Solution is also shown graphically

A solution is to be kept between $68^{o}F$ and $77^{o}F$ . What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by $F=\dfrac{9}{5}C+32?$

We have,

$F =\dfrac{9}{5}C+32\Rightarrow C=\dfrac{5}{9}(F-32)$
Now at

$F = 68^{o}, C = \dfrac{5}{9}(68-32)=\dfrac{5}{9}(36)=20$
and at

$F = 77^{o}, C=\dfrac{5}{9}(77-32)=\dfrac{5}{9}(45)=25$
Hence range of temperature in degree Celsius is

$[20^oC, 25^oC]$

A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

Let's add

$x$ liter of

$2\%$ boric acid solution.

Let's find

$x$ when final solution is

$4\%$ boric acid

Equating water content

$0.98x+.92\times 640=.96(x+640)\\ .02x=25.6\\ x=1280$

Similarly for

$6\%$ boric solution

$0.98x+.92\times 640=.94(x+640)\\ .04x=12.8\\ x=320$

How many litres of water will have to be added to $1125$ litres of the $45\%$ solution of acid so that the resulting mixture will contain more than $25\%$ but less than $30\%$ acid content?

$\textbf{Step 1: Calculate concentration of acid in solution}$

$\text{Given:}$

$\text{Total amount of solution}= 1125 \; litres$

$\text{Percentage of acid in solution}= 45 \% \; litres$

$\text{Amount of acid in solution}= 45 \% \;of\; 1125\; litres$

$\Rightarrow \dfrac{45}{100}\times 1125 \ldots(i)$

$\textbf{Step 2: Calculate amount of water to be added for lower concentration.}$

$\text{Let the amount of water added be }x$

$\text{So total solution will be }= (1125+x) \; litres$

$\text{For acid concentration more than 25}\%$

$\Rightarrow 25 \% \; of\; (1125+x)<\dfrac{45}{100}\times 1125$

$\Rightarrow \dfrac {25}{100}\times (1125+x)<\dfrac{45}{100}\times 1125$

$\Rightarrow 1125+x<45 \times45$

$\Rightarrow x< 900 \ldots(ii)$

$\textbf{Step 3: Calculate amount of water to be added for higher concentration.}$

$\text{Let the amount of water added be }x$

$\text{So total solution will be }= (1125+x) \; litres$

$\text{For acid concentration less than 30}\%$

$\Rightarrow 30 \% \; of\; (1125+x)>\dfrac{45}{100}\times 1125$

$\Rightarrow \dfrac {30}{100}\times (1125+x)>\dfrac{45}{100}\times 1125$

$\Rightarrow 1125+x>\dfrac{45 \times 1125}{30}$

$\Rightarrow x> 562.5 \ldots(iii)$

$\textbf{Step 4: Comparing the results.}$

$\text{From equation }(ii) \;and\; (iii)$

$562.5<x<900$

$\textbf{Hence the correct answer is }$

$562.5<x<900$

Find the values of $x$ , which satisfy the inequation $-2\dfrac {5}{6} < \dfrac {1}{2} - \dfrac {2x}{3} \leq 2, x \in W$ .

Given:

$-2\dfrac {5}{6} < \dfrac {1}{2} - \dfrac {2x}{3} \leq 2$

$\Rightarrow - \dfrac {17}{6} < \dfrac {3 - 4x}{6} \leq 2$
Multiplying throughout by

$6$

$\Rightarrow -17 < 3 - 4x \leq 12$

$-17 < 3 - 4x$ and

$3 - 4x \leq 12$

$\Rightarrow 4x < 3 + 17\ \Rightarrow 3 - 12 \leq 4x$

$\Rightarrow 4x < 20 \ \Rightarrow -9\leq 4x$

$\Rightarrow x < 5\ \Rightarrow -\dfrac {9}{4} < x$

$\left \{5 > x \geq \dfrac {-9}{4}\right \}$
Hence, the solution set is

$\left \{x : x \in W, -\dfrac {9}{4} \leq x < 5\right \}$

Since

$x$ is a whole number,

$\therefore$ Values of

$x$ are

$\left \{0, 1, 2, 3, 4\right \}$

Solve the following inequation and write the solution set :
$13x - 5 < 15x + 4 < 7x + 12, x \in R$
Represent the solution on a real number line.

Given :

$13x - 5 < 15x + 4 < 7x + 12$

$\Rightarrow$

$15x+4 > 13x-5$ and

$15x+4 < 7x+12$

$\Rightarrow$

$15x-13x > -5-4 \Rightarrow 15x-7x < 12 - 4$

$\Rightarrow$

$2x > -9 \Rightarrow 8x < 8$

$\Rightarrow$

$x > \dfrac{-9}{2} \Rightarrow x < 1$
Solution :

$\left\{x : \dfrac{-9}{2} < x < 1, x \in R\right\}$

Solve the following inequation and represent the solution set on the number line:
$4x-19<\dfrac{3x}{5}-2\le-\dfrac{2}{5}+x,\,\,\,\,\,\, x\in R$

Take the first inequality:

$4x-19 < \dfrac{3x}5 - 2 \Rightarrow 20x-95 < 3x -10 \Rightarrow 17x < 85$

$\Rightarrow x < 5 ... (1)$

Now tale the second inequality:

$\dfrac{3x}5 - 2 \leq -\dfrac25 +x \Rightarrow 3x -10 \leq -2+5x \Rightarrow -8 \leq 2x$

$\Rightarrow x \geq -4 ... (2)$

From

$(1)$ and

$(2)$

$-4 \leq x < 5$

Solve the following inequation, write the solution set and represent it on the number line.
$-3 (x - 7)\geq 15 - 7x > \dfrac {x + 1}{3}, x\in R$

Given:

$-3(x - 7)\geq 15 - 7x > \dfrac {x + 1}{3}$

$-3x + 21\geq 15 - 7x > \dfrac {x + 1}{3}$

$\Rightarrow -3x + 21 \geq 15 - 7x$ and

$155 - 7x > \dfrac {x + 1}{3}$

$\Rightarrow 4x \geq -6$ and

$45 - 21x > x + 1$

$\Rightarrow x \geq - \dfrac {3}{2}$ and

$44 > 22x$

$\Rightarrow x \geq \dfrac {-3}{2}$ and

$2 > x$

$\therefore$ Solution set

$= \left \{x : x \in R, \dfrac {-3}{2} \leq x < 2\right \}$

Solve the following inequation, write the solution set and represent it on the number line: $-\displaystyle\frac{x}{3}\leq \frac{x}{2}-1\frac{1}{3} < \frac{1}{6}, x\in R$ .

Given:

$-\displaystyle\frac{x}{3}\leq \frac{x}{2}-1\frac{1}{3}<\frac{1}{6}$
Taking L.C.M. of

$3, 2$ and

$6$ is

$6$ .

$-\displaystyle\frac{x}{3}\times 6\leq \frac{x}{2}\times 6-\frac{4}{3}\times 6<\frac{1}{6}\times 6$

$-2x\leq 3x-8<1$

$\Rightarrow -2x\leq 3x-8$ and

$3x-8<1$

$\Rightarrow 8\leq 3x+2x$

$\Rightarrow 3x<1+8$

$\Rightarrow 8\leq 5x$

$\Rightarrow 3x<9$

$\Rightarrow \displaystyle\frac{8}{5}\leq x$

$\Rightarrow x<3$

$\therefore$ The solution set is

$\{x: 1.6\leq x\leq 3, x\epsilon R\}$

Minimize and maximize $z=5x+10y$
subject to the constraints
$x+2y \le 120$
$x+y\ge 60$
$x-2y\ge 0$ and $x \ge 0,\, y \ge 0$ by graphical method.

$z=5x+10y$
subject to the constraints

$x+2y \le 120$

$x+y\ge 60$

$x-2y\ge 0$ and

$x \ge 0,\, y \ge 0$

We take the given inequality to fing the region bounded by these inequalities.

$x+2y = 120$

$x$	$0$	$60$	$120$
$y$	$60$	$30$	$0$

$x+y = 60$

$x$	$0$	$40$	$60$
$y$	$60$	$20$	$0$

$x-2y = 0$

$x$	$0$	$40$	$60$
$y$	$0$	$20$	$30$

The shaded region

$BDEF$ determined by linear inequalities shows the easible region

Let us evaluate the objective function

$Z$ at each point as shown below

$B(120,0),$

$Z=5\times 120+10\times 0=600$

$D(60,0),$

$Z=5\times 60+10\times 0=300$

$E(40,20),$

$Z=5\times 40+10\times 20=400$

$F(60,30),$

$Z=5\times 60 + 10\times 30=600$

Hence, Maximum value of

$Z$ is

$600$ at

$F(60,30)$ and minimum value of

$Z$ is

$300$ at

$D(60,0)$

Solve by graphical method $x+y \geq 5, x-y \leq 3$

1450563_771475_ans_a0f5dfc29a964e219104b0b60c7e9dd5.png

Indicate the polygonal region represented by the system of inequations
$x\ge 0, y≥0, x\le 4, x\ge y$ .

Solve: $5x-7 < 3(x+3), 1-\displaystyle\frac{3x}{2} \geq x-4$ .

Consider the

$5x-7<3(x+3)$

$\therefore 5x-7<3x+9$

$\therefore 2x<16$

$\therefore x<8\ \dots\ (1)$

Now, consider

$1-\dfrac{3x}{2}\geq x-4$

$\therefore x+\dfrac{3x}{2}\leq 1+4$

$\therefore \dfrac{5x}{2}\leq 5$

$\therefore x\leq 2\ \dots\ (2)$

As the solution is the joint solution of both inequations, the joint solution is

$x<8 \cap x\leq 2\implies x\leq 2$

Hence,

$x\leq 2$ is the required solution.

Solve the following inequalities.
$\displaystyle\, \frac{2 \, - \, 5x \, }{x \, + \, 1}\, > \, 2$

$\cfrac { 2-5x }{ x+1 } >2$

$\cfrac { 2-5x-2(x+1) }{ x+1 } >0$

$\cfrac { 2-5x-2x-2 }{ x+1 } >0$

$\cfrac { x }{ x+1 } <0$

$x\quad \varepsilon \quad (-1,0)$

Solve by graphical method: $y-2x\leq 1, x+y\leq 3, x\geq 0, y\geq 0$ .

$x+y\leq 3$ means that the feasible region for this condition lies below

$x+y= 3$ .

$y-2x\leq 1$ means that the feasible region for this condition lies above

$2x-y= -1$ .

Also,

$x\geq 0$ and

$y\geq 0$

The intersection point of

$x+y=3$ and

$y-2x=1$ is given by

$\left(\dfrac{2}{3},\dfrac{7}{3}\right)$ .

Intersection of above conditions is the required solution.

Hence, the feasible region is the shaded region.

Show triangle formed x + 2y = 5 and 4x - y = 5 and x-axis graphically.

1432135_1025742_ans_4e71312a4bd6456c8f685123b046ddad.png

Represent the solution of the following inequality on the real number line:
${\dfrac{2x + 5}{3}} > 3x - 3$

Given condition:

$\cfrac { 2x+5 }{ 3 } >3x-3$

$2x+5>9x-9$

$7x<14$

$x<2$

1171386_1025851_ans_e8ab9af4f11242dd88b5556f92fb0496.png

Represent $x$ on number line or find $x$
$\dfrac{3x - 2} {5} > \dfrac{4x - 3}{2}$

$\cfrac { 3x-2 }{ 5 } >\cfrac { 4x-3 }{ 2 }$

$6x-4>20x-15$

$11>14x$

$x<\cfrac { 11 }{ 14 }$

1174909_1035606_ans_8438430503c845a2bceac4e6c8e472b8.png

Solve the following equation:
$- 8 < - (3x - 5) < 13$

$-8<-(3x-5)<13$

Adding

$-5$ on both sides,

$-8-5<-3x+5-5<13-5$

$-13<-3x<8$

Multiplying with

$-1$ and reversing inequality sign,

$13>3x>-8$

Dividing by

$3$ , we have

$\cfrac { -8 }{ 3 } <x<\cfrac { 13 }{ 3 }$

$\therefore$ Solution set of

$x\epsilon \left( \cfrac { -8 }{ 3 } ,\cfrac { 13 }{ 3 } \right)$

Find the largest value of x for which $2(x - 1) \le 9 - x$ and $x \in W$ .

$2(x-1)\le 9-x$

$2x-2\le 9-x$

$3x\le 11$

$x\le 3\cfrac { 2 }{ 3 }$

$x\quad \varepsilon \quad W$

$x\quad \varepsilon \{ 0,1,2,3\}$

Largest value of

$x$ is 3.

Solve the following and find $x$
$3x + 9 \ge - x + 19$

$3x+9\ge -x+19$

$4x\ge 10$

$x\ge 2.5$

$x\quad \varepsilon [2.5,\infty )$

$x \in$ {real numbers} and $- 1 < 3 - 2x \le 7$ , evaluate x and represent it on a number line.

$(i)3-2x \le 7$

$\quad -2x\le 4$

$\quad x\ge -2$

$(ii)-1<3-2x$

$\quad 2x<4$

$\quad x<2$

996751_1025858_ans_3e387740f6594921adc5a4e5f4304654.png

Maximize $z=x+y$ subject to $x+y \le 10,3y-2x \le 15,x \le 6,x,y \ge 0$ . Find the maximum value.

From the graph,

The required feasible region is bounded by

$OABCD$ whose vertices are

$0(0, 0), A(6, 0), B(6, 4), C(3, 7),$ and

$D(0, 5)$

Objective function

$z = x + y$

As we know that, maximum or minimum value occurs at any one of the bounded points.

So, The maximum value is

$10$ and occurs at

$B(6, 4), C(3, 7)$

992701_1039427_ans_5e3c9b658773401c966fc42f5570d0f2.png

Solve the inequalities for real $x$ .
$\dfrac {x}{3}>\dfrac {x}{2}+1$

Consider the given inequality,

$\dfrac{x}{3}>\dfrac{x}{2}+1$

Multiply by $6$ on both sides, we get

$6\times \dfrac{x}{3}>6\times \dfrac{x}{2}+6\times 1$

$2x>3x+6$

$2x-3x>3x-3x+6$

$-x>6$

$x<-6$

Hence, this is the answer.

Find $x$
$\dfrac{5x}{2} + \dfrac{3x}{4} \ge \dfrac{39} {4}$

$\cfrac { 5x }{ 2 } +\cfrac { 3x }{ 4 } \ge \cfrac { 39 }{ 4 }$

$10x+3x\ge 39$

$x\ge 3\\ x\quad \varepsilon [3,\infty )$

Solve the following system of inequalities graphically
$3x+2y\le 12, x\ge 1, y\ge 2$

The set of inequalities are given as,

$3x + 2y \le 12$

$x \ge 1$

$y \ge 2$

Simplifying $3x + 2y \le 12$ by substituting different values of $x$ and $y$ .

When $x = 0$ , then, $y = 6$

When $y = 0$ , then, $x = 4$

So, the coordinates are $\left( {0,6} \right)$ and $\left( {4,0} \right)$ .

Now, if $x = 0$ and $y = 0$ , then,

$0 \le 12$

Which is true, so the graph will be shaded towards the origin.

Now, $x = 1$ will give a straight line and since $0 \ge 1$ is not true, so the graph will be shaded away the origin.

In the same way, $y = 2$ will give a straight line and since $0 \ge 2$ is not true, so the graph will be shaded away from the origin.

The common area is the required feasible region.

1004149_1064085_ans_ff0a81aa480d4a82b4aef3ebc5069770.png

Minimize $z = 6x + 2y,$ Subject to $x + 2y \ge 3, x + 4y \ge 4 , 3x + y \ge 3 , x \ge 0, y \ge 0$ .

Inequality	x+2y>=3	x+4y>=4	3x+y>=3	x>=0	y>=0
Corresponding equation	x+2y=3	x+4y=4	3x+y=3	x=0	y=0
Intersection of line with x-axis	(3,0)	(4,0)	(1,0)	-	-
Intersection of line with y-axis	(0,1.5)	(0,1)	(0,3)	-	-
Region	Non origin side	Non origin side	Non origin side	Origin side	Origin side

Z at A(0,0)=6(0)+2(0)=0 (minimum value)

Z at B(2,0.5)=6(2)+2(0.5)=13

Z at C(3,0)=6(3)+2(0)=18

Z at D(0,1)=6(0)+2(1)=2

Solve the inequalities for real $x$ .
$\dfrac { 3\left( x-2 \right) }{ 5 } \le \dfrac { 5\left( 2-x \right) }{ 3 }$

We have,

$\dfrac{3\left( x-2 \right)}{5}\le \dfrac{5\left( 2-x \right)}{3}$

$9\left( x-2 \right)\le 25\left( 2-x \right)$

$9x-18\le 50-25x$

$9x+25x\le 50+18$

$34x\le 68$

$x\le 2$

Hence, this is the answer.

Find $x$
$- \left( {x - 3} \right) + 4 < 5 - 2x$

$-(x-3)+4<5-2x$

$-x+3+4<5-2x$

$x<-2$

$x\quad \varepsilon (-\infty ,-2)$

Find $x$
$\dfrac{3} {x - 2} < 1$

$\cfrac { 3 }{ x-2 } <1$

$\cfrac { 3 }{ x-2 } -1<0$

$\cfrac { 3-x+2 }{ x-2 } <0$

$\cfrac { 5-x }{ x-2 } <0$

$\cfrac { x-5 }{ x-2 } >0$

$x\quad \varepsilon (-\infty ,2)\bigcup (5,\infty )$

Solve the inequalities for real $x$ .
$2(2x+3)-10 < 6(x-2)$

We have,

$2\left( 2x+3 \right)-10<6\left( x-2 \right)$

$4x+6-10<6x-12$

$4x-4<6x-12$

$6x-4x>12-4$

$2x>8$

$x>4$

Hence, this is the answer.

Solve the inequalities for real $x$ .
$37-(3x+5) \ge 9x-8(x-3)$

We have,

$37-\left( 3x+5 \right)\ge 9x-8\left( x-3 \right)$

$37-3x-5\ge 9x-8x+24$

$32-3x\ge x+24$

$32-24\ge x+3x$

$8\ge 4x$

$x\le 2$

Hence, this is the answer.

Solve the inequalities for real $x$ .
$\dfrac { x }{ 4 } <\dfrac { \left( 5x-2 \right) }{ 3 } -\dfrac { \left( 7x-3 \right) }{ 5 }$

Given that:

$\dfrac{x}{4}<\dfrac{5x-2}{3}-\dfrac{7x-3}{5}$

$\dfrac{x}{4}<\dfrac{25x-10-21x+9}{15}$ [Taking LCM]

$15x<16x-4$

$x<4 .$

Solve the system of in equations
$\dfrac { x }{ 2x+1 } \ge \dfrac { 1 }{ 4 } ;\dfrac { 6x }{ 4x-1 } <\dfrac { 1 }{ 2 }$

$\begin{array}{l}\frac{x}{{2x + 1}} \ge \frac{1}{4};\frac{{6x}}{{4x - 1}} < \frac{1}{2}\\4x \ge 2x + 1;12x < 4x - 1\\2x \ge 1;8x < 1\\x \ge \frac{1}{2};x < 8\\So,\\\frac{1}{2} \le x < 8\end{array}$

Solve:
$10 \le - 5(x - 2) < 20$

We have,

$10 \le - 5(x - 2) < 20$

$2 \le - (x - 2) < 4$

$-2 \ge (x - 2) >-4$

$-2+2 \ge x >-4+2$

$0 \ge x >-2$

Hence, this is the answer.

Solve the following inequation:
$(x-5)(x+9)(x-8)< 0$

$(x+9)(x-5)(x-8)<0$

Let

$f(x)=(x-5)(x+9)(x-8)$

Here the critical points are

$-9,5,8$

Plot these points on a number line. These points divide the number line into

$4$ intervals.

Now, pick a point from each interval and find the sign of

$f(x)$

$\Rightarrow f(-10)<0$

$\Rightarrow f(0)>0$

$\Rightarrow f(6)<0$

$\Rightarrow f(9)>0$

Since the inequality given in the question is

$'<'$ , the solution is the interval containing

$'-'$ sign

Thus,

$x$ lies in the interval

$(-\infty,-9)\cup (5,8)$

994457_1086128_ans_85944faac97d4fe1861c181ab1b7f3ef.jpg

Solve the inequalities for real $x$ -
$\dfrac { 3\left( x-2 \right) }{ 5 } \le \dfrac { 5\left( 2-x \right) }{ 3 }$

Consider the given inequalities.

$\dfrac{3\left( x-2 \right)}{5}\le \dfrac{5\left( 2-x \right)}{3}$

$9\left( x-2 \right)\le 25\left( 2-x \right)$

$9x-18\le 50-25x$

$9x+25x\le 50+18$

$34x\le 68$

$x\le 2$

Hence, this is the answer.

Solve the inequalities for real $x$ .
$\dfrac { \left( 2x-1 \right) }{ 3 } \ge \dfrac { \left( 3x-1 \right) }{ 4 } -\dfrac { \left( 2-x \right) }{ 5 }$

Given that:

$\dfrac{2x-1}{3}\ge \dfrac{3x-1}{4}-\dfrac{2-x}{5}$

$\dfrac{2x-1}{3}\ge \dfrac{15x-5-8+4x}{20}$

$40x-20\ge 19x-13$

$21x\ge 7$

$x\ge \dfrac{1}{3}.$

Solve $\dfrac{\left| x-3 \right|}{x^{2}-5x+6}\ge 2$

Consider the given inequality.

$\dfrac{\left| x-3 \right|}{{{x}^{2}}-5x+6}\ge 2$

So,

$\left| x-3 \right|\ge 2{{x}^{2}}-10x+12$

Now, for $x\le 3$ , we have

$3-x\ge 2{{x}^{2}}-10x+12$

$2{{x}^{2}}-9x+9\le 0$

$x\in \left[ \dfrac{3}{2},3 \right]$

Again, for $x\ge 3$ , we have

$x-3\ge 2{{x}^{2}}-10x+12$

$2{{x}^{2}}-11x+15\le 0$

$x\in \left[ \dfrac{5}{2},3 \right]$

This doesn’t satisfy $x\ge 3$ . So, we will ignore this value. Also, since, $x=3$ is a root of the equation ${{x}^{2}}-5x+6=0$ , it will make the denominator zero. So, this value will also be ignored. Therefore, the required set of values is,

$\Rightarrow x\in \left[ \dfrac{3}{2},3 \right)$

Hence, this is the required result.

Write down the three inequalities that define the unshaded region, R.

$1\le x\le 4$

Equation of line

$=x+y=4$

$x+y-4\le 0$ for

$x\ge 1$

and

$y\le 4$

1380488_1244573_ans_c9120d535a324899ba9864db45109d10.jpg

Solve $\frac{1}{2}\left( {\frac{{3x}}{5} + 4} \right) \ge \frac{1}{3}\left( {x - 6} \right)$ .

$Given:$

$\frac{1}{2}\left( {\frac{{3x}}{5} + 4} \right) \ge \frac{1}{3}\left( {x - 6} \right)$

$\Rightarrow 3 \left ( \frac{3x}{5} + 4 \right ) \geq 2 (x - 6)$

$\Rightarrow \frac{9x}{5} + 12 \geq 2x - 12$

$\Rightarrow 12 + 12 \geq 2x - \frac{9x}{5}$

$\Rightarrow 24 \geq \frac{x}{5}$

$\Rightarrow x \leq 120$

$Solution\space set$ = (

$- \infty, 120$ ]

Graph the solution sets of the following inequations :
$x - y \leq 3$ .

Given

$x-y\le 3$

Let us draw the graph of

$x-y=3$

Put

$x=0\implies 0-y=3\implies y=-3$

$(0,-3)$ is the solution set of

$x-y=3$

Put

$y=0\implies x-0=3\implies x=3$

$(3,0)$ is the solution set of

$x-y=3$

$\implies (3,0),(0,-3)$ lies on the line

$x-y=3$

$(0,0)$ lies in the left side area of

$x-y=3$

Let us check that

$(0,0)$ lies in the area of solution set of

$x-y\le 4$

$x-y=0-0=0<3$

$\implies (0,0)$ lies in the area of solution set of

$x-y\le 3$

$\therefore$ Solution set is the area which is left side of

$x-y=3$

1691144_1708359_ans_c5bcd791f4fd4914b287ea996f9247db.png

Solve: $2\left(2x+3\right)-10<6\left(x-2\right)$

$2\left(2x+3\right)-10<6\left(x-2\right)$

$\Rightarrow\,4x+6-10<6x-12$

$\Rightarrow\,4x-4<6x-12$

$\Rightarrow\,2x-2<3x-6$

$\Rightarrow \,-2+6<3x-2x$

$\Rightarrow\,4<x$

$\therefore\,x>4$

$\therefore\,x\in\left(4,\infty\right)$

Graph the solution sets of the following inequations :
$x + 2y > 1$ .

Given

$x+2 y>1$

Let us draw the graph of

$x+2 y=1$

Put

$x=0\implies 0+2 y=1\implies y=\dfrac{1}{2}$

$\left(0,\dfrac{1}{2}\right)$ is the solution set of

$x+2 y=1$

Put

$y=0\implies x+2(0)=1\implies x=1$

$(1,0)$ is the solution set of

$x+2 y=1$

$\implies (1,0),\left(0,\dfrac{1}{2}\right)$ lies on the line

$x+2 y=1$

$(0,0)$ lies in the left side area of

$x+2 y=1$

Let us check that

$(0,0)$ lies in the area of solution set of

$x+2 y< 1$

$x+2 y=0+0<1$

$\implies (0,0)$ lies in the area of solution set of

$x+2 y< 1$

$\therefore$ Solution set is the area which is left side of

$x+2 y=1$

1691172_1708362_ans_ff95a34ffc3549419a9f9205c7c63c8a.png

Graph the solution sets of the following inequations :
$x + y \geq 4$ .

Given

$x+y\ge 4$

Let us draw the graph of

$x+y=4$

Put

$x=0\implies 0+y=4\implies y=4$

$(0,4)$ is the solution set of

$x+y=4$

Put

$y=0\implies x+0=4\implies x=4$

$(4,0)$ is the solution set of

$x+y=4$

$\implies (4,0),(0,4)$ lies on the line

$x+y=4$

$(0,0)$ lies in the left side area of

$x+y=4$

Let us check that

$(0,0)$ lies in the area of solution set of

$x+y\ge 4$

$x+y=0+0<4$

$\implies (0,0)$ does not lies in the area of solution set of

$x+y\ge 4$

$\therefore$ solution set is the area which right side of

$x+y=4$

1687825_1708357_ans_54f33138fe7a4ec1b751204dc89eebc5.png

Solve the following inequalities graphically in two-dimensional plane:
$3y-5x<30$

The graphical representation of

$3y-5x=30$ is given as
dotted line in the figure above. This line divided the

$xy-$ plane in two half planes
Select a point (not on the line), which lies in one of the half
planes, do determine whether the point satisfies the given inequality or not.We select the point as

$(0,0)$
It is observed that,

$3(0)-5(0) < 30 \ or\ 0 > 30$ , which is true
Therefore, upper half plane is not the solution region of the
given inequality. Also, it is evident that any point on the
lines does not satisfy the given strict inequality.
Thus, the solution region of the given inequality is the
shaded half plane containing the point

$(0,0)$ excluding
the line.The solution region is represented by the shaded region as as
above.
1767461_1438676_ans_4416a75ac77b43d3bd0d6760a52c05f5.png

1767461_1438676_ans_4416a75ac77b43d3bd0d6760a52c05f5.png

Solve: $3x+4y\ge 12,\:x\ge 0,\:y\ge 1$ and $4x+7y\le 28$

Intersection of

$3x+4y=12$ and

$4x+7y=28$ ,

$3x+4y=12\times 4\Rightarrow 12x+16y=48$

$4x+7y=28\times 3\Rightarrow 12x+21y=84$

$\Rightarrow 5y=36\Rightarrow y=\dfrac{36}{5}$

$x=\dfrac{12}{3}-\dfrac{4}{3}\times \dfrac{36}{5}$

$\Rightarrow 4-\dfrac{48}{5}=-\dfrac{28}{5}$ .

1134147_1242865_ans_4ff2a221d5a648cf8c1acabbc1f3bc73.jpg

Maximize: $z = 3x + 5y$
Subject to: $x + 4y \leq 24, \\ 3x + y \leq 21, \\ x + y \leq 9, \\ x \geq 0, y \geq 0$

The maximize:

$z=3x+5y$

Given subjective to,

$x+4y\le 24$

$3x+y\le 21$

$x+y\le 9$ ,

$x\ge 0, y\ge 0$

Consider,

$x+4y=24$

$..........(1)$

$3x+y=21$

$..........(2)$

$x+y=9$

$..........(3)$

On solving equation (1) and (2), we get

$y=\dfrac{51}{11}$

From equation (1),

$x=\dfrac{60}{11}$

Therefore,

$(x, y)=\left(\dfrac{60}{11}, \dfrac{51}{11}\right)$

and

$z=3x+5y$ =

$\dfrac{435}{11}$ =39.54

Now,

On solving equation (2) and (3), we get

$x=6$

From equation (3),

$y=3$

Therefore,

$(x, y)=\left(6, 3\right)$

and

$z=3x+5y$ =33

Now,

On solving equation (1) and (3), we get

$y=5$

From equation (3),

$x=4$

Therefore,

$(x, y)=\left(4, 5\right)$

and

$z=3x+5y=37$

Hence,

$(x, y)=\left(\dfrac{60}{11}, \dfrac{51}{11}\right)$ ,

$z=\dfrac{435}{11}$ is the answer.

1217655_1569882_ans_b30de6337f7747e6b6f7b4a453ea824b.png

Draw the graph of the solution set of the following inequations:
$3x+2y> 6$

Given inequality,

$3x+2y>6$

$x=0,y=3$ and

$y=0,x=2$

$(0,0)$

$\implies 0>6$ (False)

So solution region lies away from the origin.

1943292_1708808_ans_f657ecfe4fa14481a00bd7db5dd5d71c.png

Graph the solution sets of the following inequations :
$2x - 3y < 4$ .

$Converting\text{ the inequality to equation}$

$2x=3y+4$

$at\,y=0$

$x=2$

$at\,y=2$

$x=5$

$at\,y=-1$

$x=1/2$

$plotting\,all\,the\,po\operatorname{int}son\,graph$

1548422_1708364_ans_f9927a49784e4d1a858a135fd63d6141.JPG

Draw the graph of the solution set of the following inequations:
$x-y\leq 3$

Given inequality,

$x-y\le 3$

$x=0,y=-3$ and

$y=0,x=3$

$(0,0)$

$0\le 3$ (true)

So solution region lies towards the origin.

1943252_1708800_ans_42f6362337154c1f90db98211c912808.png

Draw the graph of the solution set of the following inequations:
$x\geq y-2$

Given inequality,

$x\ge y-2$

$x=0,y=2$ and

$y=0,x=-2$

$(0,0)$

$0\ge -2$ (true)

SO solution region lies towards the origin.

1943269_1708803_ans_e1b2224048e947adb75c14a2199624af.png

Draw the graph of the solution set of the following inequations:
$3x+5y< 15$

Given inequality,

$3x+5y<15$

$x=0,y=3$ and

$y=0,x=5$

$(0,0)$

$\implies 0<15$ (true)

So solution region lies towards from the origin.

1943301_1708810_ans_afc7c48fc8614ddf856e6fea4434eb73.png

Graph the solution sets of the following inequations :
$x \geq y - 2$ .

$Converting\text{ the inequality to equation}$

$x=y-2$

$at\,y=0$

$x=-2$

$at\,y=2$

$x=0$

$at\,y=-1$

$x=-3$

$plotting\,all\,the\,po\operatorname{int}son\,graph$

1548424_1708365_ans_93d44330f11f43d4acf55ae617154f55.JPG

Draw the graph of the solution set of the following inequations:
$x+y\geq 4$

step 1: Convert into equation

$x + y = 4$

Step 2: Convert the equation in the slope-intercept form and find the solutions.

$y = 4 – x~~~~~~~~~~-(1)$

Putting

$x = 0$ in (1)

$y = 4$ Putting

$y = 0$ in (1)

$x = 4$

Hence shaded region including the points on the line is the solution of given inequation.

1943222_1708795_ans_5c86116cf4074223a9422e882c65e02b.png

Draw the graph of the solution set of the following inequations:
$y-2\leq 3x$

Given inequality,

$y-2\le 3x$

$x=0,y=2$ and

$y=0,x=-\dfrac{2}{3}$

$(0,0)$

$-2\le 0$ (true)

So solution region lies towards the origin.

1943259_1708802_ans_bab2b008c1bb435fa26222015a9a955a.png

Solve the following system of inequations graphically:
$x\geq 2,y\geq 3$

1947896_1708851_ans_8e47eacd312b4afd82c835e2f7331aea.jpg

Graph the solution sets of the following inequations :
$y - 2 \leq 3x$ .

$Converting\text{ the inequality to equation}$

$y=3x+2$

$at\,x=0$

$y=2$

$at\,x=1$

$y=5$

$at\,x=-1$

$y=-1$

$plotting\,all\,the\,po\operatorname{int}son\,graph$

1548426_1708366_ans_770db5c53bec4a02a3b0a32dd440077d.JPG

Find the solution set of the inequation $\dfrac{x+1}{x+2}>1$

$\dfrac{x+1}{x+2}>1$

$\Rightarrow \dfrac{x+1}{x+2}-1>0\\ \Rightarrow \dfrac{x+1-x-2}{x+2}>0\\ \Rightarrow x<-2 \\ \Rightarrow x\in (-\infty,-2)$

Show that each of the following systems of linear inequations has no solution: $3x+2y\geq 24,3x+y\leq 15,x\geq 4$

1947809_1708942_ans_ebc5fd7b0815492691d6bb06118a207e.jpg

Solve the following system of inequations graphically:
$2x+y\geq 4,x+y\leq 3,2x-3y\leq 6$

1855287_1708892_ans_500d5fe448034535bbb19369102f9db2.jpeg

Solve the following system of inequations graphically:
$3x+4y\leq 60,x+3y\leq 30,x\geq 0,y\geq 0$

1855284_1708886_ans_cee304f0f6ad4a2bbc093121acfe3b8e.jpeg

Find the solution set of the inequation $\dfrac{[x-2]}{(x-2)}>0,x\neq 2$

Clearly

$(x-2)>0\Rightarrow x>2\Rightarrow x\in (2,\infty)$

Solve the following system of inequations graphically:
$5x+4y\leq 20,x\geq 1,y\geq 2$

1854381_1708881_ans_14a2b64d6fa24b37924d50ae3fc8b420.jpeg

Show that each of the following systems of linear inequations has no solution: $2x-y\leq -2,x-2y\geq 0,x\geq 0,y\geq 0$

1860631_1708947_ans_76754d2a6a50485ab7149211f700429c.jpeg

Solve the following system of inequations graphically:
$x+y\leq 6,x+y\geq 4$

1969568_1708862_ans_89b34ed8f34e473daa604d1c3324bb80.jpg

Show that the solution set of the following system of inequations is an unbounded set: $3x+y\geq 12,x+y\geq 9,x\geq 0,y\geq 0$

1947811_1708954_ans_1841b6b5fb9d4115981c15b4b53a1b85.jpg

Find the solution set of the inequation $\dfrac{1}{x-2}<0$

$\dfrac{1}{(x-2)} <0\Rightarrow x-2<0\Rightarrow x<2\Rightarrow x\in(-\infty,2)$

Fill in the blanks with correct inequality sign $(>, <, \ge, \le )$ .
$-6x\le -18\Rightarrow x.....3$

Given,

$-6x \le -18$

Dividing both sides by

$-3$

$\Rightarrow x \ge 3$ . Dividing by negative no will change inequality sign.

Solve $|x|<4$ , where $x\in R$

${\textbf{Step 1: Use the property for }}$ ${\mathbf{\left| a \right|} \mathbf{ < x;x \in R.}}$

${\text{As we know that,}}$ $\left| a \right| < x$ ${\text{can be written as }} - a < x < a.$

${\text{It implies that, }}\left| x \right| < 4$ ${\text{can be written as }} - 4 < x < 4.$

${\textbf{Step 2: Write the value of x in interval form.}}$

$\because - 4 < x < 4$

$\therefore x \in \left( { - 4,4} \right)$

${\textbf{Hence, x}} \mathbf{\in \left( { - 4,4} \right)}{\textbf{ or it can be represent as - 4 < x < 4 is the correct answer.}}$

Solve $|x|>4$ , where $x\in R$

${\textbf{Step 1: Use the property for }}$ ${\mathbf{\left| a \right| > x;x \in R.}}$

${\text{As we know that,}}$ $\left| a \right| > x \Leftrightarrow x < - a{\text{ or }}x > a.$

${\text{It implies that, }}\left| x \right| > 4$ $\Leftrightarrow x < - 4{\text{ or }}x > 4.$

${\textbf{Step 2: Write the value of x in interval form.}}$

$\because x < - 4{\text{ or }}x > 4.$

$\therefore x \in \left( { - \infty , - 4} \right) \cup \left( {4,\infty } \right)$

${\textbf{Hence, }}$ ${\mathbf{x \in \left( { - \infty , - 4} \right) \cup \left( {4,\infty } \right)}}{\textbf{ or it can be written as }}$ ${\mathbf{\left| x \right| > 4 \Leftrightarrow x < - 4{\text{ or }}x > 4}}$ ${\textbf{ is correct answer.}}$

Solve $\dfrac{x}{x-5}>\dfrac{1}{2}$ , when $x\in R$

Given,

$\dfrac{x}{x-5}>\dfrac{1}{2}$ , when

$x\in R$

$\Rightarrow \dfrac{x}{x-5}-\dfrac{1}{2}>0\\ \Rightarrow \dfrac{2x-x+5}{2(x-5)}>0\\ \Rightarrow \dfrac{x+5}{x-5}>0$

$\therefore (x+5<0\ and\ x-5<0)$ or

$(x+5>1\ nad\ x-5>0)$

$\Rightarrow (x<-5\ adn\ x<4)$ or

$(x>-5\ adn\ x>5)$

$\Rightarrow (x>5)$ or

$(x>5)\Rightarrow x\in(-\infty,-5)\cup (5,\infty)$

Fill in the blanks with correct inequality sign $(>, <, \ge, \le )$ .
$-3x>9\Rightarrow x.....-3$

Given,

$-3x> 9$

Dividing both sides by

$-3$

$\Rightarrow x< -3$ . Dividing by negative no will change inequality sign.

Fill in the blanks with correct inequality sign $(>, <, \ge, \le )$ .
$4x>-16\Rightarrow x.....-4$

Given,

$4x>-16$

Dividing by

$4$ on both sides, we get

$\Rightarrow x>-4$ .

Solve the system of inequation $x-2\ge 0,2x-5\le 3$

$(x-2 \ge\ and\ 2x-5\le 3)$

$\Rightarrow (x\ge 2\ and\ x\le 4)\\ \Rightarrow 2\le x\le 4\\ \Rightarrow x\in [2,4]$

Solve $-4x>16$ , when $x\in R$

$-4x>16\\ \Rightarrow x<-4\\ \Rightarrow x\in[-5,-6,-7,-8,....]$

Solve $x+5>4x-10$ , when $x\in R$

$x+5>4x-10\\ \Rightarrow 3x<15\\ \Rightarrow x<4\\ \Rightarrow x\in (-\infty,5)$

Fill in the blanks with correct inequality sign $(>, <, \ge, \le )$ .
$5x<20\Rightarrow x.....4$

$5x<20\\ \Rightarrow x{<}4$ After dividing by

$5$ both sides

Fill in the blanks with correct inequality sign $(>, <, \ge, \le )$ .
$a<b$ and $c<0\Rightarrow \dfrac{a}{c}.....\dfrac{b}{c}$

Given

$c<0$

and

$a < b$

Dividing both side by

$c$

$\Rightarrow \dfrac{a}{c}> \dfrac{b}{c}$ ...... as

$c$ is negative inequality sign will change after dividing.

Fill in the blanks with correct inequality sign $(>, <, \ge, \le )$ .
$u-v=2\Rightarrow u.....v$

$u-v=2\Rightarrow u>v$

Solve the following inequations and represent the solution set on the number line.
$-2x>5$ , where $(i)x\in Z, (ii) x\in R$ .

$-2x> 5 \Rightarrow x< \dfrac{-5}{2}\Rightarrow x<-2.5$ .

(i) All integers less than

$-2.5$ are

$-3, -4, -5, -6,....$

(ii) Set of all real numbers less than

$-2.5$ are

$(-\infty, -2.5)$ .

1820335_1709328_ans_08cb9f24774241e0a99beff55fff466c.png

Solve the following inequations and represent the solution set on the number line.
$3x-4>x+6$ , where $x\in R$ .

Given,

$3x-4>x+6$ , where

$x\in R$ .

$\implies x>5$

Solution set

$=(5,\infty)$

1819349_1709336_ans_d1fe64b9c8294d17b9f1b4fe35c341e6.png

Solve the following inequations and represent the solution set on the number line.
$3-2x\ge 4x-9$ , where $x\in R$ .

$3-2x\ge 4x-9$ , where

$x\in R$ .

$3-2x\ge 4x-9\Rightarrow -6x \ge -12\Rightarrow x \le 2$ .

$\therefore$ solution set

$=(-\infty, 2]$ .

1819343_1709338_ans_07a146f3aa3f4b8b96748215f3d44ef2.png

Solve the following inequations and represent the solution set on the number line.
$\dfrac{5x}{4}-\dfrac{4x-1}{3}>1$ , where $x\in R$ .

$\dfrac{5x}{4}-\dfrac{4x-1}{3}>1$ , where

$x\in R$ .

$15x-4(4x-1)>12\\ \Rightarrow 15x-16x+4>12\\ \Rightarrow -x>8\Rightarrow x< -8$ .

$\therefore$ solution set

$=(-\infty, -8)$ .

1819317_1709346_ans_28efd497feed4b06b1876ec0d4bb9003.png

Solve the following inequations and represent the solution set on the number line.
$\dfrac{1}{4}\left(\dfrac{2}{3}x+1\right)\ge \dfrac{1}{3}(x-2)$ , where $x\in R$ .

$\dfrac{1}{4}\left(\dfrac{2}{3}x+1\right)\ge \dfrac{1}{3}(x-2)$ , where

$x\in R$ .

$\dfrac{1}{4}.\dfrac{(2x+3)}{3}\ge \dfrac{x-2}{3}\\ \Rightarrow 2x+3\ge 4x-8\Rightarrow -2x \ge -11\\ \Rightarrow x\le \dfrac{11}{2}$ .

$\therefore$ solution set

$=(-\infty, 5.5]$ .

1819327_1709348_ans_5206528986a147c19c3ab4387400eea9.png

Solve the following inequations and represent the solution set on the number line.
$\dfrac{5x-8}{3}\ge \dfrac{4x-7}{2}$ , where $x\in R$ .

$\dfrac{5x-8}{3}\ge \dfrac{4x-7}{2}$ , where

$x\in R$ .

$10x-16 \ge 12x-21\\ \Rightarrow -2x\ge -5 \Rightarrow x\le \dfrac{5}{2}$ .

$\therefore$ solution set

$=(-\infty, \dfrac{5}{2}\left.\right]$

1819334_1709343_ans_a09637ce110a45249710dae3db660c6b.png

Solve the following inequations and represent the solution set on the number line.
$3x+8>2$ , where $(i) x\in Z, (ii) x\in R$ .

$3x>-6\Rightarrow x>-2$ .

(i) All integers greater than

$-2$ are

$-1, 0, 1, 2, 3, 4....$

(ii) Set of all real numbers than

$-2$ is

$(-2, \infty)$ .

1820347_1709331_ans_88a45aebf11645ca817217518a01d9e1.png

Fill in the blanks with correct inequality sign $(>, <, \ge, \le )$ .
$x>-3\Rightarrow -2x.....6$

Given,

$x> -3$
Multiplying both side by

$-2$ (inequality sign get change)

$\Rightarrow -2x<6$ .

Solve the following inequations and represent the solution set on the number line.
$\dfrac{(2x-1)}{3}\ge \dfrac{(3x-2)}{4}-\dfrac{(2-x)}{5}$ , where $x\in R$ .

$20(2x-1) \ge 15(3x-2)-12(2-x)\\ \Rightarrow 40x-20 \ge 45x-30-24+12x$

$\Rightarrow -17x \ge -34 \Rightarrow x\le 2$ .

$\therefore$ solution set

$=(-\infty, 2]$ .

1819297_1709354_ans_b7d5393d220b43628dad01a15753fdbb.png

Solve:
$\dfrac{x-7}{x-2}\ge 0, x\in R$

$(x-7\le 0$ and

$x-2<0)$ or

$(x-7\ge 0$ and

$x-2>0)$

$\Rightarrow (x\le 7$ and

$x<2)$ or

$(x\ge 7$ and

$x>2)$

$\Rightarrow (x<2)$ or

$(x\ge 7)\Rightarrow x\in (-\infty, 2)\cup [7, \infty)$ .

Solve:
$\dfrac{1}{x-1}\le 2, x\in R$

$\dfrac{1}{x-1}-2\le 0\Rightarrow \dfrac{1-2x+2}{x-1}\le 0\Rightarrow \dfrac{3-2x}{x-1}\le 0$

$\Rightarrow (3-2x\le$ and

$x-1>0)$ or

$(3-2x\ge 0$ and

$x-1<0)$

$\Rightarrow (3\le 2x$ and

$x>1)$ or

$(-2x\le -3$ and

$x<1)$

$\left(x \ge \dfrac{3}{2}\ and\ x>1\right)$ or

$\left(x\le \dfrac{3}{2}\ and\ x<1\right)$

$\Rightarrow (x<1)$ or

$\left(x\ge \dfrac{3}{2}\right)=x\in (-\infty, 1)\cup [\dfrac{3}{2}, \infty)$ .

Solve:
$\dfrac{x-3}{x+4}>0, x\in R$

$(x-3<0$ and

$x+4<0$ or

$(x-3>0$ and

$x+4>0)$

$\Rightarrow (x<3$ and

$x<-4)$ or

$(x>3$ and

$x>-4)$

$\Rightarrow (x<-4)$ or

$(x>3)\Rightarrow x\in (-\infty, -4) \cup (3, \infty)$ .

Solve:
$\dfrac{x-3}{x+1}<0, x\in R$

$(x-3< 0$ and

$x+1>0)$ or

$(x-3>0$ and

$x+1<0)$

$\Rightarrow (x<3$ and

$x>-1$ ) or

$(x>3$ and

$x<-1)$

$\Rightarrow -1<x<3\Rightarrow x\in (-1, 3)$ .

$[x\ge 3$ and

$x<-1$ is not possible]

Solve the following inequations and represent the solution set on the number line.
$\dfrac{2x-1}{12}-\dfrac{x-1}{3}< \dfrac{3x+1}{4}$ , where $x\in R$ .

$\dfrac{2x-1}{12}-\dfrac{x-1}{3}< \dfrac{3x+1}{4}$ , where

$x\in R$ .

$(2x-1)-4(x-1)<3(3x+1)\\ \Rightarrow -2x+3<9x+3\\ \Rightarrow -11x<0\Rightarrow x>0$

$\therefore$ solution set

$=(0, \infty)$ .

1819309_1709349_ans_1374440f54c744cbae365f3b5d106dff.png

Solve:
$\dfrac{3}{x-2}> 2, x\in R$

$\dfrac{3}{x-2}-2>0\\ \Rightarrow \dfrac{3-2x+4}{x-2}>0\\ \Rightarrow \dfrac{7-2x}{x-2}>0$

$\Rightarrow (7-2x<0$ and

$x-2<0)$ or

$(7-2x>0$ and

$x-2>0)$

$\Rightarrow (7<2x$ and

$x<2)$ or

$(7>2x$ and

$x>2)$

$\left(\right.x>\dfrac{7}{2}$ and

$x<2)$ or

$\left(\right.x<\dfrac{7}{2}$ and

$x>2)$

$\Rightarrow 2<x<\dfrac{7}{2}\Rightarrow x\in \left(2, \dfrac{7}{2}\right)$ .

$[x<\dfrac{7}{2}$ and

$x>2$ cannot hold]

Solve:
$\dfrac{5x+8}{4-x}< 2, x\in R$

$\dfrac{5x+8}{4-x}-2<0\\\Rightarrow \dfrac{5x+8-8+2x}{4-x}<0\\ \Rightarrow \dfrac{7x}{4-x}<0$

$\Rightarrow (7x<0$ and

$4-x>0)$ or

$(7x>0$ and

$4-x<0)$

$\Rightarrow (x<0$ and

$4>x)$ or

$(x>0$ and

$-x<-4)$

$\Rightarrow (x<0$ and

$x<4)$ or

$(x>0$ and

$x>4)$

$\Rightarrow (x<0)$ or

$(x>4)\Rightarrow x\in (-\infty , 0)\cup (4, \infty)$ .

Solve:
$\dfrac{2x-3}{3x-7}>0, x\in R$

$(2x-3<0$ and

$3x-7<0)$ or

$(x-7\ge 0$ and

$x-2>0)$

$\Rightarrow \left(\right.x< \dfrac{3}{2}$ and

$x< \dfrac{7}{3}\left. \right)$ or

$\left(\right.x> \dfrac{3}{2}$ and

$x> \dfrac{7}{3}\left. \right)$

$\Rightarrow \left(x< \dfrac{3}{2}\right)$ or

$\left(x> \dfrac{7}{3}\right)\Rightarrow x\in \left(-\infty, \dfrac{3}{2}\right)\cup \left(\dfrac{7}{3}, \infty\right)$ .

Solve the following inequations and represent the solution set on the number line.
$\dfrac{x}{4}<\dfrac{(5x-2)}{3}-\dfrac{(7x-3)}{5}$ , where $x\in R$ .

$\dfrac{x}{4}<\dfrac{(5x-2)}{3}-\dfrac{(7x-3)}{5}$

$\Rightarrow 15x<20(5x-2)-12(7x-3)\\ \Rightarrow 5x<100x-40-84x+36$

$\Rightarrow 15x-16x<-4\Rightarrow -x<-4\Rightarrow x>4$ .

$\therefore$ solution set

$=(4, \infty)$ .

1819307_1709352_ans_f55950f232d149bc9c8b55f9c9f11bca.png

Solve the inequation, $3x 11 < 3$ where $x \in \{1, 2, 3,, 10\}$ . Also, represent its solution on a number line.

Given inequation is,

$3x - 11 < 3$

$\Rightarrow3x < 3 + 11$

$\Rightarrow 3x < 14$

$\Rightarrow x < \dfrac{14}3$

But, $x \in \{1, 2, 3,……, 10\}$

Hence, the solution set is $\{1, 2, 3, 4\}$ .

Representation of the solution on a number line is shown in the image.

1662675_1782889_ans_c69beda15e514036a51a319e37901b34.png

Solve:
$|5-2x| \le 3, x\in R$ .

Using

$|x| \le a \Leftrightarrow -a\le x\le a$ , we get

$|5-2x| \le 3\Rightarrow -3\le 5-2x\le 3$

$\Rightarrow 2x\le 8$ and

$2x\ge 2\Rightarrow x\le 4$ and

$x\ge 1$

$\Rightarrow 1\le x\le 4\Rightarrow x\in [1, 4]$ .

Solve:
$|4x-5| \le \dfrac{1}{3}, x\in R$ .

Using

$|x| \le a\Leftrightarrow -a \le x\le a$ , we get

$|4x-5| \le \dfrac{1}{3}\Rightarrow -\dfrac{1}{3}\le 4x-5\le \dfrac{1}{3}\Rightarrow \dfrac{14}{3}\le 4x$ and

$4x\le \dfrac{16}{3}$

$\Rightarrow \dfrac{7}{6}\le x$ and

$x\le \dfrac{4}{3}\Rightarrow x\in \left[\dfrac{7}{6}, \dfrac{4}{3}\right]$ .

List the solution set of $30 4 (2x 1) < 30,$ given that $x$ is a positive integer.

Given inequation,

$30 - 4 (2x - 1) < 30$

$\Rightarrow 30 -8x + 4 < 30$

$\Rightarrow 34 - 8x < 30$

$\Rightarrow -8x < 30 - 34$

$\Rightarrow -8x < -4$

$\Rightarrow 8x > 4\quad$ [On multiplying both sides by -1, the inequality reverses]

$\Rightarrow x > 4/8$

$\Rightarrow x > 1/2$

As $x$ is a positive integer

The solution set is $\{1, 2, 3, … \}$

Solve: $\dfrac{(2x – 3)}4 ≥ \dfrac12 , x \in \{0, 1, 2,…,8\}$

Given inequation, $\dfrac{(2x - 3)}4 ≥ \dfrac12$

$\Rightarrow 2x - 3 ≥ 4 \times \dfrac12$

$\Rightarrow 2x - 3 ≥ 2$

$\Rightarrow 2x ≥ 2+3$

$\Rightarrow x ≥ \dfrac52$

As $x\in \{0,1,2,...,8\}$

The solution set is $\{3, 4, 5, 6, 7, 8\}$ .

Solve $5 -4x > 2 - 3x, x \in W$ . Also represent its solution on the number line.

Given inequation,

$5 - 4x > 2 - 3x$

$\Rightarrow - 4x + 3x > 2 - 5$

$\Rightarrow -x > -3$

On multiplying both sides by $-1$ , the inequality reverses

$\Rightarrow x < 3$

Since, $x ∈ W$

The solution set is $\{0, 1, 2\}$

Representation the solution on a number line is as shown in the image.

1662700_1782893_ans_1c17b030c01b41938980628cdecb4556.png

Solve:
$|3x-7| > 4, x\in R$ .

Using

$|x|>a\Rightarrow x < -a$ or

$x>a$ , we get

$|3x-7|> 4\Rightarrow 3x-7<-4$ or

$3x-7>4$

$\Rightarrow 3x<3$ or

$3x>11\Rightarrow x<1$ or

$x>\dfrac{11}{3}$

$\Rightarrow x\in (-\infty, 1)\cup \left(\dfrac{11}{3}, \infty\right)$ .

If $x$ is a negative integer, find the solution set of $\dfrac23 + \dfrac13 (x + 1) > 0$ .

Given inequation, $\dfrac23 + \dfrac13 (x + 1) > 0.$

$\Rightarrow \dfrac23 + \dfrac x3 + \dfrac13 > 0$

$\Rightarrow \dfrac x3 + 1 > 0$

$\Rightarrow \dfrac x3 > -1$

$\Rightarrow x > -3$

As $x$ is a negative integer

The solution set is $\{-2, -1\}.$

Solve: $2 (x – 2) < 3x – 2, x \in \{– 3, – 2, – 1, 0, 1, 2, 3\}.$

Given inequation is,

$2 (x – 2) < 3x – 2$

$\Rightarrow 2x – 4 < 3x – 2$

$\Rightarrow2x – 3x < -2 + 4$

$\Rightarrow -x < 2$

$\Rightarrow x > -2\quad$ {on multiplying by $-1$ , the inequality sign reverses]

But, $x \in \{– 3, – 2, – 1, 0, 1, 2, 3\}$

Hence, the solution set is $\{– 1, 0, 1, 2, 3\}.$

If $x\epsilon Z$ , solve : $2 + 4x < 2x - 5 \leq 3x$ . Also, represent its solution on the real number line.

Given inequation is,

$2+4x<2x-5\le3x$

$\Rightarrow 2+4x<2x$ and

$2x-5\le 3x$

$\Rightarrow 4x-2x<-2$ and

$-5\le 3x-2x$

$\Rightarrow 2x<-2$ and

$-5\le x$

$\Rightarrow x<-1$ and

$-5\le x$

$\Rightarrow -5\le x<-1$

$x \in Z$

The solution set is

$\{-5,-4,-3,-2\}$ .

Representation on number line is as shown in the image.

1663906_1782929_ans_fda61900fb0b413088076f933e5e10ee.jpg

Given that $x \in I$ , solve the inequation and graph the solution on the number line: $3 \ge \dfrac{(x 4)}2 + \dfrac x3 \ge 2$

Given inequation is,

$3 ≥ \dfrac{(x – 4)}2 +\dfrac x3 ≥ 2$

$\Rightarrow 3 ≥ \dfrac{3(x – 4)+2x}6 ≥ 2$

$\Rightarrow 18 ≥ 5x-12≥ 12\qquad$ [multiplying by $6$ ]

$\Rightarrow 18+12 ≥ 5x≥ 12+12$

$\Rightarrow 30 ≥ 5x≥ 24$

$\Rightarrow 6 ≥ x≥ \dfrac{24}5$

As $x\in I$

Solution set of $x = \{5, 6\}.$

Representation of the solution on a number line is shown in the adjoining figure

1663768_1782924_ans_2abeab08c94b413b8de6a77e1e3a2b28.jpg

Given $A = \{x: x \in I, – 4 ≤ x ≤ 4\}$ , solve $2x - 3 < 3$ where $x$ has the domain A. Graph the solution set on the number line.

Given equation,

$2x - 3 < 3$

$\Rightarrow 2x < 6$

$\Rightarrow x < 3$

But

$x$ has a the domain

$A = \{x: x ∈ I, – 4 ≤ x ≤ 4\}$

i.e.,

$A = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$

Hence, the solution set is

$\{-4, -3, -2, -1, 0, 1, 2\}$ .

Representation of the solution on a number line is shown in the adjoining image

1662597_1782912_ans_805919e15833477d90d5adc4d6238804.png

If $x\in W$ , find the solution set of $\dfrac35 x - \dfrac{2x 1}3 > 1.$ Also graph the solution set on the number line, if possible.

Given inequation is,

$\dfrac35 x - \dfrac{2x – 1}3 > 1$

$\Rightarrow \dfrac9{15} x - \dfrac{5(2x – 1)}{15} > 1 \quad$ [Taking L.C.M]

$\Rightarrow 9x - 5(2x -1) > 15\quad$ [Multiplying by

$15$ on both sides]

$\Rightarrow 9x -10x + 5 > 15$

$\Rightarrow -x > 15 - 5$

$\Rightarrow -x > 10$

$\Rightarrow x < -10$

But,

$x \in W$

Hence, the solution set is a null set.

Thus, it can’t be represented on the number line.

Solve $x 3 (2 + x) > 2 (3x 1), x { 3, 2, 1, 0, 1, 2, 3}.$ Also represent its solution on the number line.

Given inequation, $x- 3 (2 + x) > 2 (3x - 1)$

$\Rightarrow x - 6 - 3x > 6x - 2$

$\Rightarrow -2x - 6 > 6x - 2$

$\Rightarrow -6x - 2x > -2 + 6$

$\Rightarrow -8x > 4$

$\Rightarrow 8x <-4\quad$ [On multiplying by $-1$ , the inequality sign reverses]

$\Rightarrow x < -\dfrac48$

$\Rightarrow x < -\dfrac12$

But, $x\in \{ – 3, – 2, – 1, 0, 1, 2, 3\}$

Hence, the solution set is $\{-3, -2, -1\}$

Find the values of $x$ , which satisfy the inequation :
$-2 \leq \dfrac{1}{2}-\dfrac{2x}{3} \leq1\dfrac{5}{6}$ , $x \in N.$ Graph the solution set on the number line.

Given inequation is

$-2 \leq \dfrac{1}{2}-\dfrac{2x}{3} \leq1\dfrac{5}{6}$

$\Rightarrow -2 ≤ \dfrac{3 – 4x} 6 ≤ \dfrac{11}6$

$\Rightarrow-12 ≤ 3 – 4x ≤ 11$

$\Rightarrow-12 – 3 ≤ -4x ≤ 11 – 3$

$\Rightarrow-15 ≤ -4x ≤ 8$

$\Rightarrow \dfrac{-15}4 ≤ -x ≤ \dfrac84$

$\Rightarrow \dfrac{15}4 ≥ x ≥ -2$

As $x ∈ N,$

The solution set is $\{1, 2, 3\}.$

Representation of the solution on a number line is given in the adjoining image.

1662052_1782919_ans_8407b0a44d1b4cfba6391e8933cdc270.png

Given $x \in \{1, 2, 3, 4, 5, 6, 7, 9\}$ solve $x - 3 < 2x - 1$ .

Given inequation, $x - 3 < 2x - 1$

$\Rightarrow x - 2x < -1 + 3$

$\Rightarrow -x < 2$

$\Rightarrow x > -2\qquad$ [On multip;ying by $-1$ , the inequality sign reverses]

But, $x \in \{1, 2, 3, 4, 5, 6, 7, 9\}$
Hence, the solution set is $\{1, 2, 3, 4, 5, 6, 7, 9\}.$

Solve:
(i) $\dfrac x2 + 5 \le \dfrac x3 + 6$ , where $x$ is a positive odd integer.
(ii) $\dfrac{(2x + 3)}3 ≥ \dfrac{(3x – 1)}4$ , where $x$ is positive even integer.

(i) Given inequation is

$\dfrac x2 + 5 \le \dfrac x3 + 6$

$\Rightarrow \dfrac{(x + 10)}2 \le \dfrac{(x + 18)}3 \qquad$ [Taking L.C.M on both sides]

$\Rightarrow 3 (x + 10) \le 2 (x + 18)\qquad$ [On cross-multiplying]

$\Rightarrow 3x + 30 \le 2x + 36$

$\Rightarrow 3x -2x \le 36 - 30$

$\Rightarrow x \le 6$

$x$ is a positive odd integer.

Hence, the solution set is

$\{1, 3, 5\}.$

(ii) Given inequation is ,

$\dfrac{(2x + 3)}3 \ge \dfrac{(3x – 1)}4$

$\Rightarrow 4 (2x + 3) \ge 3 (3x – 1)\qquad$ [On cross-multiplying]

$\Rightarrow 8x + 12 \ge 9x - 3$

$\Rightarrow -9x + 8x \ge -12 - 3$

$\Rightarrow -x \ge-15$

$\Rightarrow x \le 15$

$x$ is positive even integer.

Hence, the solution set is

$\{2, 4, 6, 8, 10, 12, 14\}.$

Given $x \in \{1, 2, 3, 4, 5, 6, 7, \},$ find the values of $x$ for which $-3 < 2x – 1 < x + 4.$

Given inequation is,

$-3 < 2x – 1 < x + 4$

$\Rightarrow -3+1<2x<4+1$

$\Rightarrow -4<2x<5$

$\Rightarrow -2<x<\dfrac52$

$x \in \{1, 2, 3, 4, 5, 6, 7, 9\}$

The solution set is

$\{1, 2, 3, 4\}$ .

List the solution set of the inequation
$\dfrac12 + 8x > 5x -\dfrac32, x \in Z$

Given inequation, $\dfrac12 + 8x > 5x -\dfrac32$

$\Rightarrow 8x - 5x > -\dfrac32 - \dfrac12$

$\Rightarrow 3x > -\dfrac42$

$\Rightarrow x > -\dfrac23$

As $x \in Z$

The solution set is $\{0, 1, 2, 3, 4, 5, …\}$

Solve the following inequation and represent the solution set on the number line:
$-3<-\dfrac{1}{2}-\dfrac{2x}{3}\leq \dfrac{5}{6},xR$

Given in equation,

$-3<-\dfrac{1}{2}-\dfrac{2x}{3}\leq \dfrac{5}{6},x∈R$

$-3 < -(3 + 4x)/6 ≤ 5/6 [Taking L.C.M]$

$-18 < -3 – 4x ≤ 5$ [Multiplying by $6]$

$-18 + 3 < -4x ≤ 5 + 3$

$-15 < -4x ≤ 8$

$-15/4 < -x ≤ 8/4$

$-2 ≤ x < 15/4$

Hence, the solution set is {x : x ∈ R, -2 ≤ x < 15/4}

Representing the solution on a number line:

1664020_1782943_ans_c5214ec2f8dd4ea6b2f0372bcdfd31b1.png

Solve the inequation $12 + 1\dfrac{5}{6}x 5 + 3x, x R$ . Represent the solution on a number

Given inequation,

$12 + 1\dfrac{5}{6}x ≤ 5 + 3x$

$12 + 11/6 x ≤ 5 + 3x$

$72 + 11x ≤ 30 + 18x$ [Multiplying by 6 on both sides]

$11x – 18x ≤ 30 – 72$

$-7x ≤ -42$

$-x ≤ -6$

$x ≥ 6$

As $x ∈ R$

The solution set is ${x : x ∈ R, x ≥ 6}$

Representing the solution on a number line:

1663816_1782931_ans_37644762d0094c10b60e1fc4455eff32.png

$A = \{x \ ;\ 11x - 5 > 7x + 3, x \in R\}$ and $B = \{x\ ;\ 18x - 9 ≥ 15 + 12x, x \in R\}$ . Find the range of set $A \cap B$ and represent it on a number line.

Given,

$A=\{x\ ;\ 11x-5>7x+3,x\in R\}$ and

$B = \{x\ ;\ 18x -9 ≥ 15 + 12x, x \in R\}$

Solving for $A$ ,

$11x - 5 > 7x + 3$

$\Rightarrow 11x - 7x > 3 + 5$

$\Rightarrow 4x > 8$

$\Rightarrow x > 2$

Hence, $A = \{x\ ;\ x > 2, x \in R\}$

Next, solving for $B$ ,

$\Rightarrow 18x- 9 ≥ 15 + 12x$

$\Rightarrow 18x - 12x ≥ 15 + 9$

$\Rightarrow 6x ≥ 24$

$\Rightarrow x ≥ 4$

Hence, $B = \{x : x ≥ 4, x \in R\}$

Thus, $A \cap B = \{x\ ;\ x ≥ 4, x \in R\}$

Representing the solution on a number line:

1664110_1782951_ans_0c491e2d0de3439d92c107aee9e93337.png

If $P$ is the solution set of $-3x + 4 < 2x - 3, x \in N$ and $Q$ is the solution set of $4x - 5 < 12, x \in W$ , find
(i) $P \cap Q$
(ii) $Q – P$

Given inequations are,

$-3x + 4 < 2x – 3$ where

$x ∈ N$ and

$4x – 5 < 12$ where

$x ∈ W$

Solving

$-3x + 4 < 2x - 3$

$\Rightarrow -3x - 2x < -3 - 4$

$\Rightarrow -5x < -7$

$\Rightarrow x > \dfrac75$

$x\in N$

The solution set

$P$ is

$\{2, 3, 4, 5, …\}$

And

$4x - 5 < 12$

$\Rightarrow 4x < 12 + 5$

$\Rightarrow 4x < 17$

$\Rightarrow x < \dfrac{17}4$

$x\in W$

The solution set

$Q$ is

$\{0, 1, 2, 3, 4\}$

Therefore,

(i)

$P \cap Q = \{2, 3, 4\}$

(ii)

$Q - P = \{0, 1\}$

Solve $2 2x 3 5, x R$ and mark it on a number line.

Given inequation, $2 ≤ 2x – 3 ≤ 5$

$\Rightarrow 2 + 3 ≤ 2x ≤ 5 + 3$

$\Rightarrow 5 ≤ 2x ≤ 8$

$\Rightarrow \dfrac52 ≤ x ≤ 4$

Hence, the solution set is $\{x: x ∈ R, \dfrac52 ≤ x ≤ 4\}$

Representing the solution on a number line:

1663895_1782938_ans_55cc2adc7465402c9ad0c132cfc2b3ea.png

Given that $x \in R$ , solve the following inequation and graph the solution on the number line: $-1 \leq 3 + 4x < 23$

Given inequation is

$-1 \le 3 + 4x < 23$

$\Rightarrow -1 -3 \le 4x < 23 - 3$

$\Rightarrow -4 \le 4x < 20$

$\Rightarrow -\dfrac44 ≤ x < \dfrac{20}4$

$\Rightarrow -1 \le x < 5$

Hence, the solution set is

$\{-1 ≤ x < 5; x ∈ R\}$

Representation of the solution on a number line is shown in the adjoining image

1662523_1782939_ans_1a41a2b394af4bfbb14b5815a7f0a38e.png

Solve: (4x 10)/3 (5x 7)/2, x R and represent the solution set on the number line.

Given inequation, $(4x – 10)/3 ≤ (5x – 7)/2$

$2 (4x – 10) ≤ 3 (5x – 7)$ [On cross-multiplying]

$8x – 20 ≤ 15x – 21$

$8x – 15x ≤ -21 + 20$

$-7x ≤ -1$

$-x ≤ -1/7$

$x ≥ 1/7$

As $x ∈ R$

Hence, the solution set is ${x: x ∈ R, x ≥ 1/7}$

Representing the solution on a number line:

1663833_1782932_ans_1b5d38cfafb541bf8b184977d343dd93.png

Solve $\dfrac{2x+1}{2}+2(x-3)\geq 7,xR$ . Also graph the solution set on the number line

Given inequation,

$\dfrac{2x+1}{2}+2(x-3)\geq 7,x∈R$

$[2x + 1 + 4(3 – x)]/2 ≥ 7 [Taking L.C.M]$

$2x + 1 + 12 – 4x ≥ 14$

$-2x + 13 ≥ 14$

$-2x ≥ 14 – 13$

$-2x ≥ 1$

$-x ≥ 1/2$

$x ≤ -1/2$

Hence, the solution set is ${x : x ∈ R, x ≤ -1/2}$

Representing the solution on a number line:

1664050_1782944_ans_8bd2cf8acf354c1c858449d0f2893b60.png

Solve 3x/5 (2x 1)/3 > 1, x R and represent the solution set on the number line.

Given inequation, $3x/5 – (2x – 1)/3 > 1$

$(9x – 10x + 5)/15 > 1$ [Taking L.C.M]

$-x + 5 > 15$

$-x > 15 – 5$

$-x > 10$

$x < -10$

As $x ∈ R$

Hence, the solution set is ${x: x ∈ R, x < -10}$

Representing the solution on a number line:

Solve the following inequation and graph the solution on the number line.
$-2\dfrac{2}{3}\leq x+\dfrac{1}{3}<3+\dfrac{1}{3} , x \in R$

Given inequation is,

$-2\dfrac{2}{3}\leq x+\dfrac{1}{3}<3+\dfrac{1}{3} , x \in R$

$\Rightarrow \dfrac{-8}3 ≤ \dfrac{(3x + 1)}3 < \dfrac{10}3$

$\Rightarrow-8 ≤ 3x + 1 < 10\qquad[$ Multiplying by

$3]$

$\Rightarrow-8 - 1 ≤ 3x < 10 - 1$

$\Rightarrow-9 ≤ 3x < 9$

$\Rightarrow-3 ≤ x < 3\qquad[$ Dividing by

$5]$

Thus, the solution set is

$\{x: x \in R, -3 ≤ x < 3\}$

Representation of the solution on a number line is shown in the image.

1664032_1782942_ans_787b8896e7b441f4ab9235fde7475b10.png

$2x-5>11$ is an equation

1902931_1789788_ans_c99f138a472740af804633c34208464b.jpg

Solve the inequation:
$\dfrac{5x+1}{7}-4\left(\dfrac{x}{7}+\dfrac{2}{5}\right)\leq 1\dfrac{3}{5}+\dfrac{3x-1}{7},xR$

Given inequation,

$\dfrac{5x+1}{7}-4\left(\dfrac{x}{7}+\dfrac{2}{5}\right)\leq 1\dfrac{3}{5}+\dfrac{3x-1}{7},x∈R$

$(5x + 1)/7 – 4 (5x + 14)/35 ≤ 8/5 + (3x – 1)/7$

$[5(5x + 1) – 4 (5x + 14)]/ 35 ≤ [56 + 5 (3x – 1)]/ 35 [Taking L.C.M]$

$(25x + 5 – 20x – 56) ≤ 56 + 15x – 5$

$5x – 51 ≤ 51 + 15x$

$5x – 15x ≤ 51 + 51$

$-10x ≤ 102$

$-x ≤ 102/10$

$x ≥ – 51/5$

Hence, the solution set is {x : x ∈ R , x ≥ – 51/5}

Solve the inequation: $5x 2 3 (3 x)$ where $x \left\{-2, -1, 0, 1, 2, 3, 4\right\}$ . Also represent its solution on the number line.

Given inequation, $5x - 2 ≤ 3 (3 - x)$

$5x - 2 ≤ 9 - 3x$

$\Rightarrow 5x + 3x ≤ 9 + 2$

$\Rightarrow 8x ≤ 11$

$\Rightarrow x ≤ \dfrac{11}8$

As $x \in \{-2, -1, 0, 1, 2, 3, 4\}$

The solution set is $\{-2, -1, 0, 1\}$

Representing the solution on a number line:

1663493_1782967_ans_00016b8c51094c05874109e47f140385.png

Find the solution set of the inequation x + 5 2x + 3; x R
Graph the solution set on the number line.

Given inequation, $x + 5 ≤ 2x + 3$

$x – 2x ≤ 3 – 5$

$-x ≤ -2$

$x ≥ 2$

As $x ∈ R$

Thus, the solution set is ${2, 3, 4, 5, …}$

Representing the solution on a number line:

1663448_1782969_ans_5db07a0c603c4859ab738907174def07.png

Solve the inequation: 6x 5 < 3x + 4, x I

Given inequation, 6x–5<3x+4

$6x – 3x < 4 + 5$

$3x < 9$

$x < 9/3$

$x < 3$

As $x ∈ I$

The solution set is ${2, 1, 0, -1, -2, …}$

Given 20 – 5 x < 5 (x + 8), find the smallest value of x, when
$(i) x ∈ I$
$(ii) x ∈ W$
$(iii) x ∈ N.$

Given inequation, $20 – 5 x < 5 (x + 8)$

$20 – 5x < 5x + 40$

$-5x – 5x < 40 – 20$

$-10x < 20$

$-x < 20/10$

$x > -2$

Thus,

(i) For x ∈ I, the smallest value = -1

(ii) For x ∈ W, the smallest value = 0

(iii) For x ∈ N, the smallest value = 1

One-third of a bamboo pole is buried in mud, one-sixth of it is in water and the part above the water is greater than or equal to 3 metres. Find the length of the shortest pole.

Let’s assume the length of the shortest pole = x metre

Now,

Length of the pole which is buried in mud = $x/3$

Length of the pole which is in the water =$$ x/6$4

Then according to the given condition, we have

$x – [x/3 + x/6] ≥ 3$

$x – [(2x + x)/6] ≥ 3$

$x – 3x/6 ≥ 3$

$x – x/2 ≥ 3$

$x/2 ≥ 3$

$x ≥ 6$ [Multiplying by 6]

Therefore, the length of the shortest pole is 6 metres.

Find the greatest integer which is such that if $7$ is added to its double, the resulting number becomes greater than three times the integer.

$\textbf{Step 1 : Find the greatest integer value}$

$\text{Let’s consider the greatest integer to be}$ $x$

$\text{Then according to the given condition, we have}$

$2x + 7 > 3x$

$\Rightarrow 2x – 3x > -7$

$\Rightarrow -x > -7$

$\Rightarrow x < 7 , x ∈ R$

$\textbf{Hence, the greatest integer value is 6}$

Solve the inequation and represent the solution set on the number line.
$-3+x\leq \dfrac{8x}{3}+2 <\dfrac{14}{3}+2x,\text{ where }x\in I$

Given inequation,

$-3+x\leq \dfrac{8x}{3}+2 <\dfrac{14}{3}+2x,\text{ where }x∈I$

So, we have

$-3 + x ≤ \dfrac{8x}3 + 2$ and $\dfrac{8x}3 + 2 ≤ \dfrac{14}3 + 2x$

$\Rightarrow x - \dfrac{8x}3 ≤ 2 + 3$ and $\dfrac{ 8x}3 - 2x ≤\dfrac { 14}3 -2$

$\Rightarrow \dfrac{(3x - 8x)}3 ≤ 5$ and $\dfrac{(8x - 6x)}3 ≤ \dfrac{(14 - 6)}3$ [Taking L.C.M]

$\Rightarrow \dfrac{-5x}3 ≤ 5$ and $2x ≤ 8$

$\Rightarrow -5x ≤ 15$ and $x ≤ \dfrac{8}2$

$\Rightarrow -x ≤ 3$ and $x ≤ 4$

$\Rightarrow x ≥ -3$ and $x ≤ 4$

$\Rightarrow -3 ≤ x ≤ 4$

Thus, the solution set is $\{-3, -2, -1, 0, 1, 2, 3, 4\}$

Representing the solution on a number line:

1663570_1782961_ans_bbbd6a3fee8349c188d311603f5e51c2.png

Given: $P= \{x : 5 < 2x - 1 ≤ 11, x \in R\}$ and $Q =\{x \ :\ - 1 ≤ 3 + 4x < 23, x \in I\}$ where R = (real numbers), I = (integers)
Represent $P$ and $Q$ on the number line. Write down the elements of $P \cap Q$ .

Given, $P=\{x\ : \ 5 < 2x - 1 ≤ 11, x \in R\}$ and $Q =\{x\ :\ - 1 ≤ 3 + 4x < 23, x \in I\}$

Solving for $P$ ,

$5 < 2x -1 ≤ 11$

$\Rightarrow 5 + 1 < 2x ≤ 11 + 1$

$\Rightarrow 6 < 2x ≤ 12$

$\Rightarrow 3 < x ≤ 6$

Hence, $P = \{x\ :\ 3 < x ≤ 6, x \in R\}$

The solution $P$ is represented on number line (a).

Next, solving for $Q$

$-\Rightarrow 1 ≤ 3 + 4x < 23$

$\Rightarrow -1- 3 ≤ 4x < 23 - 3$

$\Rightarrow -4 ≤ 4x < 20$

$\Rightarrow -1 ≤ x < 5$

As $Q\in I$

Hence, solution $Q = \{-1, 0, 1, 2, 3, 4\}$

The solution $Q$ is represented on number line (b).

Therefore, $P \cap Q = \{4\}$

1663932_1782952_ans_9e0ac2e7264945b0b7ea59c5148002a9.png

Solve the inequality $2\le 3x-4\le 5$

$2\le 3x-4\le 5$

$\Rightarrow \ 2+4\le 3x-4+4\le 5+4$

$\Rightarrow \ 6\le 3x\le 9$

$\Rightarrow \ 2\le x\le 3$

Thus, all the real numbers,

$x$ which are greater than or equal to but less than or equal to

$3$ , are the solutions of the given inequality. The solution set for the given inequality is

$[2,3]$ .

Solve the inequality $-15 < \dfrac {3(x-2)}{5}\le 0$

$-15 < \dfrac {3(x-2)}{5}\le 0$

$\Rightarrow \ -75 < 3(x-2)\le 0$

$\Rightarrow \ -25 < x-2\le 0$

$\Rightarrow \ -25+2< x\le 2$

$\Rightarrow \ -23< x \le 2$

Thus, the solution for the given inequality is

$[-23,2]$ .

Solve the inequalities
$2(x-1)< x+5, 3(x+2) > 2-x$

$2(x-1)< x+5\Rightarrow \ 2x-2< x+5\Rightarrow \ 2x-x< 5+2$

$\Rightarrow \ x< 7...(1)$

$3(x+2)> 2-z\Rightarrow \ 3x+6 > 2-x\Rightarrow \ 3x+x > 2-6$

$\Rightarrow \ 4x >-4$

$\Rightarrow \ x > -1....(2)$

From

$(1)$ and

$(2)$ it can be concluded that the solution set for the given system of inequalities is

$(-1,7)$ .

Solve the inequality $-12 < 4-\dfrac {3x}{-5}\le 2$

$-12 < 4-\dfrac {3x}{-5}\le 2$

$\Rightarrow \ -12-4< \dfrac {-3x}{-5}\le 2-4$

$\Rightarrow \ -16< \dfrac {3x}{5}\le -2$

$\Rightarrow \ -80 < 3x \le -10$

$\Rightarrow \ \dfrac {-80}{3}< x \le \dfrac {-10}{3}$

Thus, the solution for the given inequality is

$\left(\dfrac {-80}{3}, \dfrac {-10}{3}\right]$ .

Solve the inequalities
$3x-7 > 2(x-6), 6-x > 11-2x$

$3x-7 > 2(x-6)\Rightarrow \ 3x-7> 2x-12\Rightarrow \ 3x-2x> -12+7$

$\Rightarrow \ x> -5...(1)$

$-6x > 11-2x\Rightarrow \ -x+2x > 11-6$

$\Rightarrow \ x >5...(2)$

From

$(1)$ and

$(2)$ it can be concluded that the solution set for the given system of inequalities is

$(5,\infty)$ .

Solve the inequality $7\le \dfrac {(3x+11)}{2}\le 11$

$7\le \dfrac {(3x+11)}{2}\le 11$

$\Rightarrow \ 14\le 3x+11\le 22$

$\Rightarrow \ 14-11\le 3x \le 22-11$

$\Rightarrow \ 3\le 3x \le 11$

$\Rightarrow \ 1\le x\le \dfrac {11}{3}$

Thus, the solution for the given inequality is

$\left[1,\dfrac {11}{3}\right]$

Solve the inequality $6\le -3(2x-4) < 12$

$6\le -3(2x-4) < 12$

$\Rightarrow \ 2\le -(2x-4)< 4$

$\Rightarrow \ 2\ge 2x-4 >-4$

$\Rightarrow \ 4-2 \ge 2x > 4-4$

$\Rightarrow \ 2\ge 2x > 0$

$\Rightarrow \ 1\ge x > 0$

Thus, the solution set for the given inequality is

$[0,1]$ .

Solve the inequality $-3\le 4-\dfrac {7x}{2} \le 18$

$-3\le 4-\dfrac {7x}{2} \le 18$

$\Rightarrow \ -3-4\le -\dfrac {7x}{2} \le 18-4$

$\Rightarrow \ 7\le -\dfrac {7x}{2}\le 14$

$\Rightarrow \ 7\ge \dfrac {7x}{2}\ge -14$

$\Rightarrow \ 1 \ge \dfrac x2 \ge -2$

$\Rightarrow \ 2\ge x \ge -4$

Thus, the solution set for the given inequality is

$[-4,2]$

Solve the inequalities
$5(2x-7)-3(2x+3)\le 0, 2x+19\le 6x+47$

$5(2x-7)-3(2x+3)\le 0\Rightarrow \ 10x-35-6x-9\le 0 \Rightarrow \ 4x-44\le 0 \Rightarrow \ 4x\le 44$

$\Rightarrow \ x\le 11...(1)$

$2x+19 \le 6x+47\Rightarrow \ 19-47\le 6x-2x\Rightarrow \ -28\le 4x$

$\Rightarrow \ -7\le x...(2)$
From

$(1)$ and

$(2)$ it can be concluded that the solution set for the given system of inequalities is

$(-7,11)$ .

Solve the inequality
$5x+1 > -24, 5x-1 < 24$

$5x+1 > -24, 5x-1 < 24$

$\Rightarrow \ x >-5....(1)$

$5x-1< 24 \Rightarrow \ 5x < 25$

$\Rightarrow \ x < 5...(2)$

From

$(1)$ and

$(2)$ it can be concluded that the solution set for the given system of inequalities is

$(-5,5)$ .

Solve the given inequality graphically in two-dimensional plane: $2x-3y > 6$

The graphical representation of

$2x-3y=6$ is given as
dotted line in the figure above. This line divided the

$xy-$ plane in two half planes
Select a point (not on the line), which lies in one of the half
planes, do determine whether the point satisfies the given inequality or not.We select the point as

$(0,0)$
It is observed that,

$2(0)-3(0) > 6 \ or\ 0 > 6$ , which is false
Therefore, upper half plane is not the solution region of the
given inequality. Also, it is clear that any point on the lines
does not satisfy the given strict inequality.
Thus, the solution region of the given inequality is the
shaded half plane containing the point

$(0,0)$ including
the line.The solution region is represented by the shaded region as as
above.
1767466_1796270_ans_7e14be4e2bde4b77935a1a9040476e76.png

1767466_1796270_ans_7e14be4e2bde4b77935a1a9040476e76.png

Solve the given inequality graphically in two-dimensional plane: $y < -2$

The graphical representation of

$y=-2$ is given as dotted
line in the figure above. This line divided the

$xy-$ plane in two half planes
Select a point (not on the line), which lies in one of the half
planes, do determine whether the point satisfies the given inequality or not.We select the point as

$(0,0)$
It is observed that,

$0 < -2$ , which is false
Also, it is evident that any point on the lines does not satisfy
the given strict inequality.
Hence, every point below the line,

$y=-2$ (excluding all the points on the line determine the solution of the given
inequality.The solution region is represented by the shaded region as as
above.
1767472_1796277_ans_3dd32620eb7a4010ad834791a5f20949.png

1767472_1796277_ans_3dd32620eb7a4010ad834791a5f20949.png

Solver the following system of inequalities graphically : $3x+2y \ge 12, x\ge 1, y \ge 2$

$3x+2y \le 12$ .....(1)

$x \ge 1$ ......(1)

$y \ge 2$ ......(2)
The graph of the lines,

$3x+2y=12, x=1$ and

$y=2$ , are drawn in the figure.
Inequality (1) represents the region below the line,

$3x+2y=12$ ( including the lines

$3x+2y=12$ ), and inequality (2) represents the region on the right side of the line,

$x=1$ ( including the line

$x=1$ ).
Inequality (3) represents the region above the line,

$y=2$ ( including the line

$y=2$ ).
Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows.
1767480_1796325_ans_2ca5b03d35584309b0e41a0fa0cdc4a7.png

1767480_1796325_ans_2ca5b03d35584309b0e41a0fa0cdc4a7.png

Solve the given inequality graphically in two-dimensional plane: $-3x+2y \ge -6$

The graphical representation of

$-3x+2y=-6$ is given as
dotted line in the figure above. This line divided the

$xy-$ plane in two half planes
Select a point (not on the line), which lies in one of the half
planes, do determine whether the point satisfies the given inequality or not.We select the point as

$(0,0)$
It is observed that,

$-3(0)+2(0) \ge -6 \ or\ 0 \ge -6$ , which is true
Therefore, lower half plane is not the solution region of the
given inequality. Also, it is evident that any point on the
lines does not satisfy the given strict inequality.
Thus, the solution region of the given inequality is the
shaded half plane containing the point

$(0,0)$ including
the line.The solution region is represented by the shaded region as as
above.
1767470_1796273_ans_fbb8cdca9c4e4edf98500427be3668bd.png

1767470_1796273_ans_fbb8cdca9c4e4edf98500427be3668bd.png

Solve the following system of inequalities graphically : $4x+3y \le 60, y \ge 2x, x \ge 3, x,x \ge 0$

$4x+3y \le 60$ .......(1)

$y \ge 2x$ .....(2)

$x \ge 3$ .....(3)
The graph of the lines,

$4x+3y=60, y=2x, x=3$ , are drawn in the figure below.
Inequality (1) represents the region below the line,

$4x+3y=60$ ( including the line

$4x+3y=60$ ). Inequality (2) represents the region above the line,

$y=2x$ ( including the line

$y=2x$ ). Inequality (3) represents the region on the right hand side of the line,

$x=3$ ( including the line

$x=3$ ).
Hence, the solution of the given system of the linear inequalities is represented by the common shaded region including the points on the respective lines as follows.
1768264_1796347_ans_c339373b6ab944cdb2b7f2066c64e790.png

1768264_1796347_ans_c339373b6ab944cdb2b7f2066c64e790.png

Solve the following system of inequalities graphically : $x-2y \le 3, 3x +4y \ge 12, x \ge 0, y \ge 1$

$x-2y \le 3$ ......(1)

$3x+4y \ge 12$ .......(2)

$y \ge 1$ .........(3)
The graph of the lines,

$x=2= 3, 3x+4y=12$ and

$y=1$ , are drawn in the figure below.
Inequality (1) represents the region above the line,

$x-2y=3$ ( including the line

$x-2y=3$ ).
Inequality (2) represents the region above the line,

$3x+4y=12$ ( including the line

$3x+4y=12$ ).
Inequality (3) represents the region on the right and side of

$y-$ axis ( including

$y-$ axis).
Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines and

$y$ - axis as follows.
1767494_1796345_ans_bbc24591b5ee48a3a2c815482e940d7d.png

Solve the following system of inequalities graphically : $x+y \ge 4, 2x-y > 0$

$x+y \ge 4$ .... (1)

$2x-y > 0$ .....(2)
The graph of the lines,

$x+y =4$ and

$2x-y=0$ , are drawn in the figure below. Inequality (1) represents the region above the line,

$x+y=4$ ( including the line

$x+y=4$ ). It is observed that

$(1, 0)$ satisfies the inequality

$2x-y > 0. [2(1) -0=2 >0]$
Therefore, inequality (2) represents the half plane corresponding to the line,

$2x-y=0$ , containing the point

$(1, 0)$ satisfies the inequality,

$2x-y > 0]$ .
Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines

$x+y=4$ and excluding the points on line

$2x-y=0$ as follows.
1767486_1796332_ans_656390d8d2f146469f7871fcfdfa789a.png

Solve the given inequality graphically in two-dimensional plane: $x > -3$

The graphical representation of

$x=-3$ is given as dotted
line in the figure above. This line divided the

$xy-$ plane in two half planes
Select a point (not on the line), which lies in one of the half
planes, do determine whether the point satisfies the given inequality or not.We select the point as

$(0,0)$
It is observed that,

$0 < -3$ , which is true
Also, it is evident that any point on the lines does not satisfy
the given strict inequality.
Hence, every point below the line,

$x=-3$ (excluding all
the points on the line determine the solution of the given
inequality.The solution region is represented by the shaded region as as
above.
1767477_1796279_ans_698324a2e916448e9b8b9538e0a621c1.png

1767477_1796279_ans_698324a2e916448e9b8b9538e0a621c1.png

Solver the following system of inequalities graphically : $2x-y > 1, x-2y < -1$

$2x-y > 1$ .... (1)

$x-2y < -1$ .....(2)
The graph of the lines,

$2x-y =1$ and

$x-2y=-1$ , are drawn in the figure below. Inequality (1) represents the region below the line,

$2x-y=1$ ( excluding the line

$2x-y=1$ ). and inequality (2) represents the region above the line,

$x-2y=-1$ , ( excluding the line

$x-2y=-1$ ).
Hence, the solution of the given system of linear inequalities is represented by the common shaded region excluding the points on line as follows.
1767489_1796335_ans_db402faa41004a4b9f58b34e1bff5ef4.png

1767489_1796335_ans_db402faa41004a4b9f58b34e1bff5ef4.png

Solve the given inequality graphically in two-dimensional plane: $3x+4y \ge 12$

$3x+4y \ge 12$
The graphical representation of

$3x+4y=12$ is given as
dotted line in the figure above. This line divided the

$xy-$ plane in two half planes, $$I\
and\ II$$.
Select a point (not on the line), which lies in one of the half
planes, do determine whether the point satisfies the given inequality or not.We select the point as

$(0,0)$
Thus, the solution region of the given inequality is the
shaded half plane

$I$ including the points on the line.This can be represented as above.
1767464_1796268_ans_3c4b7f27c6e64e1284f1d361a4ae638f.png

Solve the following system of inequalities graphically : $x+2y \le 10, x+y \ge 1, x-y \le 0 , x\ge 0, y \ge 0$

$x+2y \le 10$ .......(1)

$x+y \ge 1$ .....(2)

$x -y \le 0$ .....(3)
The graph of the lines,

$x+2y=10, x+y=1$ ,and

$x-y=0$ , are drawn in the figure below.
Inequality (1) represents the region below the line,

$x+2y=10$ ( including the line

$x+2y=10$ ). Inequality (2) represents the region above the line,

$x+y=1$ ( including the line

$x+y=1$ ). Inequality (3) represents the region above the line,

$x-y=0$ ( including the line

$x-y=0$ ).
Since

$x \ge 0$ and

$y \ge 0$ , every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities.
1765535_1796350_ans_fe5b2c3beda246c382dc1cb8ea9ec4b0.png

1765535_1796350_ans_fe5b2c3beda246c382dc1cb8ea9ec4b0.png

Solve $24x < 100$ , when
$x$ is an integer.

The given inequality is

$24x < 100$

$\Rightarrow \dfrac{24x}{24} < \dfrac{100}{24}$

$\Rightarrow x < \dfrac{25}{6}$

The integers less than

$\dfrac{25}{6}$ are

$... -3, -2, 1, 0, 2, 3, 4$ .

Thus, when

$x$ is an integer, the solutions of the given inequality are

$... -3, -2, -1, 0, 1, 2, 3, 4$

Hence, in this case, the solution set is

$\left\{.........-3, -2, -1, 0, 1, 2, 3, 4\right\}$

Solve $5x-3 < 7$ , when
$x$ is a real number

The given inequality is

$5x-3 < 7$ .

$5x-3 < 7$

$\Rightarrow 5x-3+3 < 7+3$

$\Rightarrow 5x < 10$

$\Rightarrow \dfrac{5x}{5} < \dfrac{10}{5}$

$\Rightarrow x < 2$

When

$x$ is this case the solution of the given inequality are given by

$x < 2$ , that is, all real numbers

$x$ which are less than

$2$ .

Thus, the solution set of the given inequality is

$x\in (-\infty, 2)$ .

Solve $-12x > 30$ , when
$x$ is an integer

The given inequality is

$-12x > 30$ .

$\Rightarrow \dfrac{-12x}{-12} < \dfrac{30}{-12}$ [Dividing both sides by same negative number]

$\Rightarrow x < -\dfrac{5}{2}$

The integer less than

$\left(-\dfrac{5}{2}\right)$ are

$..... -5, -4, -3$ . Thus, when

$x$ is an integer, the solution of the given inequality are

$...., -5, -4, -3$

Hence, in this case, the solution set us

$\left\{...., -5, -4, -3\right\}$

Solve $-12x > 30$ , when
$x$ is a natural number

The given inequality is

$-12x > 30$ .

$\Rightarrow \dfrac{-12x}{-12} < \dfrac{30}{-12}$ [Dividing both sides by same negative number]

$\Rightarrow x < -\dfrac{5}{2}$
There is no natural number less than

$\left(-\dfrac{5}{2}\right)$
Thus, when

$x$ is a natural number, there is no solution of the given inequality.

Solve the following system of inequalities graphically : $3x+2y \le 150, x+4y=80, x \le 15, y \ge 0, x \ge 0$

$3x+2y \le 150$ .......(1)

$x+4y =80$ .....(2)

$x \le 15$ .....(3)
The graph of the lines,

$3x+2y=150, x+4y=80$ ,and

$x=15$ , are drawn in the figure below.
Inequality (1) represents the region below the line,

$3x+2y=150$ ( including the line

$3x+2y=150$ ). Inequality (2) represents the region below the line,

$x+4y=80$ ( including the line

$x+4y=80$ ). Inequality (3) represents the region on the left hand side of the line,

$x=15$ ( including the line

$x=15$ ).
Since

$x \ge 0$ and

$y \ge 0$ , every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities.
1768267_1796348_ans_111dae0ee982441894d325a1b9a4f995.png

1768267_1796348_ans_111dae0ee982441894d325a1b9a4f995.png

Solve $5x-3 < 7$ , when
$x$ is an integer

The given inequality is

$5x-3 < 7$ .

$5x-3 < 7$

$\Rightarrow 5x-3+3 < 7+3$

$\Rightarrow 5x < 10$

$\Rightarrow \dfrac{5x}{5} < \dfrac{10}{5}$

$\Rightarrow x < 2$

The integers less than

$2$ are

$...., -4, -3, -2, -1, 0,1$

Thus, when

$x$ is an integer, the solution of the given inequality are

$...., -4, -3, -2, -1, 0,1$ .

Hence, in this case, the solution set is

$\left\{....., -4, -3, -2, -1, 0,1\right\}$ .

Solve $3x+8 > 2$ , when
$x$ is an integer

The given inequality is

$3x+8 > 2$

$\Rightarrow 3x+8-8 > 2-8$

$\Rightarrow 3x >-6$

$\Rightarrow \dfrac{3x}{3} > \dfrac{-6}{3}$

$\Rightarrow x > -2$

The integers greater than

$-2$ are

$-1,0, 1, 2,...$

Thus, when

$x$ is an integer the solutions of the given inequality are

$-1, 0, 1, 2,....$

Hence, in this case, the solution set is

$\left\{-1, 0, 1, 2,....\right\}$

Solve $24x < 100$ , when
$x$ is a natural number

The given inequality is

$24x < 100$

$\Rightarrow \dfrac{24x}{24} < \dfrac{100}{24}$

$\Rightarrow x < \dfrac{25}{6}$
It is evident that

$1,2,3$ and

$4$ are the only natural numbers less than

$\dfrac{25}{6}$
Thus, when

$x$ is a natural number, the solution of the given inequality are

$1,2,3$ and

$4$ .
Hence, in this case, the solution set is

$\left\{1,2,3,4\right\}$

Solve $3x+8 > 2$ , when
$x$ is a real number

The given inequality is

$3x+8 > 2$

$\Rightarrow 3x+8-8 > 2-8$

$\Rightarrow 3x >-6$

$\Rightarrow \dfrac{3x}{3} > \dfrac{-6}{3}$

$\Rightarrow x > -2$

When

$x$ is a real number, the solution of the given inequality are all the real numbers, which are greater than

$-2$ .

Thus, in this case, the solution set is

$(-2, \infty)$ .

Solve the given inequality for real $x:3x-7> 5x-1$

$3x-7 > 5x-1$

$\Rightarrow 3x-7+7 > 5x-1+7$

$\Rightarrow 3x > 5x+6$

$\Rightarrow 3x-5x > 5x+6-5x$

$\Rightarrow -2x > 6$

$\Rightarrow \dfrac{-2x}{-2}< \dfrac{6}{-2}$

$\Rightarrow x < -3$

Thus, all real numbers

$x$ , which are less than

$-3$ , are the solutions of the given inequality.

Hence, the solution set of the given inequality is

$(-\infty, -3)$

Solve the given inequality for real $x:3(2-x)\ge (1-x)$

$3(2-x)\ge 2(1-x)$

$\Rightarrow 6-3x\ge 2-2x$

$\Rightarrow 6-3x+2x\ge 2-2x+2x$

$\Rightarrow 6-x\ge 2$

$\Rightarrow 6-x-6\ge 2-6$

$\Rightarrow -x\ge -4$

$\Rightarrow x\le 4$

Thus, all real numbers

$x$ , which are greater than or equal to

$4$ , are the solution of the given inequality.

Hence, the solution set of the given inequality is

$(-\infty, 4]$ .

Solve the given inequality for real $x:\dfrac{1}{2}\left(\dfrac{3x}{5}+4\right)\ge \dfrac{1}{3}(x-6)$

$\dfrac{1}{2}\left(\dfrac{3x}{5}+4\right)\ge \dfrac{1}{3}(x-6)$

$\Rightarrow \dfrac{9x}{5}+12\ge 2x-12$

$\Rightarrow 12+12\ge 2x-\dfrac{9x}{5}$

$\Rightarrow 24\ge \dfrac{10x-9x}{5}$

$\Rightarrow 24\ge \dfrac{x}{5}$

$\Rightarrow 120\ge x$

Thus, all real numbers

$x$ , which are less than or equal to

$120$ , are the solutions of the given inequality.

Hence, the solution set of the given inequality is

$(-\infty, 120]$ .

Solve the given inequality for real $x:37-(3x+5)\ge 9x-8(x-3)$

$37-(3x+5)\ge 9x-8(x-3)$

$\Rightarrow 37-3x-5\ge 9x-8x+24$

$\Rightarrow 32-3x\ge x+24$

$\Rightarrow$ 32-24\ge x+3x$$

$\Rightarrow 8\ge 4x$

$\Rightarrow 2\ge x$

Thus all the ral numbers

$x$ , which are less than or equal to

$2$ , are the solutions of the given inequality.

Hence, the solution set of the given inequality is

$(-\infty, 2]$

Solve the given inequality for real $x:\dfrac{3(x-2)}{5}\le \dfrac{5(2-x)}{3}$

$\dfrac{3(x-2)}{5}\le \dfrac{5(2-x)}{3}$

$\Rightarrow 9(x-2)\le 25(2-x)$

$\Rightarrow 9x-18+25x\le 50$

$\Rightarrow 34x-18\le 50$

$\Rightarrow 34x\le 50+18$

$\Rightarrow 34x\le 68$

$\Rightarrow \dfrac{34x}{34}\le \dfrac{68}{34}$

$\Rightarrow x\le 2$

Thus, all real numbers,

$x$ , which are less than or equal to

$2$ , are the solutions of the given hence the solution set of the given inequality is

$(-\infty, 2]$ .

Solve the given inequality $x:2(2x+3)-10 < 6(x-2)$

$2(2x+3)-10 < 6(x-2)$

$\Rightarrow 4x+6-10<6x-12$

$\Rightarrow 4x-4<6x-12$

$\Rightarrow -4+12<6x-4x$

$\Rightarrow 8<2x$

$\Rightarrow 4<x$

Thus, all real numbers

$x$ , which are greater than

$4$ , are the solution of the given inequality.

Hence, the solution set of the given inequality is

$)4, \infty)$ .

Solve the given inequality for real $x:\dfrac{x}{3} > \dfrac{x}{2}+1$

$\dfrac{x}{3} > \dfrac{x}{2}+1$

$\Rightarrow \dfrac{x}{3}-\dfrac{x}{2} > 1$

$\Rightarrow \dfrac{2x-3x}{6} > 1$

$\Rightarrow \dfrac{x}{6} > 1$ $

$\Rightarrow -x > 6$

$\Rightarrow x < -6$

Thus, all real numbers

$x$ , which are less than

$-6$ , are the solutions of the given inequality.

Hence, the solution set of the given inequality is

$(-\infty, -6)$

Solve the given inequality for real $x:3(x-1) \le 2(x-3)$

$3(x-1)\le 2(x-3)$

$\Rightarrow 3x-3\le 2x-6$

$\Rightarrow 3x-3+3\le 2x-6+3$

$\Rightarrow 3x\le 2x-3$

$\Rightarrow 3x-2\le 2x-3-2x$

$\Rightarrow x\le -3$

Thus, all real numbers

$x$ , which are less than or equal to

$-3$ , are the solutions of the given inequality.

Hence, the solution set of the given inequality is

$(-\infty, -3]$ .

Solve the given inequality for real $x:4x+3 < 5x+7$

$4x+3 < 5x+7$

$\Rightarrow 4x+3-7 < 5x+7-7$

$\Rightarrow 4x-4 < 5x$

$\Rightarrow 4x-4-4x < 5x -4x$

$\Rightarrow -4 < x$

Thus, all real numbers

$x$ , which are greater than

$-4$ , are the solution of the given inequality.

Hence, the solution set of the given inequality is

$(-4,\infty)$ .

Solve the given inequality for real $x:x+\dfrac{x}{2}+\dfrac{x}{3} < 11$

$x+\dfrac{x}{2}+\dfrac{x}{3} < 11$

$\Rightarrow x\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right) < 11$

$\Rightarrow x\left(\dfrac{6+3+2}{6}\right) < 11$

$\Rightarrow \dfrac{11x}{6} < 11$

$\Rightarrow \dfrac{11x}{6\times 11} < \dfrac{11}{11}$

$\Rightarrow \dfrac{x}{6} < 1$

$\Rightarrow x < 6$
Thus, all real numbers

$x$ , which are less than

$6$ m are the solutions of the given inequality.
Hence, the solution set of the given inequality is

$(-\infty, 6)$ .

Solve the given inequality for real $x:\dfrac{(2x-1)}{3} \ge \dfrac{(3x-2)}{4}-\dfrac{(2-x)}{5}$

$\dfrac{(2x-1)}{3}\ge \dfrac{(3x-2)}{4}-\dfrac{(2-x)}{5}$

$\Rightarrow \dfrac{(2x-1)}{3}\ge \dfrac{5(3x-2)-4(2-x)}{20}$

$\Rightarrow \dfrac{(2x-1)}{3}\ge \dfrac{15x-10-8+4x}{20}$

$\Rightarrow \dfrac{(2x-1)}{3}\ge \dfrac{19x-18}{20}$

$\Rightarrow 20(2x-1)\ge 3(19x-18)$

$\Rightarrow 40x-20\ge 57x-54$

$\Rightarrow -20+54\ge 57x-40$

$\Rightarrow 34\ge 17x$

$\Rightarrow 2\ge x$

Thus, all real numbers

$x$ , which are less than or equal to

$2$ , are the solution, of of the given inequality.

Hence, the solution set of the given inequality is

$(-\infty, 2]$

Solve the given inequality and show the graph of the solution on number line:
$5x-3\ge 3x-5$

$5x-3\ge 3x-5$

$\Rightarrow 5x-3x\ge -5+3$

$\Rightarrow 2x\ge -2$

$\Rightarrow \dfrac{2x}{2}\ge \dfrac{-2}{2}$

$\Rightarrow x\ge -1$
The graphical representation of the solution of the given inequality is as follows:
1765585_1796446_ans_fc11bb5f91584a068202daf0a68fcf2c.png

1765585_1796446_ans_fc11bb5f91584a068202daf0a68fcf2c.png

To receive Grade $A$ in a course, one must obtain an average of $90$ marks or more in five examinations (each of $100$ marks). If Sunita's marks in first four examination are $87, 92, 94$ and $95$ , find minimum marks that sunita must obtain in fifth examination to get grade $'A'$ in the course.

Let

$x$ be the marks obtained by Sunita in the fifth examination.
Therefore

$\dfrac{87+92+94+95+x}{5}\ge 90$

$\Rightarrow \dfrac{368+x}{5}\ge 90$

$\Rightarrow 368+x\ge 450$

$\Rightarrow x\ge 450-368$

$\Rightarrow x\ge 82$
Thus, Sunita must obtain greater than or equal to

$82$ marks in the fifth examination.

Find all pairs of consecutive even positive integers, both of which are larger than $5$ such that their sum is less than $23$ .

Let

$x$ be the smaller of the two consecutive even positive integers. Then, the other integer is

$x+2$

Since both the integers are larger than

$5$ ,

$x > 5$ .... (1)

Also, the sum of the two integers is less than

$23$

$x+(x+2) < 23$

$\Rightarrow 2x+2 < 23$

$\Rightarrow 2x < 23-2$

$\Rightarrow 2x < 21$

$\Rightarrow x < \dfrac{21}{2}$

$\Rightarrow x < 10.5$ ..... (2)

From (1) and (2), we obtain

$5 < x < 10.5$

Since

$x$ is an even number,

$x$ can take the values

$6,8$ and

$10$ .

Thus, the required possible pairs are

$(6,8), (8,10)$ and

$(10,12)$

Solve the given inequality and show that graph of the solution on number line:
$\dfrac{x}{2} > \dfrac{(5x-2)}{3}-\dfrac{(7x-3)}{5}$

$\dfrac{x}{2}\ge \dfrac{(5x-2)}{3}-\dfrac{(7x-3)}{5}$

$\Rightarrow \dfrac{x}{2}\ge \dfrac{5(5x-2)-3(7x-3)}{15}$

$\Rightarrow \dfrac{x}{2}\ge \dfrac{25x-10-21x+9}{15}$

$\Rightarrow \dfrac{x}{2}\ge \dfrac{4x-1}{15}$

$\Rightarrow 15x\ge 2(4x-1)$

$\Rightarrow 15\ge 8x-2$

$\Rightarrow 15-8x\ge 8x-2-8x$

$\Rightarrow 7x\ge -2$

$\Rightarrow x\ge -\dfrac{2}{7}$
The graphical representation of the solution of the given inequality is as follows.
1765643_1796452_ans_3918c86797294f078e868b6c38b786c4.png

1765643_1796452_ans_3918c86797294f078e868b6c38b786c4.png

Solve the given inequality and show the graph of the solution on number line:
$3(1-x) < 2 (x+4)$

$3(1-x) < 2(x+4)$

$\Rightarrow 3-3x < 2x+8$

$\Rightarrow 3-8 < 2x+3x$

$\Rightarrow -5x < 5x$

$\Rightarrow \dfrac{-5}{5} < \dfrac{5x}{5}$

$\Rightarrow -1 < x$
The graphical representation of the solutions of the given inequality is as follows:
1765642_1796448_ans_730fe092bed6418dac73a97fcc52321a.png

1765642_1796448_ans_730fe092bed6418dac73a97fcc52321a.png

Find all pairs of consecutive odd positive integers both which are smaller than $10$ such that their sum is more than $11$ .

$\textbf{Step 1: Analysing the question given}$

$\text{The odd numbers smaller than 10 are}$

$1,3,5,7,9$

$\textbf{Step 2: Finding possible solutions}$

$9+7>11$

$9+5>11$

$9+3>11$

$7+5>11$

$\textbf{Hence, the pair of consecutive odd integers both smaller than 10 such that}$

$\textbf{their sum is more than 11 are (5,7), (7,9)}$ .

Ravi obtained $70$ and $75$ marks in the first two-unit test. Find the minimum marks he should get in the third test to have an average of at least $60$ marks.

Let

$x$ be the marks obtained by Ravi in the third unit test.

Since the student should have an average of at least

$60$ marks.

$\dfrac{70+75+x}{3}\ge 60$

$\Rightarrow 145+x\ge 180$

$\Rightarrow x\ge 180-145$

$\Rightarrow x\ge 35$

Thus, the student must obtain a minimum of

$35$ marks to have an average of at least

$60$ marks.

Solve the given inequality for real $x:\dfrac{x}{4} < \dfrac{(5x-2)}{3}-\dfrac{(7x-3)}{5}$

$\dfrac{x}{4}<\dfrac{(5x-2)}{3}-\dfrac{(7x-3)}{5}$

$\Rightarrow \dfrac{x}{4}<dfrac{5(5x-2)-3(7x-3)}{15}$

$\Rightarrow \dfrac{x}{4} < \dfrac{25x-10-21x+9}{15}$

$\Rightarrow \dfrac{x}{4} < \dfrac{4x-1}{15}$

$\Rightarrow 15x < 4(4x-1)$

$\Rightarrow 15x < 16x-4$

$\Rightarrow 4 < 16x-15x$

$\Rightarrow 4 < x$

Thus, all real numbers

$x$ , which are greater than

$4$ , are the solution of the given inequality.

Hence, the solution set of the given inequality is

$(4, \infty)$ .

Solve the given inequality and show the graph of the solution on number line:
$3x-2 < 2x+1$

$3x-2 < 2x+1$

$\Rightarrow 3x-2x < 1+2$

$\Rightarrow x < 3$
The graphical representation of the solutions of the given inequality is as follows:
1765580_1796440_ans_5575e2290fe743bbbc1f4d22d85c7179.png

1765580_1796440_ans_5575e2290fe743bbbc1f4d22d85c7179.png

Solve $-7< 4x + 1 \leq 23, x \, \epsilon\, l$

Given,

$-7 < 4x + 1 \leq 23$
We take

$-7 < 4x + 1 \leq 23$
Subtracting -1 from all sides,

$-7 -1 < 4x + 1 -1 \leq 23 - 1$

$-8 < 4x \leq 22$

$\dfrac{-8}{4} < \dfrac{4x}{4} \leq \dfrac{22}{4}$ ( Divding by 4)

$-2 < x \leq 5.5$
As

$x\, \epsilon\, l$ , the solution set= {-1, 0, 1, 2, 3, 4, 5}

Represent the solution of the following inequalities graphically :
$x \leq 4 , x\, \varepsilon \, N$

Given

$x \leq 4 , x\, \varepsilon \, N$

The solution set

$= \left \{ 1 , 2 , 3 , 4 \right \}$

These four numbers are shown indicating with thick dot on the number line.

1707003_1808995_ans_bf0d1de0f612406a8b75dac49ee54a17.PNG

Solve $3 - 4x < x - 12, x\, \epsilon\, {-1, 0, 1, 2, 3, 4, 5, 6, 7}$

Given

$3 - 4x < x - 12$
Subtracting 3 from both sides

$\implies\, -3 + 3 - 4x< x - 12 - 3$

$\implies\, -4x < x - 15$
Subtracting x from both sides

$\implies\, -4x -x < x - x -15$

$\implies\, -5x < -15$

$\implies\, x > 3$
(Dividing by -5 and reverse the symbols)
As

$x\, \epsilon { -1, 0, 1, 2, 3, 4, 5, 6, 7}$
The solution set= { 4, 5, 6, 7}

Solve the following inequations and represent their solution on a number line:
(i) $\dfrac{2x - 1}{3} \leq 2 \dfrac{1}{2}, x \epsilon W$
(ii) $-1 < \dfrac{2x}{3} + 1 \leq 4, x\, \epsilon\, l$

(i) Given

$\dfrac{2x - 1}{3} \leq 2 \dfrac{1}{2}$

$\implies\,\dfrac{2x - 1}{3} \leq \dfrac{5}{2}$

Multiplying both sides by 6, we get

$\dfrac{6(2x -1)}{3} \leq \dfrac{5 \times 6}{2}$

$4x -2 \leq 15$

$\implies\, 4x \leq 15 + 2$

$4x \leq 17$

$\implies\,x \leq \dfrac{17}{4}$

$x\, \epsilon\, W$ , the solution set =

${0, 1, 2, 3, 4}$

(ii)

$-1 < \dfrac{2x}{3} + 1 \leq 4, x\, \epsilon\, l$

Given

$-1 < \dfrac{2x}{3} + 1 \leq 4$

Multiplying by 3,

$-3 < 2x + 3 \leq 12$

$-3 -3 < 2x + 3 -3 \leq 12 -3$

( Subtracting -3)

$-6 < 2x \leq 9$

$\dfrac{-6}{2} < x \leq \dfrac{9}{2}$ (Dividing by 2)

$-3 < x \leq \dfrac{9}{2}$

As x

$\epsilon$ l, the solution set

${-2, -1, 0, 1, 2, 3, 4}$

Represent the solution of the following inequalities graphically :
$x < 5 , x \, \varepsilon\, W$

Given

$x < 5 , x \, \varepsilon\, W$

The solution set

$= \left \{ 0 , 1 , 2, 3 , 4 \right \}$

These five numbers are shown indicating with thick dots on the number line.

1707525_1808996_ans_6cceb9d9ce924c569627d8b1109933f2.PNG

A man wants to cut three from a single piece of board of length $91\ cm$ , The second length is to be $3\ cm$ longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at lest $5\ cm$ longer than the second?

Let us assume the length of the shortest piece be

$x\ cm$

∴ According to the question, length of the second piece

$= (x + 3)\ cm$

And, length of the third piece

$= 2x\ cm$

As all the three lengths are to be cut from a single piece of the board having a length of

$91\ cm$

$∴ x + (x + 3) + 2x ≤ 91\ cm$

$= 4x + 3 ≤ 91$

$= 4x ≤ 88$

$=\dfrac{4x}4=\dfrac{88}4$

$= x ≤ 22 …(i)$

Also, it is given in the question that, the third piece is at least 5 cm longer than the second piece

$∴ 2x ≥ (x+3) + 5$

$2x ≥ x + 8$

$x ≥ 8 …(ii)$

Thus, from equation (i) and (ii) we have:

$8 ≤ x ≥ 22$

Hence, it is clear that the length of the shortest board is greater than or equal to

$8\ cm$ and less than or equal to

$22\ cm$

Represent the following inequations graphically:
(i) $x \leq 3, x \epsilon N$
(ii) $x<4, x \epsilon W$
(iii) $-2 \leq x< 4, x \epsilon l$
(iv) $-3 \leq x \leq 2, x \epsilon l$

(i) Given

$x \leq 3, x \epsilon\, N$
The solution set = {1,2,3}
The solution set is shown by thick dots on the number line.

(ii)

$x < 4, x \epsilon\, W$
The solution set = {0, 1, 2, 3}
The solution set is shown by thick dots on the number line.

(iii)

$-2 \leq x <4, x \epsilon\, l$
The solution set = {-2, -1, 0, 1, 2, 3}
The graph of the solution set is shown by thick dots on the number line.

(iv)

$-3 \leq x \leq 2, x \epsilon\, l$
The solution set ={-3, -2, -1, 0, 1, 2}
The graph of the solution set is shown by thick dots on the number line.

Solve the following inequations.
(i) $4 - x > -2, x\, \epsilon\, N$
(ii) $3x + 1 \leq 8, x\, \epsilon\, W$
Also represent their solutions on the number line.

(i) Given, 4 - x > -2
Subtract 4 from both sides

$\implies\, -4 + 4 - x > -2 -4 - x > -6$

$\implies\, x < 6$ (Reverse the symbols)
As

$x\, \epsilon\, N$ , the solution set= {1, 2, 3, 4, 5}
The graph of the solution set

(ii) Given

$3x+ 1 \leq 8$
Subtracting -1 from both sides,

$3x + 1 -1 \leq 8 - 1$

$3x \leq 7$
Dividing both sides by 3

$\implies\, x \leq \dfrac{7}{3}$
As x= W, the solution set= {0, 1, 2}
The graph of the solution set

Solve the following inequations and graph their solutions on a number line
$-1 < \dfrac{x}{2} + 1 \leq 3 , x \, \varepsilon\, I$

Given

$-1 < \dfrac{x}{2} + 1 \leq 3$

By subtracting

$-1$ , we get ,

$-1 -1 < \left ( \dfrac{x}{2} + 1 \right ) - 1 \leq 3 - 1$

$-2 < \dfrac{x}{2} \leq -2$

Multiplying by

$2$ , we get,

$-2 \times 2 < \dfrac{x}{2} \times 2 \leq - 2 \times 2$

$-4 < x \leq - 4$

$x \, \varepsilon\, I$ ,

Therefore solution set

$= \left \{-3 , -2 , -2 , 0 , 1 , 2 , 3 , 4 \right \}$

The graphical representation for this solution set is shown indicating with thick dots on the number line.

1707629_1809031_ans_87700f7ae4bb4506a39b8e4f39694219.PNG

If the replacement set is $\left \{ - 3 , -2 , -1 , 0 , 1 , 2 , 3 \right \}$ solve the inequation $\dfrac{3x - 1}{2} < 2$ . Represent its solution on the number line.

Given

Replacement set

$= \left \{-3 , -2 , -1 , 0 , 1 , 2 , 3 \right \}$ and

Inequation

$= \dfrac{3x - 1}{2} < 2$

$3x - 1 < 2 \times 2$

$3x - 1 < 4$

$3x < 4 + 1$

$3x < 5$

We get ,

$x < \dfrac{5}{3}$

Therefore , solution

$= \left \{ -3 , -2 , -1 , 0 ,1 \right \}$

The graphical representation for this solution set is shown indicating with thick dots on the number line.

1707684_1809027_ans_d9cc024e28e2465e888735a351fffbbb.PNG

Solve the following inequation :
$\dfrac{2}{3}p + 5 < 9 , p\, \varepsilon \, W$

Given

$\dfrac{2}{3}p + 5 < 9 \\$

$\dfrac{2}{3}p < 9 - 5 \\$

$\dfrac{2}{3}p < 4 \\$

$2p < 4 \times 3\\$

$2p < 12$

We get,

$p < \dfrac{12}{2} \\$

$p < 6$

$p\, \varepsilon \, W$ ,

Therefore , solution set

$= \left \{0 , 1 , 2 , 3 , 4 , 5 \right \}$

Solve the following inequations and graph their solutions on a number line
$-4 \leq 4x < 14 , x \, \varepsilon \, N$

Given

$-4 \leq 4x < 14$

Dividing by

$4$ , we get,

$\dfrac{-4}{4} \leq \dfrac{4x}{4} < \dfrac{14}{4}$

$-1 \leq x \leq \dfrac{7}{2}$

$x \, \varepsilon \, N$ ,

Therefore solution set

$= \left \{ 1 , 2, 3 \right \}$

The graphical representation for this solution set is shown indicating with thick dots on the number line.

1707641_1809030_ans_40648923a71d479f93fc0445af1df3dc.PNG

Solve $\dfrac{x}{3} + \dfrac{1}{4} < \dfrac{x}{6} + \dfrac{1}{2} , x \, \varepsilon\, W$ . Also represent its solution on the number line.

Given :

$\dfrac{x}{3} + \dfrac{1}{4} < \dfrac{x}{6} + \dfrac{1}{2}$

$\dfrac{x}{3} - \dfrac{x}{6} < \dfrac{1}{2} - \dfrac{1}{4}$

$\dfrac{2x - x}{6} < \dfrac{2 - 1}{4}$

$\dfrac{x}{6} < \dfrac{1}{4}$

$x < \dfrac{6}{4}$

We get ,

$x < \dfrac{3}{2}$

$x \, \varepsilon\, W$ ,

Hence solution set

$= \left \{ 0 , 1 \right \}$

The graphical representation for this solution set is shown indicating with thick dots on the number line.

1707665_1809029_ans_ed523fd80dd7445daa026a86a4860f96.PNG

Represent the solution of the following inequalities graphically :
$-3 \leq x < 3 , x \, \varepsilon \, 1$

Given

$-3 \leq x < 3 , x \, \varepsilon \, 1$

The solution set

$= \left \{-3 , -2 , 1 , 0 , 1 , 2 \right \}$

These six numbers are shown indicating with thick dots on the number line.

1707557_1808998_ans_eeb32de60b714ab28f63e798b3f48344.PNG

If the replacement set is $\left \{-6 , -4 , -2 , 0 , 2 , 4 , 6 \right \}$ ; then represent the solution set of the inequality $-4 \leq x < 4$ graphically

Given

Replacement set

$= \left \{-6 , -4 , -2 , 0 , 2 , 4 , 6 \right \}$

Inequality

$= -4 \leq x < 4$

Solution set

$= \left \{ -4 , -2 , 0 , 2 \right \}$

The graphical representation for this solution set is shown indicating with thick dots on the number line.

1707688_1808999_ans_0e382825cac34bcdb7d1a9b621b4d474.PNG

If $25-4 x \leq 16,$ find the smallest value of $\mathbf{x},$ when $\mathbf{x}$ is a real number.

Given,

$25-4 x \leq 16$

$-4 x \leq 16-25$

$-4 x \leq-9$

$\mathrm{x} \geq 9 / 4$

$x \geq 2.25$

Now, The smallest value of

$x$ , when

$x$ is a real number is

$2.25 .$

Represent the following inequalities on real number line:
$8 \geq x>-3$

$8 \geq x>-3$

Solution on the number line is shown in figure.

1792015_1827944_ans_934e8b15a31a4e6783b6d6f9b37b655b.png

Represent the following inequalities on the real number line:
$2 x-1<5$

Given,

$2 x-1<5$

$2 x<6$

$x<3$

Solution on the number line is shown in figure.

1791994_1827925_ans_b319e45ad0ba4a2385685848796701eb.png

Represent the following inequality on real number line:
$-4<x<4$

$-4<x<4$

Solution on the number line is shown in figure.

1792009_1827939_ans_a791de6de885477a8bd5ef5786709cbc.png

Represent the following inequality on real number line:
$3 x+1 \geq-5$

Given,

$3 x+1 \geq-5$

$3 x \geq-5-1$

$3 x \geq-6$

$x \geq-2$

Solution on the number line is shown in figure.

1791998_1827926_ans_0ae8eef88ba44c7f8b5f39b6255df19c.png

Solve the inequation:
$3-2 \mathbf{x} \geq \mathbf{x}-12$ given that $\mathbf{x} \in \mathbf{N}$

Given,

$3-2 x \geq x-12$

$-2 x-x \geq-12-3$

$-3 x \geq-15$

$x \leq 5$

$As, \mathrm{x} \in \mathrm{N}$

Thus, the solution set

$=\{1,2,3,4,5\}$

For given graph alongside, write an inequation taking $\mathbf{x}$ as the variable:

For given graph an inequation is

$x \leq -1 ,x \in \mathrm{R}$

If $25-4 x \leq 16,$ find the smallest value of $x$ , when $x$ is an integer.

Given,

$25-4 x \leq 16$

$-4 x \leq 16-25$

$-4 x \leq-9$

$\mathrm{x} \geq 9 / 4$

$x \geq 2.25$

Now,

The smallest value of

$x$ , when

$x$ is an integer is

$3 .$

Represent the following inequalities on real number line:
$-2 \leq x<5$

$-2 \leq x<5$

Solution on the number line is shown in figure.

1792011_1827940_ans_2b0548e23d694bb9a17dfcffc88c7b08.png

Represent the following inequality on real number line:
$2(2 x-3) \leq 6$

Given,

$2(2 x-3) \leq 6$

$4 x-6 \leq 6$

$4 x \leq 12$

$x \leq 3$

Solution on the number line is shown in figure.

1792002_1827938_ans_43b57d297d954aa6bb81c9a6f1fc441d.png

For the given graph alongside, write an inequation taking $\mathbf{x}$ as the variable.

For given graph an inequation is

$-1 < x \leq 5 , x \in \mathrm{R}$

For the given graph alongside, write an inequation taking $\mathbf{x}$ as the variable.

For given graph an inequation is

$x \geq 2, x \in \mathrm {R}$

Represent the solution of each of the following inequations on the real number line:
$x+3 \leq 2 x+9$

Given,

$x+3 \leq 2 x+9$

$x-2 x \leq-3+9$

$-x \leq 6$

$x \geq-6$

The solution on number line is shown in figure.

1793392_1828172_ans_9a3ac5c78f264984890ee449d8d71a3a.png

For the following inequation, graph the solution set on the real number line:
$-4 \leq 3 x-1<8$

Given,

$-4 \leq 3 x-1<8$

$-4 \leq 3 x-1 \quad$ and

$\quad 3 x-1<8$

$-3 \leq 3 x \quad$ and

$\quad 3 x<9$

$-1 \leq 3 x \quad$ and

$\quad x<3$

Solution on the number line is shown in figure.

1792573_1827971_ans_3198196f4183496bb87e45b441924046.png

For the following inequation, graph the solution set on the real number line:
$x-1<3-x \leq 5$

Given,

$x-1<3-x \leq 5$

$\begin{array}{lll}\mathrm{x}-1<3-\mathrm{x} & \text { and } & 3-\mathrm{x} \leq 5 \\ 2 \mathrm{x}<4 & \text { and } & -\mathrm{x} \leq 2 \\ \mathrm{x}<2 & \text { and } & \mathrm{x} \geq 2\end{array}$

Solution on the number line is shown in figure.

1792580_1827973_ans_c682e769b8c4422896a0c03784577518.png

Represent the solution of each of the following inequations on the real number line:
$7-x \leq 2-6 x$

Given,

$7-x \leq 2-6 x$

$6 x-x \leq 2-7$

$5 x \leq-5$

$x \leq-1$

Solution on the number line is shown in figure.

1792588_1828168_ans_2a78ac2f755c4a17b44479048af4adaf.png

Represent the solution of each of the following inequations on the real number line:
$4 x-1>x+11$

Given,

$4 x-1>x+11$

$4 x-x>1+11$

$3 x>12$

$x>4$

The solution on number line is shown in figure.

1793388_1828155_ans_a64bd8b877d74df9bc4ca7eaa1e6ec2a.png

Represent the solution of each of the following inequations on the real number line:
$1+x \geq 5 x-11$

Given,

$1+x \geq 5 x-11$

$12 \geq 4 x$

$x \leq 3$

The solution on number line is shown in figure.

1793402_1828177_ans_561cef051d8044238318cac3dfb07bb4.png

For the given graph alongside, write an inequation taking $\mathbf{x}$ as the variable.

For given graph an inequation is

$-4 \leq x < 3, x \in \mathrm{R}$

Represent the solution of each of the following inequations on the real number line:
$2-3 x>7-5 x$

Given,

$2-3 x>7-5 x$

$5 x-3 x>7-2$

$2 x>5$

$x>5 / 2$

$x>2.5$

The solution on number line is shown in figure.

1793397_1828175_ans_c7c0ea1d42d94e4fade78c3b6e82382e.png

Solve and graph the solution set of:
$x+5 \geq 4(x-1)$ and $3-2 x<-7, x \in R$

$x+5 \geq 4(x-1)$ and

$\quad 3-2 x<-7$

$9 \geq 3 x$ and

$\quad-2 \mathrm{x}<-10$

$3 \geq x$ and

$\quad x>5$

Hence, the solution set = Empty set (as there is no cross section)

Find the range of values of $x$ which satisfies
$-2 \dfrac{2}{3} \leq x+\dfrac{1}{3}<3 \dfrac{1}{3} ; x \in R$
Graph these values of $x$ on the number line.

Given,

$-2 \dfrac{2}{3} \leq x+\dfrac{1}{3}<3 \dfrac{1}{3} ; x \in R$

$-2 \dfrac{2}{3} \leq x+\dfrac{1}{3}$ and

$x+\dfrac{1}{3}<3 \dfrac{1}{3}$

$\Rightarrow-\dfrac{8}{3} \leq x+\dfrac{1}{3}$ and

$x+\dfrac{1}{3}<\dfrac{10}{3}$

$\Rightarrow-\dfrac{8}{3}-\dfrac{1}{3} \leq x$ and

$x<\dfrac{10}{3}-\dfrac{1}{3}$

$\Rightarrow-\dfrac{9}{3} \leq x$ and

$x<\dfrac{9}{3}$

$\Rightarrow-3 \leq x$ and

$x<3$

Therefore,

$-3 \leq x<3$

And the required graph of the solution set is given in figure.

1793875_1828196_ans_c88fcfa91a02406ab7b407b0db2cee99.png

$\mathrm{x} \in\{\text { real numbers }\}$ and $-1<3-2 x \leq 7$ , evaluate $x$ and represent it on a number line.

Given,

$-1<3-2 x \leq 7$

$-1<3-2 x \quad$ and

$\quad 3-2 x \leq 7$

$2 x<4 \quad$ and

$\quad-2 x \leq 4$

$\mathrm{x}<2 \quad$ and

$\quad \mathrm{x} \geq-2$

The solution set

$=\{-2 \leq x<2, x \in R\}$

Hence, the solution can be represented on a number line as shown in figure.

1793888_1828188_ans_7408909f01f84d2685c4d078c969d41a.png

Solve and graph the solution set of:
$2 x-9<7$ and $3 x+9 \leq 25, x \in R$

${\textbf{Step -1: Write the two solution set given}}{\textbf{.}}$

$2{\text{x}} - 9 < 7{\text{ and 3x}} + 9 \leqslant 25$

${\textbf{Step -2: Solve the given solution set}}{\textbf{.}}$

$2{\text{x > 16 and 3x}} \leqslant {\text{16}}$

${\text{x < 8 and x}} \leqslant {\text{ 5}}\frac{1}{3}$

$\therefore {\text{Solution set = }}\left\{ {{\text{x}} \leqslant {\text{5}}\frac{1}{3},{\text{ x}} \in {\text{R}}} \right\}$

${\text{The required graph on number line is:}}$

$\textbf{Therefore, given equations are solved }$

Given $\mathbf{x} \in\{\text { real numbers }\}$ , find the range of values of $\mathbf{x}$ for which $-5 \leq 2 x-3<x+2$ and represent it on a real number line.

Given inequation,

$-5 \leq 2 x-3<x+2$

$-5 \leq 2 x-3 \quad$ and

$\quad 2 x-3<x+2$

$-2 \leq 2 x \quad$ and

$\quad x<5$

$-1 \leq x$ and

$\quad x<5$

Thus, the required range is

$-1 \leq x<5$

And the required graph is given figure.

1794180_1828210_ans_8ced69d4707147f2858129a3ab984156.png

Solve and graph the solution set of:
$2 x-9 \leq 7$ and $3 x+9>25, x \in \mathbf{I}$

Given,

$2 x-9 \leq 7$ and

$3 x+9>25$

$2 x \leq 16$ and

$3 x>16$

$x \leq 8$ and

$x>16 / 3$

Thus, the solution set

$=\{16 / 3<\mathrm{x} \leq 8, \mathrm{x} \in \mathrm{I}\}=\{6,7,8\}$

And the required graph on number line is as shown in figure.

1793180_1828222_ans_9fe45be114cc47b48f9ac5ba277eccb2.png

Solve and graph the solution set of:
$3 x-2>19$ or $3-2 x \geq-7, x \in R$

Given,

$\begin{array}{lll} 3 \mathrm{x}-2>19 & \text { or } & 3-2 \mathrm{x} \geq-7\end{array}$

$3 x>21 \quad$ or

$\quad-2 x \geq-10$

$x>7 \quad$ or

$\quad x \leq 5$

The graph of solution set of

$x>7$ or

$x \leq 5$ is equal to the graph of points which belong to

$x>7$ or

$x \leq 5$ or both.

Thus, the graph of the solution set is as shown in figure.

1793169_1828726_ans_d8f558aa815c40b6b913c5a5af62d60d.png

Solve the following inequation and graph the solution set on the number line:
$2 x-3<x+2 \leq 3 x+5, x \in R$

Given inequation,

$2 x-3<x+2 \leq 3 x+5$

$2 x-3<x+2$ and

$x+2 \leq 3 x+5$

$x<5 \quad$ and

$\quad-3 \leq 2 x$

$x<5$ and

$\quad-1.5 \leq x$

So, the solution set

$=\{-1.5 \leq x<5\}$

And the solution set be graphed on the number line is shown in figure.

1794186_1828216_ans_dabd9d4acd7b42708b2aaeb2c7a2238f.png

Find the values of $x$ , which satisfy the equation:
$-2 \leq \dfrac{1}{2}-\dfrac{2 x}{3} \leq 1 \dfrac{5}{6}, x \in N$
Graph the solution on the number line.

Given inequation,

$-2 \leq \dfrac{1}{2}-\dfrac{2 x}{3} \leq 1 \dfrac{5}{6}, x \in N$

$\Rightarrow-2-\dfrac{1}{2} \leq \dfrac{1}{2}-\dfrac{2 x}{3}-\dfrac{1}{2} \leq \dfrac{11}{6}-\dfrac{1}{2}$

$\Rightarrow-\dfrac{5}{2} \leq-\dfrac{2 x}{3} \leq \dfrac{8}{6}$

$\Rightarrow-\dfrac{5}{2} \leq-\dfrac{2 x}{3}$ and

$-\dfrac{2 x}{3} \leq \dfrac{8}{6}$

$\Rightarrow-15 \leq-4 x$ and

$-2 x \leq 4$

$\Rightarrow 15 \geq 4 x$ and

$2 x \geq-4$

$\Rightarrow \dfrac{15}{4} \geq x$ and

$x \geq-2$

$\Rightarrow 3.75 \geq x$ and

$x \geq-2$

Hence, the solution set is

$\{x \in N:-2 \leq x \leq 3.75\}$

And as

$\mathrm{x} \in \mathrm{N},$ the values of

$\mathrm{x}$ are 1,2,3

The required graph of the solution on the number line is shown in figure.

1794628_1828204_ans_e6e9a074a5604fe7a4fc0acf02e9fb67.png

Represent the solution of each of the following inequations on the real number line:
$(2 x+5) / 3>3 x-3$

Given,

$(2 x+5) / 3>3 x-3$

$2 x+5>3(3 x-3)$

$2 x+5>9 x-9$

$9+5>9 x-2 x$

$7 x<14$

$x<2$

The solution on number line is shown in figure.

1794176_1828180_ans_c1a33b4748264eedb43945f53ea343c3.png

Solve and graph the solution set of:
$5>\mathrm{p}-1>2$ or $7 \leq 2 \mathrm{p}-\leq 17, \mathrm{p} \in \mathrm{R}$

Given,

$5>p-1>2 \quad$ or

$7 \leq 2 p-1 \leq 17$

$6>p>3$

$8 \leq 2 p \leq 18$

$6>\mathrm{p}>3 \quad$ or

$\quad 4 \leq \mathrm{p} \leq 9$

Now, we have to understand that:

Graph of solution set of

$6>\mathrm{p}>3$ or

$4 \leq \mathrm{p} \leq 9$

= Graph of points which belong to

$6>\mathrm{p}>3$ or

$4 \leq \mathrm{p} \leq 9$ or both

= Graph of points which belong to

$3<\mathrm{p} \leq 9$

Thus, the graph of the solution set is as shown in figure.

1793173_1828732_ans_9c26c6deda254a5fa33d87aedf9207d5.png

Draw graph of : $x \geq 0$

${\textbf{Step-1: Describe}}$

$x \geqslant 0.$

${\text{Consider the line whose equation is}}$

$x =0.$

${\text{To find the solution set, we have to check any point other than origin.}}$

${\text{Let us check the point (1, 1).}}$

${\text{Here}}$

$x = 1 \Rightarrow x \geqslant 0.$

$\therefore$

${\text{(1, 1) lies in the required region.}}$

${\textbf{Step-2: Draw a graph and show required part.}}$

${\text{Therefore, the solution set is the Y-axis and the right side of the Y-axis which is shaded in the graph.}}$

${\textbf{Hence above graph is required graph }}$

Solve graphically : $y \leq 0$

Consider the line whose equation is

y=0y=0. This represents the X-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1 ,1)
When

y=1,y0y=1,y⪇0.

∴∴ (1,1) does not lie in the required region.
Therefore, the solution set is the X-axis and below the X-axis which is shaded in the graph.
1954182_1847820_ans_22aba83b49de4673bffc9fbdf818ea18.png

1954182_1847820_ans_22aba83b49de4673bffc9fbdf818ea18.png

Solve the graph the solution set on a number line:
$x-5 < -2; x\in N$

$x-5 < -2$

$x < 3$

$\therefore$ The required graph is shown in figure.

1778839_1839017_ans_7c191f7c344b4acd917a0bae12391a05.png

Draw graph of : $y \geq 0$

${\textbf{Step-1: Describe}}$

$y \geqslant 0.$

${\text{Consider the line whose equation is}}$

$y =0.$

${\text{To find the solution set, we have to check any point other than origin.}}$

${\text{Let us check the point (1, 1).}}$

${\text{Here}}$

$y = 1 \Rightarrow y \geqslant 0.$

$\therefore$

${\text{(1, 1) lies in the required region.}}$

${\textbf{Step-2: Draw a graph and show required part.}}$

${\text{Therefore, the solution set is the X-axis and the right side of the X-axis which is shaded in the graph.}}$

${\textbf{Hence above graph is required graph}}$

Solve graphically : $x \leq 0$

Consider the line whose equation is

x=0x=0.
This represents the Y-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1,1).
When

x=1,x0x=1,x⪇0

∴∴ (1, 1) does not lie in the required region.
Therefore, the solution set is the Y-axis and the left side of the Y-axis is shaded in the graph.
1954179_1847793_ans_8f8677afa1b344efbaa61ce9cb09fb18.png

1954179_1847793_ans_8f8677afa1b344efbaa61ce9cb09fb18.png

Solve: $3+x < 2, x\in N$

Given

$3+x < 2, x\in N$

$x < 2-(-3)$

$x < 2+3$

$x < 5$

$\therefore x=1,2,3,4$

$(\because x\in N)$

$\therefore$ Solution set

$=\left\{1,2,3,4\right\}$

Solve the inequation $8-2x > x-5; x\in N$

Given

$8-2x\ge x-5; x\in N$

$8+5\ge 2x+x$

$13 > 3x\Rightarrow 3x \le 13$

$x\le \dfrac{13}{3}=4\dfrac{1}{3}$

Solution set

$=\left\{1,2,3,4\right\}$

Solve the inequality $18-3(2x-5) > 12; x\in W$

Given

$18-3(2x-5)>12; x\in W$

$18-6x+15>12\\ 33-12 >6x$

$21 > 6x$

$6x < 21\Rightarrow x < \dfrac{21}{6} +\dfrac{7}{2}=3\dfrac{1}{2}$

But

$x\in W, x=0,1,2,3$

$\therefore$ Solution set

$=\left\{0,1,2,3\right\}$

Solve: $\dfrac{2x+1}{3}+15 < 17; x\in W$

Given

$\dfrac{2x+1}{3}+15\le 17; x\in W$

$\dfrac{2x+1}{3}\le 17-15=2$

$2x+1\le 6\Rightarrow 2x\le 5$

$x\le \dfrac{5}{2}=2\dfrac{1}{2}$

$\therefore x=0, 1, 2$

$\therefore$ Solution set is

$=\left\{0,1,2\right\}$

Solve graphically : $3x + 4 \leq 0$

Consider the line whose equation is

$3x + 4 = 0, i.e. x = - \dfrac{4}{3}.$
This represents a line parallel to Y-axis passing through the point

$(- \dfrac{4}{3},0)$ .
Draw the line

$x = - \dfrac{4}{3}.$
To the solution set, we have to check the position of the origin (0,0).
When

$x = 0, \;3x + 4 = 3 * 0 + 4 = 4 \nleqslant 0$

$\therefore$ the coordinates of the origin does not satisfy the given inequality.

$\therefore$ the solution set consists of the line

$x = - \dfrac{4}{2}$ the non-origin side of the line which is shaded in the graph.
1954190_1847930_ans_7bbc7c5b9d984494959f71810c26b46b.png

Solve graphically : $2x - 5y \geq 10$

Consider the line whose equation is

$2x - 5y = 10$ . To find the points of intersection of this line with the coordinate axes.
Put

$y = 0,$ we get

$x = 10, \;i.e. x = 5,$

$\therefore A = (5, 0)$ is apoint on the line.
Put

$x = 0,$ we get

$- 5y = 10, \;i.e. y = -2$

$\therefore B = (0, -2)$ is another point on the line.

Draw the line AB joining these points. This line divide the plane in two parts.
1. Origin side
2. Non-origin side
To find the solution set, we have to check the position of the origin (0, 0) with respect to the line.
when

$x = 0, y = 0, \;then \;2x - 5y = 0$ which is neither greater non equal to 10.

$\therefore 2x - 5y \ngeqslant 10$ in the case.
Hence (0, 0) will not lie in the required region,
Therefore, the given inequality is the non-origin side, which is shaded in the graph.
This is the solution set of

$2x - 5y \geq 10.$
1954198_1848061_ans_30f1b007897746449f93840f0831b589.png

Solve graphically : $x \geq 0$ and $y \geq 0$

Consider the lines whose equations are

x=0,y=0x=0,y=0. These represents the equations of Y-axis and X=axis respectively, Which divide the plane into four parts.
Since

x≥0,y≥0.x≥0,y≥0. the solution set is in the first quadrant which is shaded in the graph.
1954183_1847849_ans_9c3a9f468d5e4353a18d9c8093e721cd.png

Solve graphically : $2y - 5 \geq 0$

Consider the line whose equation is

$2y - 5 = 0, i.e. y = \dfrac{5}{2}$
This represents a line parallel to X-axis passing 5 through the point

$(0, \dfrac{5}{2})$
Draw the line

$y = \dfrac{5}{2}$ .
To find the solution set, we have to check the position of the origin (0, 0).
When

$y = 0, \;2y - 5 = 2x - 5 = -5 \ngeqslant 0$

$\therefore$ the coordinates of the origin does not satisfy the given inequality.

$\therefore$ the solution set consists of the line

$y = \dfrac{5}{2}$ and the non-origin side of the line which is shaded in the graph.
1954189_1847911_ans_72b1e4cb865b4185bdb337b943623663.png

Solve graphically : $x \leq 0$ and $y \leq 0$

Consider the lines whose equations are

x=0,y=0x=0,y=0. These represents the equations of Y-axis and X-axis respectively, which divide the plane into four parts.
Since

x≤0,y≤0,x≤0,y≤0, the solution set is in the third quadrant which is shaded in the graph.
1954185_1847882_ans_936bb07de00b4d6f9ebed248fa48a64b.png

Solve graphically : $x + 2y \leq 6$

Consider the line whose equation is

$x + 2y \leq 6$ . To find the points of intersection of this line with the coordinate axes.
Put

$y = 0,$ we get

$x = 6.$

$\therefore A = (6, 0)$ is a point on the line.
Put

$x = 0,$ we get

$2y = 6, \;i.e. y = 3$

$\therefore B = (0, 3)$ is another point on the line.

Draw the line AB joining these points. This line divide the line into two parts.
1. Origin side
2. Non-origin side
To find the solution set, we have to check the position of the origin (0, 0) with respect to the line.
When

$x = 0, y = 0, \;then \;x + 2y = 0$ which is less than 6.

$\therefore x + 2y \leq 6$ in this case.
Hence, origin lies in the required region. therefore, the given inequality is the origin side which is shaded in the graph.
This is the solution set of

$x + 2y \leq 6.$
1954197_1847999_ans_f3a79d90cbd743fb8946e343260844bf.png

Solve graphically : $2x - 3 \geq 0$

Consider the line whose equation is

$2x - 3 \geq 0, i.e. x = \dfrac{3}{2}$
This represents a line parallel to Y-axis passing 3 through the point

$(\dfrac{3}{2}, 0).$
Draw the line

$x = \dfrac{3}{2}$ .
To find the solution set, we have to check the position of the origin (0, 0).
When

$x = 0, 2x - 3 = 2 * 0 - 3 = -3 \ngeqslant 0$

$\therefore$ coordinates of the origin does not satisfy the given inequality.

$\therefore$ the solution set consists of the line

$x = \dfrac{3}{2}$ and the 2 non - origin side of the line which is shaded in the graph
1954188_1847902_ans_ab5ea82709684961b67206dc9ce684a5.png

Solve graphically : $5y + 3 \leq 0$

Consider the line whose equation is

$5y + 3 \leq 0, i.e. y = \dfrac{-3}{5}$
This represents a line parallel to X-axis passing through the point

$(0, \dfrac{-3}{5})$
Draw the line

$y = \dfrac{-3}{5}.$
To find the solution set, we have to check the position of the origin (0, 0).
When

$y = 0, 5y + 3 = 5 * 0 + 3 = 3 \nleqslant 0$

$\therefore$ the coordinates of the origin does not satisfy the given inequality.

$\therefore$ the solution set consists of the line

$y = \dfrac{-3}{5}$ and the non-origin side of the line which is shaded in the graph.
1954195_1847954_ans_831b424bc00342f7b77624e86a02ae2e.png

Solve graphically : $5x - 3y \leq 0$

Consider the line whose equation is

$5x - 3y = 0.$ The constant term is zero, therefore this line is passing through the origin.

$\therefore$ one point on the line is the origin O = (0, 0).
To find the other point, we can give any value of x and get the corresponding value of y.
Put

$x = 3,$ we get

$15 - 3y = 0, \;i.e. y = 5$

$\therefore A = (3, 5)$ is another point on the line . Draw the line OA.
To find the solution set, we cannot check O(0, 0), as it is already on the line. We can check any point which is not on the line.
Let us check the point (1, 1).
When

$x = 1, y = -1$ then

$5x - 3y = 5 + 3 = 8$ which is neither less now equal to zero.

$\therefore 5x - 3y \nleqslant 0$ in this case.
Hence (1, -1) will not lie in the required region.
Therefore the required region is the upper side which is shaded in the graph.

This is the solution set of

$5x - 3y \leq 0.$

1954200_1848425_ans_38ae84f5171d45dc848d1e4ed867e545.png

Solve graphically : $3x + 2y \geq 0$

Consider the line whose equation is

$3x + 2y = 0$ . The constant term is zero, therefore this line is passing through the origin.

$\therefore$ one point on the line is O = (0, 0).
To find the another point, we can give any value of x and get the corresponding value of y.
Put

$x = 2,$ we get

$6 + 2y = 0 \;i.e. y = -3$

$\therefore A = (2, -3),$ is another point on the line. Draw theline OA.
To find the solution set, we cannot check (0, 0) as it is already on the line.
We can check any other point which is not on the line.
Let us check the point (1, 1).

When

$x = 1, y = 1, \;then \;3x + 2y = 3 + 2 = 5$ which is greater than zero.

$\therefore 3x + 2y > 0$ in this case.
Hence (1, 1) lies in the required region.
Therefore, the required region is the upper side which is shaded in the graph.
This is the solution set of

$x + 2y > 0.$
1954199_1848405_ans_06547cff4b9043238a42ecd4fb3cbe8b.png

Solve graphically : $x - y \leq 2$ and $x + 2y \leq 8$

Line	Equation	Points on the X- axis	Points on the Y- axis	Sign	Region
AB	$x-y \leq 2$	A(2, 0)	B(0, -2)	$\leq$	origin side of line AB
CD	$x+2y \leq 8$	C(8, 0)	D(0, 4)	$\leq$	origin side of line CD

The solution set of the given system of inequalities is shaded in the graph.

1954203_1848475_ans_f300b3108739429e9b6c298271cf3ffa.png

Find the feasible solution for the following system of linear inequations:
$0 \leq * \leq 3, 0 \leq y \leq 3, x + y \leq 5, 2x + y \geq 4$

First we draw the lines AB, CD, EF and GH whose equations are x + y = 5, 2x + y = 4, x =3 and y = 3 respectively.

Line	Equation	Points on the X- axis	Points on the Y- axis	Sign	Region
AB	x + y = 5	A(5,0)	B(0,5)	$\leq$	origin side of line AB
CD	2x + y = 4	C(2,0)	D(0,4)	$\geq$	non-origin side of line CD
EF	x = 3	E(3,0)		$\leq$	origin side of line EF
GH	y = 3		H(0,3)	$\leq$	origin side of line GH

The feasible solution is CEPQRC which is shaded in the graph.
1954211_1848570_ans_e697b90fbe824283bc7501eb6a7253fd.png

Solve graphically : $2x + y \geq 2$ and $x - y \leq 1$

First we draw the line AB and AC whose equations are

$2x + y = 2$ and

$x - y = 1$ respectively.

Line	Equation	Points on the X- axis	Points on the X- axis	Sign	Region
AB	$2x-y=2$	A(1, 0)	B(0, 2)	$\geq$	non-origin side of line AB
AC	$x-y=1$	A(1, 0)	C)0, -1)	$\leq$	origin side of the line AC

The solution set of the given system of inequalities is shaded in the graph.
1954201_1848454_ans_7e1f6c47e3144125be630fb05d1a34bb.png

Find the feasible solution of the following inequation:
$2x + 3y \leq 6, x + y \geq 2, x \geq 0, y \geq 0$

First we draw the lines AB and CA whose equations are 2x + 3y = 6 and x + y = 2 respectively.

Line	Equation	Points on the X- axis	Points on the Y- axis	Sign	Region
AB	2x + 3y = 6	A(3, 0)	B(0, 2)	$\leq$	origin side of line AB
CD	x + y = 2	C(2, 0)	D(0, 2)	$\geq$	non-origin side of line CB

The feasible solution is

$\Delta ABC$ which is shaded in the graph.
1954209_1848561_ans_15947a8ef3bd42ca805930b2a39ef778.png

Solve graphically : $2x + y \geq 5$ and $x - y \leq 1$

First we draw the lines AB and CD Whose equations are 2x + y = 5 and x - y = 1 respectively.

Line	Equation	Points on the X- axis	Points on the X- axis	Sign	Region
AB	2x + y = 5	A(2.5, 0)	B(0, 5)	$\geq$	non-origin side of line AB
CD	x - y = 1	C(1, 0)	D(0, -1)	$\leq$	origin side of line CD

The solution set of the given system of inequalities is shaded in the graph.
1954206_1848544_ans_0365f879899a4ecdbaa8af8e6a5151c2.png

Find the feasible solution of the following inequations:
$x - 2y \leq 2, x + y \geq 3, -2x + y \leq 4, x \geq 0, y \geq 0$

First we draw the lines AB, CD and EF whose equations are x - 2y = 2, x + y = 3 and -2x + y = 4 respectively.

Line	Equation	Points on the X- axis	Points on the Y- axis	Sign	Region
AB	x - 2y = 2	A(2,0)	B(0,-1)	$\leq$	origin side of line AB
CD	x + y = 3	C(3,0)	D(0,3)	$\geq$	non-origin side of line AB
EF	-2x + y = 4	E(-2,0)	F(0,4)	$\leq$	origin side of line EF

The feasible solution is shaded in the graph.
1954212_1848572_ans_86634b2f25e1408fb1e059ee19902b51.png

Find the feasible solution of the following inequation:
$3x + 4y \geq 12, 4x + 7y \leq 28, y \leq 1, x \leq 0$

First we draw the lines AB, CD and EF whose equations are 3x + 4y = 12 and 4x + 7y = 28 and y = 1 respectively.

Line	Equation	Points on the X- Axis	Points on the Y- Axis	Sign	Region
AB $3x+4y=12$		A(4,0)	B(0,3)	$\geq$	non-origin side of lineAB
CD	$4x+ 7y = 28$	C(7, 0)	D(0, 4)	$\leq$	origin side of line CD
EF	$y = 1$		F(0, 1)	$\geq$	non-origin side of line EF

The feasible solution is PQDB which is shaded in the graph.

1993106_1848563_ans_0024439bb2164fe98fb5d58d884892f9.png

Solve graphically : $2x + 3y \leq 6$ and $x + 4y \geq 4$

First we draw the lines AB and CD whose equations are 2x + 3y = 6 and x + 4y = 4 respectively.

Line	Equation	Points on the X- axis	Points on the Y- axis	Sign	Region
AB	2x + 3y = 6	A(3, 0)	B(0, 2)	$\leq$	origin side of line AB
CD	x + 4y = 4	C(4, 0)	D(0, 1)	$\geq$	non-origin side of line CD

The Solution set of the given system of inequalities shaded in the graph.
1993108_1848518_ans_ec0c1ace22b84fed806456000ac07eca.png

Find the feasible solution of the following inequation:
$x + 4y \leq 24, 3x + y \leq 21, x + y \leq 9, x \geq 0, y \geq 0$

First we draw the lines AB,CD and EF whose equations are x + 4y = 42, 3x + y = 21 and x + y = 9 respectively.

Line	Equation	Points on the X- axis	Points on the y- axis	Sign	Regio
AB	x + 4y =24	A(24, 0)	B(0, 6)	$\leq$	origin side of line AB
CD	3x + y = 21	C(7, 0)	D(0, 21)	$\leq$	origin side of line CD
EF	x + Y = 9	E(9,0)	F90,9)	$\leq$	origin side of line EF

The feasible solution is OCPQBO which is shaded in the graph.
1954210_1848565_ans_c2b31922bdbf47449053dcd46d9f9097.png

Solve graphically : $x + y \geq 6$ and $x + 2y \leq 10$

First we draw the lines AB and CD whose equations are

$x + 2y = 10$ respectively.

Line	Equation	Points on the X- axis	Points on the Y- axis	Sign	Region
AB	$x+y=6$	A(6, 0)	B(0, 6)	$\geq$	non-origin side of line AB
CD	$x+2y=1$	C(0, 5)	D(0, 5)	$\leq$	origin side of the line CD

The solution set of the given system of inequalities is shaded in the graph.
1954205_1848501_ans_65f7acc814364bec9144aa2dc2b59250.png

$5y -12 \geq 0$

Consider the line
whose equation is $5y -12 \geq 0$ i.e. $y = \dfrac{12}{5}$ $\dfrac{12}{5})$

Draw the line $y = \dfrac{12}{5}$

To find the solution set, we have to check the position of
the origin (0, 0)

when y = 0, 5y -12 = 5(0) - 12 = - 12 > 0

$\therefore$ the coordinates of the origin does not satisfy
the given inequality.

$\therefore$ the solution set consists of the line $y = \dfrac{12}{5}$ and the non-origin side of the line which is shaded in the graph.

1954231_1850171_ans_1bdc8a2a48564a3099c4f3cdcf508c53.png

x is a natural number.

Given: 30x < 200

$x<\cfrac{200}{30} = 6.6$
x is a natural number:
Solution set = {1, 2, 3, 4, 5, 6}

Find graphical solution for the following system of linear in equation:

$3x + 4y \leq 12, x - 2y \geq 2, y \geq -1$

Ref. image 1

First we draw the lines AB,CD and ED whose equations are 3x + 4y = 12, x - 2y = 2 and y = -1 respectively.
Ref. image 2

The solution set of given system of inequation is shaded in the graph.

1954234_1850224_ans_a5e74155e79044348803f110a59bd055.png

x is an integer.

Given: 30x < 200

$x<\cfrac{200}{30} = 6.6$
x is a real number:

$\therefore$ Solution set = {... -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}

$4x - 18 \geq 0$

Consider thrline whose equation is

$4x - 18 \geq 0$ i.e.

$x = \dfrac{18}{4} = \dfrac{9}{2} = 4.5$
This represents a line parallel to Y-axis passing 3 through the point (4.5, 0)
Draw the line x = 4.5
To find the solution set, we have to check the position of the origin (0, 0).
When x 0, 4x - 18 = 4 * 0 -18 = -18>0

$\therefore$ the coordinates of the origin does not satisfy the given inequality.

$\therefore$ the solution set consists of the line x = 4.5 and the non-origin side of the line which is shaded in the graph.
1954224_1850126_ans_e1abcf33ccc7415d9f8da1ce01fd1de4.png

1954224_1850126_ans_e1abcf33ccc7415d9f8da1ce01fd1de4.png

$y \leq -3.5$

Consider the line
whose equation is

$y \leq -3.5$ i.e. y = - 3.5

This represents a line parallel to X-axis passing 3 through the
point (0, -3.5)

Draw the line y = - 3.5

To find the solution set, we have to check the position of
the origin (0, 0)

$\therefore$ the coordinates of the origin does not satisfy
the given inequality.

$\therefore$ the solution set consists of the line y = - 3.5
and the non-origin side of the line which is shaded in the graph.
1954232_1850185_ans_bd7465bd13bf4be49bfeab6239733fd1.png

1954232_1850185_ans_bd7465bd13bf4be49bfeab6239733fd1.png

A company produces two types of articles A and B which require silver and gold. Each unit of A requires 3 gm of silver and 1 gm of gold, while each unit of B requires 2 gm of silver and 2 gm of gold. The company has 6 gm of silver and 4 gm of gold. Construct the inequations and find feasible solution graphically.

Let the company produces x units of articles A and y units of article B. The given data can be tabulated as:

	Article A (x)	Article B (y)	Availability
Gold	1	2	4
Silver	3	2	6

Inequations are:

$x + 2y \leq 4$ and

$3x + 2y \leq 6$

$x$ and

$y$ are number of items,

$x \geq 0, y \geq 0$
First we draw the lines AB and CD whose equations are

$x + 2y = 4$ and

$3x + 2y = 6$ respectively.

Line	Equation	Point on the X- axis	Point on the Y- axis	Sign	Region
AB	x + 2y + 4	A(4,0)	B(0,2)	$\leq$	origin side of line AB
CD	3x + 2y = 6	C(2,0)	D(0,3)	$\leq$	origin side of line CD

The feasible solution id OCPBO which is shaded in the graph.
1954213_1848586_ans_cf01a357546d4826a0c39fea8216f957.png

$- 11x - 55 \leq 0$

Consider the line whose equation is

$- 11x - 55 \leq 0$ i.e. x = -5
This represents a line parallel to Y-axis passing through the point (-5,0)
Draw the line x = - 5
To find the solution set, we have to check the position of the origin (0, 0)
when x = 0, -11x -55 = -11(0) - 55 = -55> 0

$\therefore$ the coordinates of the origin does not satisfy the given inequality.

$\therefore$ the solution set consists of the line x = -5 and the non-origin side of the line which is shaded in the graph.

1954230_1850137_ans_a0fe159434b54ca4a6609c1d05aa7d20.png

1954230_1850137_ans_a0fe159434b54ca4a6609c1d05aa7d20.png

$x$ is a real number

Given:

$3x + 8 > 2$

$\Rightarrow 3x > 2 - 8 = -6$

$\therefore x>-\cfrac{6}{3} = -2$
When

$x$

$\epsilon$

$M$ , solution set = (

$-2, \infty$ )

$x$ $\epsilon$ $N$

Given: -12x > 30

$\Rightarrow x<\cfrac{30}{-12} = -2.5$
When

$x$

$\epsilon$

$N$ , solution set

$= \varphi$

$x$ is an integer

Given:

$5x - 3 < 3x + 1$

$\Rightarrow 5x - 3x < 1 + 3$

$\Rightarrow 2x < 4$

$\therefore x>-\cfrac{4}{2} = 2$
When

$x$

$\epsilon$

$Z$ , solution set = {...., -3, -2, -1, 0, 1}

x is an integer

Given: 5x - 3 < 7

$\Rightarrow 5x < 7 + 3 = 10$

$\therefore x<\cfrac{10}{5} = 2$
If x is an integer,
Solution set = {..., -3, -2, -1, 0, 1}

$x$ is a real number.

Given:

$5x - 3 < 3x + 1$

$\Rightarrow 5x - 3x < 1 + 3$

$\Rightarrow 2x < 4$

$\therefore x>-\cfrac{4}{2} = 2$
When

$x$

$\epsilon$

$M$ , solution set = (-

$\infty$ , 2).

Solve: $4x + 3 < 6x + 7$

$x$ $\epsilon$ $Z$

Given: -12x > 30

$\Rightarrow x<\cfrac{30}{-12} = -2.5$
When

$x$

$\epsilon$

$Z$ , solution set

$= \{..., -5, -4, -3 \}$

Solve: $3x - 7 > 5x - 1$

$Given:$

$3x - 7 > 5x - 1$

$\Rightarrow 1 - 7 > 5x - 3x$

$\Rightarrow -6 > 2x$

$\Rightarrow \therefore x > \frac{-6}{2} = -3$

$\therefore$

$Solution\space set\space$ = (

$-3, \infty$ )

x is a real number.

Given: 5x - 3 < 7

$\Rightarrow 5x < 7 + 3 = 10$

$\therefore x<\cfrac{10}{5} = 2$
If x is a real number,
Solution set = (

$- \infty, 2$ )

Solve: $37 - (3x + 5) \geq 9x - 8(x - 3)$

Given:

$37 - (3x + 5) \geq 9x - 8(x - 3)$

$\Rightarrow 32 - 24 \geq x + 3x$

$\Rightarrow 8 > 4x$

$\Rightarrow x < 2$

$\therefore$

$Solution$

$set$

$= (- \infty, 2)$

Solve: $3(2 - x) > 2(1 - x)$

$Given:$

$6 - 3x > - 2 + 3x$

$\Rightarrow 6 -3x > 2-2x$

$\Rightarrow4 > x$

$\therefore$

$Solution\space set\space = (-\infty,4)$

Solve : $x+\cfrac{x}{2}+ \cfrac{x}{3}<11$

Given:

$\frac{6x + 3x + 2x}{6} < 11$

$11x < 11 \times 6 = 66$

$\therefore x < \frac{66}{11} = 6$

Solve:
$\cfrac{2 x-1}{3} \geq \cfrac{3 x-2}{4}- \cfrac{2-x}{5}$

Given:

$\dfrac{2x - 1}{3} \geq \dfrac{15x - 10 - 8 + 4x}{20}$

$\Rightarrow 20 (2x - 1) \geq 3(19x - 18)$

$\Rightarrow 40x - 20 \geq 57x - 54$

$\Rightarrow 54 - 20 \geq 57x - 40x$

$\Rightarrow 34 \geq 17x$

$\Rightarrow x \leq \frac{34}{17} = 2$

$\therefore$

$Solution$

$set$ = (

$- \infty , 2$ ]

Solve:
$\cfrac{x}{3}>\cfrac{x}{2} + 1$

$Given:$

$\frac{x}{3} > \frac{x}{2} + 1$

$\Rightarrow -1 > \frac{x}{2} - \frac{x}{3}$

$\Rightarrow -1 > \frac{3x - 2x}{6}$

$\Rightarrow -6 > x$

$\therefore$

$Solution\space set$

$= (- \infty, - 6)$

$3x - 2 < 2x + 1$

$\textbf{Step 1: Apply relevant property of inequality and simplify.}$

$\text{We have, 3x - 2 < 2x + 1}$

$\Rightarrow$

$\text{3x - 2x < 2 + 1}$

$\Rightarrow$

$\text{x < 3}$

$\therefore$

$\text{x }$

$\boldsymbol{\in} (-\infty, 3)$

$\textbf{Hence, solution set is}$

$\boldsymbol{(-\infty, 3).}$

Solve: $\frac{3(x - 2)}{5} \leq \frac{5(2 - x)}{3}$

$Given$ :

$\frac{3(x - 2)}{5} \leq \frac{5(2 - x)}{3}$

$\Rightarrow 9(x - 2) \leq 25(2 - x)$

$\Rightarrow 9x - 18 \leq 50 - 25x$

$\Rightarrow 25x + 9x \leq 50 + 18$

$\Rightarrow 34x \leq 68$

$\Rightarrow x \leq \frac{68}{34} = 2$

$\therefore$ Solution set = (-

$\infty, 2$ ]

Solve: $3(x - 1) < 2 (x - 3)$

$Given:$

$3(x - 1) < 2(x - 3)$

$\Rightarrow 3x - 3 < 2x - 6$

$\Rightarrow 3x - 2x < -6 + 3$

$\Rightarrow x < -3$

$\therefore$

$Solution \space set\space=(-\infty,3)$

Solve: $\cfrac{5 - 2 x}{3} \leq \cfrac{x}{6} - 5$

$Given:$

$\frac{5 - 2x}{3} \leq \frac{x}{6} - 5$

$\Rightarrow 2(5 - 2x) \leq x - 30$

$\Rightarrow 10 - 4x \leq x - 30$

$\Rightarrow 10 + 30 \leq x + 4x$

$\Rightarrow \therefore 40 \leq 5x$

$\Rightarrow x \geq \frac{40}{5} = 8$

$\therefore$

$Solution\space set$ = [

$8, \infty$ )

Solve:
$\cfrac{x}{4} <\cfrac{5 x - 2}{3} - \cfrac{7 x-3}{5}$

$\cfrac{x}{4}<\cfrac{5(5 x-2) - 3(7 x-3)}{15}$

$\Rightarrow 15x < 16x - 4$

$\Rightarrow 4 < 16x - 15x$

$\Rightarrow x > 4$

$\therefore$

$Solution$

$set$

$(4, \infty)$

x > -3

Consider the equation x = -3
Draw a graph of x = -3 by dotted line, which is parallel to y-axis, to the left of y-axis at distance
Put x = 0 in the given inequation x > -3, we get 0 > -3 which is true.

$\therefore$ Solution region consists of (0,0).

$\therefore$ The shaded region is the solution region.
1955271_1864032_ans_9b427bb505444a958a1289bcb1746ecc.png

Solve the following system of inequalities graphically.
$x \geq 3, y \geq 2$

$x \geq 3$ .................. (1)
First draw the line x = 3 by thick line.
Put x = 0 in (1), we get 0 > 3 is not true.

$\therefore$ Solution set of (1) is not containing the origin.

$y \geq 2$ .................. (2)
First draw the line y = 2 by thick line.
Put y = 0 in (2), we get 0 > 2 is not true.

$\therefore$ Solution set of (2) is not containing the origin.

$\therefore$ Shaded region is the required solution origin.
1955274_1864045_ans_646e7afb23534ced8eb71e3e67f35297.png

$5x - 3 \geq 3x - 5$

$\Rightarrow 2x \geq - 2$

$\Rightarrow x \geq -1$

$\therefore$ Solution set [-1,

$\infty$ )
1955264_1864006_ans_9310bdc42d0e45e6b882dcbfacb945fc.png

$7x + 3 < 5x + 9$

$\Rightarrow 2x < 6$

$\Rightarrow x < 3$

$\therefore$ Solution set

$= (- \infty, 3)$
1955265_1864007_ans_643e74f0eb6447439c5094a63c196876.png

y < -2

y < -2 ............(1)

Consider the equation y = -2

Draw a graph of y = -2 by dotted line, which is parallel to x-axis, below x-axis and at distance 2.

Put y = 0 in (1), 0 > -2, which is not true.

$\therefore$ Solution region is not containing (0,0).

$\therefore$ The shaded region is the solution region.

1938645_1864033_ans_db0b22b89fcd4b52a820e3edbd63b666.png

y > 2

y > 2 ..............(1)
Draw the graph of y = 2 by dotted line. Put y = 0 in (1) we get 0 > 2 which is false.

$\therefore$ Solution region doesn't contain the origin.

$\therefore$ The shaded region is the solution region.
1955273_1864044_ans_4d2e9ad1d4e64c0c9f9adf2b41f6969b.png

3x + 2y > 6

3x + 2y > 6 .........(1)
Draw the graph of 3x + 2y = 6 by dotted line.
It passes through (2, 0) and (0, 3). Join these points.
Put x = 0 and y = 0 in (1), we get 0 + 0 > 6 which is not true.

$\therefore$ The solution region does not contain the origin.

$\therefore$ The shaded region is the solution region.
1955272_1864039_ans_6f2a8dcbdfd54829b1952b1f79f03dd0.png

$\cfrac{3 x-4}{2} \geq \cfrac{x+1}{4} - 1$

$\Rightarrow 2(3x - 4) \geq x + 1 - 4$

$\Rightarrow 6x - x \geq -3 + 8$

$\Rightarrow x \geq 1$

$\therefore$ Solution set = [1,

$\infty$ )
1955268_1864009_ans_156ad97bc83a4f38a1e5152bb21cd396.png

$3(1 - x) < 2(x + 4)$

$3 - 3x < 2x + 8$

$\Rightarrow -5 < 5x$

$\Rightarrow x > - \frac{5}{8} = -1$

$\therefore$ Solution set

$= (-1, \infty)$
1955267_1864008_ans_3c780b28605e4698a53cae23966007b8.png

$\cfrac{x}{2} \geq \cfrac{5 x-2}{3} - \cfrac{7 x-3}{5}$

$\Rightarrow \frac{25x - 10 - 21x + 9}{15}$

$\Rightarrow 15x \geq 2(4x - 1) = 8x - 2$

$\Rightarrow 7x \geq - 2$

$\Rightarrow x \geq - \frac{2}{7}$

$\therefore$ Solution set = [ -

$\frac{2}{7}, \infty$ ).

1955270_1864010_ans_41310da9ba3344298084cb54fcf26faa.png

Solve the following system of inequalities graphically.
$2x + y \geq 8, x + 2y \geq 10$

Consider

$2x + y \geq 8$ ................(1)
Draw the graph of

$2x + y = 8$ by thick line.
It passes through (4, 0) and (0, 8).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 \geq 8$ which is false.

$\therefore$ Solution set of (1) is not containing the origin.
Consider

$x + 2y > 10$ ................(2)
Draw the graph of

$x + 2y = 10$ by thick line.
It passes through (10, 0) and (0, 5).
Join these points, put x = 0 and y = 0 in (2), we get

$0 + 0 > 10$ which is false.

$\therefore$ Solution set of (2) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of inequations.

2x + y > 6

2x + y > 6...........(1)
Draw the graph of 2x + y = 6 by thick line.
It passes through (3,0) and (0,6).
Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 > 6 which is false.

$\therefore$ Solution region does not contain the origin.

3x + 4y < 12

3x + 4y < 12 ...........(1)
Draw the graph of 3x + 4y = 12 by thick line.
It passes through (4, 0) and (0, 3).
Join these points, put x = 0 and y = 0 in (1), we get

$0 + 0 \leq 12$ which is true.

$\therefore$ The solution region contains the origin.

$\therefore$ The shaded region is the solution region

Solve the following system of inequalities graphically.
$x + y \geq 4, 2x - y > 0$

Consider

$x + y \geq 4$ ................(1)
Draw the graph of

$x + y = 4$ by thick line.
It passes through (4, 0) and (0, 4).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 > 4$ which is false.

$\therefore$ Solution set of (1) is not containing the origin.
Consider

$2x - y > 0$ ................(2)
Draw the graph of

$2x - y > 0$ by dotted line.
It passes through (0, 0) and (1, 2).
Join these points put x = 2 and y = 1 in (2), we get

$2(2) - 1 > 0$ which is true.

$\therefore$ Solution set of (2) is containing the point (2, 1)
Hence, the shaded region represents the solution set of the given system of inequations.

Solve the following system of inequalities graphically.
$2x + y \geq 6, 3x + 4y\leq 12$

Consider

$2x + y \geq 6$ ................(1)
Draw the graph of

$2x + y = 6$ i.e.

$\frac{x} {3} + \frac {y} {6} = 1$ by thick line.
It passes through (3, 0) and (0, 6).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 \geq 6$ which is false.

$\therefore$ Solution set of (1) is not containing the origin.
Consider

$3x + 4y\leq 12$ ................(2)
Draw the graph of

$3x + 4y\leq 12$ i.e.

$\frac{x} {4} + \frac {y} {3} = 1$ by thick line.
It passes through (4, 0) and (0, 3).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 < 6$ which is true.

$\therefore$ Solution set of (2) is containing the origin.
Hence, the shaded region represents the solution set of the given system of inequalities.
1955278_1864048_ans_aa66fc2ee38841feb147eead45e6eff7.png

1955278_1864048_ans_aa66fc2ee38841feb147eead45e6eff7.png

Solve the following system of inequalities graphically.
$x + y \leq 6, x + y \geq 4$

Consider

$x + y \leq 6$ ................(1)
Draw the graph of

$x + y = 6$ by thick line.
It passes through (6, 0) and (0, 6).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 < 6$ which is true.

$\therefore$ Solution set of (1) is containing the origin.
Consider

$x + y \geq 4$ ................(2)
Draw the graph of

$x + y = 4$ by thick line.
It passes through (4, 0) and (0, 4).
Join these points, put x = 0 and y = 0 in (2), we get

$0 + 0 \geq 4$ which is false.

$\therefore$ Solution set of (2) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of inequations.

Solve the following system of inequalities graphically.
$2x - y > 1, x - 2y < -1$

Consider

$2x - y > 1$ ................(1)
Draw the graph of

$2x - y > 1$ by dotted line.
It passes through (

$\frac {1} {2}$ , 0) [ and (0, -1).
Join these points put x = 0 and y = 0 in (1), we get

$0 - 0 > 1$ which is false.

$\therefore$ Solution set of (1) is not containing the origin.
Consider

$x - 2y < -1$ ................(2)
Draw the graph of

$x - 2y < -1$ by dotted line.
It passes through (-1, 0) and (-1, 0) and (0,

$\frac {1} {2}$ )
Join these points, put x = 0 and y = 0 in (2), we get

$0 - 0 < -1$ which is false.

$\therefore$ Solution set of (2) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of inequations.

3x - 6 > 0

$3x - 6 \geq 0$ .......(1)
Draw the graph of 3x - 6 = 0 i.e., x = 2 by thick line
Put x = 0 in (1), we get -6 > 0 which is false.

$\therefore$ Solution region does not contains the origin.
The shaded region is the solution region.
1955275_1864046_ans_39e727552dab4facab1398b47ba2f606.png

1955275_1864046_ans_39e727552dab4facab1398b47ba2f606.png

Solve the following system of inequalities graphically.
$8x + 3y \leq 100, x \geq 0, y \geq 0$

Consider

$8x + 3y \leq 100$ ................(1)
Draw the graph of

$8x + 3y = 100$ by thick line.
It passes through (

$\frac{25} {2}$ , 0) and (0,

$\frac{100} {3}$ ).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 \leq 100$ which is true.

$\therefore$ Solution set of (1) is containing the origin.
Since

$x \geq 0, y \geq 0$ , every point in the shaded region in the first quadrant, including the points on the line and the axes, represents the solution of the given system of inequations.

Solve the following system of inequalities graphically.
$x+y \geq 3, x-y \geq 2$

$x + 3 \geq 5$ .................. (1)
Draw thegraph of x+y = 5 by thick line.
It passes through (5, 0) and (0,5)
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 \geq 5$ which is false

$\therefore$ Solution set of (1) is not containing the origin.
Consider

$x - y \leq 3$ .................. (2)
Draw the graph of x-y = 3 by thick line.
It passes through (3, 0) and (0,-3)
Join these points put x = 0 and y = 0 in (2), we get

$0 - 0 < 3$ which is true.

$\therefore$ Solution set of (2) is containing the origin.
Hence, the shaded region represents the solution set of the given system of inequalities.
1955277_1864047_ans_0a873458b8c34988ab5b7e2ae7508e01.png

1955277_1864047_ans_0a873458b8c34988ab5b7e2ae7508e01.png

Solve the following system of inequalities graphically.
$5x + 4y \leq 20, x \geq 1, y \geq 2$

Consider

$5x + 4y \leq 20$ ................(1)
Draw the graph of

$5x + 4y = 20$ by thick line.
It passes through (4, 0) and (0, 5).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 < 20$ which is true.

$\therefore$ Solution set of (1) is containing the origin.
Consider

$x > 1$ ......... (2)
Draw the graph of x = 1
Clearly solution set of (2) is not containing the origin.
Consider

$y > 2$ ......... (3)
Draw the graph of y = 2
Clearly solution set of (3) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of inequations.

Solve the following system of inequalities graphically.
$x + 2y \leq 8, 2x + y \leq 8, x \geq 0, y \geq 0$

Consider

$x + 2y \leq 8$ ................(1)
Draw the graph of

$x + 2y = 8$ by thick line.
It passes through (8, 0) and (0, 4).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 < 8$ which is true.

$\therefore$ Solution set of (1) is containing the origin.
Consider

$2x + y \leq 8$ ......... (2)
Draw the graph of

$2x + y = 8$ by thick line.
It passes through (4, 0) and (0, 8).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 < 8$ which is true.

$\therefore$ Solution set of (2) is containing the origin.
Since

$x > 0, y > 0$ , every point in the shaded region in the first quadrant, including the points on the lines, represents the solution of the given system of inequations.

Solve the following system of inequalities graphically.
$3x + 4y \leq 60, x + 3y \leq 30, x \geq 0, y \geq 0$

Consider

$3x + 4y \leq 60$ ................(1)
Draw the graph of

$3x + 4y = 60$ by thick line.
It passes through (20, 0) and (0, 15).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 < 60$ which is true.

$\therefore$ Solution set of (1) is containing (0, 0).
Consider

$x + 3y < 30$ ......... (2)
Draw the graph of

$x + 3y = 30$ .
It passes through (30, 0) and (0, 10).
Join these points put x = 0 and y = 0 in (2), we get

$0 + 0 \leq 30$ is true.

$\therefore$ Solution set of (2) is containing (0, 0)
Since

$x \geq 0, y \geq 0$ , every point in the shaded region in the quadrant, including the points on the lines, represents the solution of inequations.

Solve the following system of inequalities graphically.
$3x + 2y \leq 12, x \geq 1, y \geq 2$

Consider

$3x + 2y \leq 12$ ................(1)
Draw the graph of

$3x + 2y = 12$ by thick line.
It passes through (4, 0) and (0, 6).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 \leq 12$ which is true.

$\therefore$ Solution set of (1) is containing the origin.
Consider

$x \geq 1$ ......... (2)
Draw the graph of x = 1
Clearly solution set of (2) is not containing the origin.
Consider

$y \geq 2$ ......... (3)
Clearly solution set of (3) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of inequations.

Solve the following system of inequalities graphically.
$5x + 4y \leq 40, x \geq 0, y \geq 0$

Consider

$5x + 4y \leq 40$ ................(1)
Draw the graph of

$5x + 4y = 40$ by thick line.
It passes through (8, 0) and (0, 10).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 \leq 40$ which is true.

$\therefore$ Solution set of (1) is containing the origin.
Consider

$x \geq 2$ ......... (2)
Draw the graph of x = 2.
Clearly solution set of (2) is not containing the origin.
Consider

$y \geq 3$ ......... (3)
Draw the graph of y = 3.
Clearly solution set of (3) is not containing the origin.
Hence, the shaded region represents the solution set of the given system of inequations.

x - y < 2

$x - y \leq 2$ ......(1)
Draw the graph of

$x - y \leq 2$ by thick line
It passes through (2, 0) and (0, -2)
Join these points, put x = 0 and y = 0 in (1), we get 0 - 0 < 2 which is true.

$\therefore$ The solution region contains the origin.

$\therefore$ The shaded region is the solution region.
1955299_1864075_ans_2378792294964efdabc66b6d0f58874d.png

Solve the following system of inequalities graphically.
$2x + y \geq 4, x + y \leq 3, 2x - 3y \leq 6$

Consider

$2x + y > 4$ ................(1)
Draw the graph of

$2x + y = 4$ by thick line.
It passes through (2, 0) and (0, 4).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 \leq 4$ is not true.

$\therefore$ Solution set of (1) does not contain (0, 0).
Consider

$x + y \leq 3$ ................(2)
Draw the graph of

$x + y = 3$ , by thick line.
It passes through (3, 0) and (0, 3).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 < 4$ which is true.

$\therefore$ Solution set of (2) contains (0, 0).
Consider

$2x - 3y \leq 6$ ................(3)
Draw the graph of

$2x - 3y= 6$ , by thick line.
It passes through (3, 0) and (0, -2).
Join these points put x = 0 and y = 0 in (1), we get

$0 - 0 < 6$ which is true.

$\therefore$ Solution set of (3) contains (0, 0).
Hence, the shaded region represents the solution set of the given system of inequations.
1955297_1864070_ans_1dbe1f661c9342c8bf8b2d396e6389eb.png

1955297_1864070_ans_1dbe1f661c9342c8bf8b2d396e6389eb.png

Solve the following system of inequalities graphically.
$x + y \leq 9, y > x, x \geq 0$

Consider

$x + y \leq 9$ ................(1)
Draw the graph of

$5x + 4y = 20$ by thick line.
It passes through (9, 0) and (0, 9).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 < 9$ which is true.

$\therefore$ Solution set of (1) is containing the origin.
Consider

$y > x$ ......... (2)
Draw the graph of y = x by dotted line.
Clearly solution set of (2) does not contain (1, 0).
Since

$x \geq 0$ , every point in the shaded region in the first quadrant, including the points on the line

$x + y = 9$ , excluding the points on the line

$y \geq x$ and y-axis, represents the solution of the given system of inequations.

y + 8 > 2x

y + 8 > 2x..............(1)
Draw the graph of y + 8 = 2 by thick line
It passes through (4, 0) and (0, -8)
Join these points, put x = 0 and y = 0 in (1), we get

$0 + 8 \leq 0$ which is true.

$\therefore$ The solution region contains the origin

$\therefore$ The shaded region is the solution region.

Solve the following system of inequalities graphically.
$x - 2y \leq 3, 3x + 4y \geq 12, x \geq 0, y \geq 0$

Consider

$x - 2y \geq 3$ ................(1)
Draw the graph of

$x - 2y = 3$ by thick line.
It passes through (3, 0) and (0,

$-\frac{3} {2}$ ).
Join these points put x = 0 and y = 0 in (1), we get

$0 - 0 < 3$ which is true.

$\therefore$ Solution set of (1) contains the origin.
Consider

$3x + 4y > 12$ ................(2)
Draw the graph of

$3x + 4y = 12$ , by thick line.
It passes through (4, 0) and (0, 3).
Join these points put x = 0 and y = 0 in (1), we get

$0 + 0 > 12$ which is false.

$\therefore$ Solution set of (2) does not contain the origin.
Consider

$y > 1$ ................(3)
Draw the graph of y = 1, clearly solution set of (3) does not contain (0, 0).
Since

$x > 0$ , every point the shaded region in the first quadrant, including the points on the lines, represents the solution of the given system of inequations.
1955298_1864073_ans_cf02a87665d34d92823f7e7f9b20989a.png

1955298_1864073_ans_cf02a87665d34d92823f7e7f9b20989a.png

Solve the following system of inequalities graphically.
$x + 2y \leq 10, x + y \geq x - y \leq 0, x \geq 0, y \geq 0$

Consider

$x + 2y \leq 10$ ................(1)
Draw the graph of

$x + 2y = 10$ .
It passes through (10, 0) and (0, 5) put x = 0 and y = 0 in (1), we get

$0 + 0 < 10$ which is true.

$\therefore$ Solution set of (1) contains (0, 0)
Consider

$x + y > 1$ ................(2)
Draw the graph of

$x + y = 1$ .
It passes through (1, 0) and (0, 1) put x = 0 and y = 0 in (2), we get

$0 + 0 > 1$ , which is false.

$\therefore$ Solution set of (2) is not containing the origin.
Consider

$x - y < 0$ ................(3)
Draw the graph of

$x - y = 0$
It passes through (0, 0) and (1, 1) put x = 2 and y = 0 in (3), we get

$2 - 0 < 0$ , which is false.

$\therefore$ Solution set of (3) does not contain (2, 0)
Since

$x > 0, y > 0$ , every point in the shaded region in the first quadrant, including the points on the lines, represents the solution of the given system of inequations.
1955302_1864080_ans_f0a52bcd9c3d439c8ade1565e626c0e1.png

1955302_1864080_ans_f0a52bcd9c3d439c8ade1565e626c0e1.png

Solve the following system of inequalities graphically.
$3x + 2y \leq 150, x + 4y \leq 80, x \leq 15, y \geq 0, x \geq 0$

Consider

$3x + 2y \leq 150$ ................(1)
Draw the graph of

$3x + 2y = 150$ .
It passes through (50, 0) and (0, 75) put x = 0 and y = 0 in (1), we get

$0 + 0 < 150$ which is true.

$\therefore$ Solution set of (1) contains (0, 0)
Consider

$x + 4y < 80$ ................(2)
Draw the graph of

$x + 4y = 80$ .
It passes through (80, 0) and (0, 20).
Join these points, put x = 0 and y = 0 in (2), we get

$0 + 0 < 80$ , which is true.

$\therefore$ Solution set of (2) contains (0, 0)
Consider

$x < 15$ ................(3)
Clearly solution set of (3) does not contain the origin.
Since

$x \geq 0, y \geq 0$ , every point the shaded region in the first quadrant, including the points on the lines, represents the solution of the given system of inequations.
1955301_1864079_ans_0f39caaa8e90436085a24b5d9bf90180.png

1955301_1864079_ans_0f39caaa8e90436085a24b5d9bf90180.png

Solve the following system of inequalities graphically.
$4x + 3y \leq 60, y \geq 2x, x \geq 3, y \geq 0$

Consider

$4x + 3y \leq 60$ ................(1)
Draw the graph of

$4x + 3y= 60$ by the thick line.
It passes through (15, 0) and (0,20).
Join these points put x = 0 and y = 0 in (1), we get

$0 - 0 \leq 60$ which is true.

$\therefore$ Solution set of (1) contains (0, 0)
Consider

$y \geq 2x$ ................(2)
Draw the graph of

$y = 2x$ , by thick line.
It passes through (0, 0) and (1, 2).
Join these points, clearly solution set of (2) is not containing (1, 1)
Consider

$x \geq 3$ ................(3)
Draw the graph of x = 1, clearly solution set of (3) does not contain the origin.
Since

$y > 0$ , every point the shaded region in the first quadrant, including the points on the lines, represents the solution of the given system of inequations.
1955300_1864077_ans_26923c753337484abc8ba0e01c00ef31.png

1955300_1864077_ans_26923c753337484abc8ba0e01c00ef31.png

Solve the equations.
$-5 \leq \frac{5 - 3x} {2} \leq 8$

$-5 \leq \frac{5 - 3x} {2} \leq 8$

$\Rightarrow -5(-2) \geq 3x - 5 \geq 8 (-2)$

$\Rightarrow - 16 \leq 3x - 5 \leq 10$

$\therefore -16 + 5 \leq x < \frac{15} {3} = 5$

$\Rightarrow - \frac{11} {3} \leq x \leq \frac{15} {3} = 5$

$\therefore x \space \epsilon \left[-\frac{11} {3}, 5 \right]$

-3x + 2y > -6

-3x + 2y > -6............(1)
Draw a graph of -3x + 2y = -6 by thick line.
It passes through (2, 0) and (0, -3)
Join these points, put x = 0 and y = 0 in (1), we get 0 + 0 > -6 which is true.

$\therefore$ The solution region contains the origin.

$\therefore$ The shaded region is the solution region.
1955303_1864083_ans_f5a378b173da445d8355e73c6cc8ea03.png

Solve the equations.
$6 < -3 \left(2JC - 4 \right) < 12$

$6 < -3 \left(2JC - 4 \right) < 12$

$\Rightarrow \frac{6} {-3} \geq 2x - 4 \geq \frac{12} {-3}$

$\Rightarrow - 4 < 2x - 4 \leq -2$

$\Rightarrow -4 + 4 < 2x \leq -2 + 4$

$\Rightarrow \frac{0} {2} < x < \frac{2} {2}$

$\Rightarrow 0 < x \leq 1 \therefore x \space \epsilon \left (0, 1 \right]$

Solve the equations.
$-15 \leq \frac{3 \left(x - 2 \right)} {5} \leq 0$

Given:

$-15<\dfrac{3(x-2)}{5} \leq 0$

$\Rightarrow-15\left(\dfrac{5}{3}\right)<x-2 \leq 0\left(\dfrac{5}{3}\right)$

$\Rightarrow-25+2<x \leq 2$

$\Rightarrow-23<x \leq 2$

$\therefore x \in(-23,2]$

Solve the equations.
$-3 \leq 4 - \frac{7x} {2} \leq 18$

$-3 \leq 4 - \frac{7x} {2} \leq 18$

$\Rightarrow -18 \leq \frac{7x} {2} - 4 \leq 3$

$\Rightarrow - 18 + 4 \leq \frac{7x} {2} \leq 3 + 4$

$\Rightarrow - 14 \times \frac{2} {7} \leq x < 7 \times \frac{2} {7}$

$\Rightarrow - 4 \leq x < 2$

$\therefore x \space \epsilon \left[-4, 2 \right]$

Solve the equations.
$2 \leq 3x - 4 \leq 5$

$2 \leq 3x - 4 \leq 5$

$\Rightarrow 2 + 4 \leq 3x \leq 5 + 4$

$\Rightarrow \frac{6} {3} \leq x \leq \frac{9} {3}$

$\Rightarrow 2 \leq x \leq 3$

$\therefore x \space \epsilon \left[2, 3 \right]$

Solve the equations.
$-8 \leq 5x - 3 < 7$

−8≤5x−3<7−8≤5x−3<7

A solution of $8 \%$ boric acid to be diluted by. adding a $2 \%$ boric acid solution to it. The resulting mixture is to be more than $4 \%$ but less than $6 \%$ boric acid. If we have 640 liters of the $8 \%$ solution, how many liters of the $2 \%$ solution will have to be added?

Let

$x$ be the number of liters of

$2 \%$ boric acid solution.

$\therefore$ Total mixture

$=(640+x)$ liters.

$\Rightarrow \dfrac{2 x}{100}+\dfrac{8}{100}(640)>\dfrac{4}{100}(640+x)$

$\Rightarrow 2 x+5120>2560+4 x$

$\Rightarrow 5120-2560>4 x-2 x$

$\Rightarrow 2 x<2560$

$\Rightarrow x<1280 \ldots \ldots \ldots .(1)$

$\mathrm{Also}, 2 \%$ of

$\mathrm{x}+8 \%$ of

$640<6 \%$ of

$(640+\mathrm{x})$

$\Rightarrow \dfrac{2 x}{100}+\dfrac{8}{100}(640)<\dfrac{6}{100}(640+x)$

$\Rightarrow 2 x+5120<3840+6 x$

$\Rightarrow 5120-3840<6 x-2 x$

$\Rightarrow 1280<4 x$

$\Rightarrow 320<x \ldots \ldots \ldots \ldots \ldots . .$ (2)
From (1) and

$(2),$ we get

$320<x<1280$

Solve the inequalities and represent the solution graphically on the number line:
$5(2 x-7)-3(2 x+3) \leq 0,2 x+19 \leq 6 x+47$

$5(2 x-7)-3(2 x+3) \leq 0$

$\Rightarrow 10 x-35-6 x-9 \leq 0$

$\Rightarrow 4 x-44 \leq 0$

$\Rightarrow x \leq \dfrac{44}{4}$

$\therefore x \leq 11 \ldots(1)$
Now, consider

$2 x+19 \leq 6 x+47$

$\Rightarrow 19-47 \leq 6 x-2 x$

$\Rightarrow \quad-28 \leq 4 x$

$\Rightarrow \quad x \geq-7 \quad \ldots(2)$
From ( 1 ) and

$(2),$ we get

$-7 \leq x \leq 11$ .

$\therefore x \in[-7,11]$
1955345_1865101_ans_774f90167332419aa968d7babd7a866c.png

Solve the inequalities and represent the solution graphically on the number line:
$3 x-7>2(x-6), 6-x>11 .-2 x$

Consider

$3 x-7>2(x-6)$

$\Rightarrow 3 x-2 x>-12+7$

$\Rightarrow \quad x>-5 \ldots(1)$
Now consider

$6-x>11-2 x$

$\Rightarrow 2 x-x>11-6$

$\Rightarrow \quad x>5 \ldots(2)$
From(1) and (2) we get the solution set

$x>5$ i.e.,

$x \in(5, \infty)$
1955341_1865099_ans_93c0f5198b584b918ccb3bbf62ed06ca.png

Solve the inequalities and represent the solution graphically on the number line:
$5 x+1>-24,5 x-1<24$

Consider

$5 x+1>-24$

$\Rightarrow 5 x>-25$

$\therefore x>-5 \ldots(1)$
Consider

$5 x-1<24$

$\Rightarrow 5 x<25$

$\therefore x<5 \ldots(2)$
From ( 1 ) and (2) we get

$-5<x<5$

$\therefore x \in(-5,5)$
1955337_1865096_ans_b7ef7581e2e34fb58135f3b72f9d8219.png

A manufacturer has 600 liters of $12 \%$ solution of acid. How many liters of a $30 \%$ acid solution must be added to it so that acid content in the resulting mixture will be more than $15 \%$ but less than $18 \%?$

Let x liters of

$30 \%$ acid solution is required to be added. Then, total mixture

$=(x+600)$ liters.

$\therefore 30 \%$ of

$x+12 \%$ of

$600>15 \%$ of

$(x+600)$

$\Rightarrow \dfrac{30 x}{100}+\dfrac{12}{100}(600)>\dfrac{15}{100}(x+600)$

$\Rightarrow 30 x+7200>15 x+9000$

$\Rightarrow 15 x>9000-7200=1800$

$\Rightarrow x>120 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .(1)$
Also

$30 \%$ of

$x+12 \% 600<18 \%$ of

$(x+600)$

$\Rightarrow \dfrac{30 x}{100}+\dfrac{12}{100}(600)<\dfrac{18}{100}(x+600)$

$\Rightarrow 30 x+7200<18 x+10800$

$\Rightarrow 12 x<10800-7200=3600$

$\Rightarrow x<300 \ldots \ldots \ldots \ldots .$ (2)
From (1) and (2) we get

$120<x<300$
Thus, the number of liters of the

$30 \%$ solution of acid will have to be more than 120 liters but less than 300 liters.

Solve the inequalities and represent the solution graphically on the number line:
$2(x-1)<x+5,3(x+2)>2-x$

Consider

$2(x-1)<x+5$

$\Rightarrow 2 x-x>2-6$

$x<7 \ldots(1)$
Now consider

$3(x+2)>2-x$

$\Rightarrow 3 x+x>2-6$

$\Rightarrow 4 x>-4$

$\Rightarrow x>-1 \ldots(2)$
From (1) and (2) we get the solution set

$-1<x<7$

$\therefore x \in(-1,7)$
1955342_1865100_ans_3cf6d971cf2345d2bec7f5e35522bee9.png

Solve the inequalities and represent the solution graphically on the number line:
$3 x-7<5+x, 11-5 x \leq 1$

Consider

$3 x-7<5+x$

$\Rightarrow 3 x-x<5+7$

$\Rightarrow x<\dfrac{12}{2}=6$

$x<6 \ldots(1)$
Consider

$11-5 x \leq 1$

$\Rightarrow 10 \leq 5 x$

$\Rightarrow \dfrac{10}{2} \leq x$

$\therefore 2 \leq x \ldots(2)$
From (1) and (2) we get

$2 \leq x<6$

$\therefore x \in[2,6)$
1955340_1865098_ans_94697b6d539346cdbb0d54cbf050e967.png

Solve the equations.
$-12<4-\dfrac{3 x}{-5} \leq 2$

$Given:$

$-12<4-\dfrac{3 x}{-5} \leq 2$

$\Rightarrow-12<4+\dfrac{3 x}{5} \leq 2$

$\Rightarrow-16<\dfrac{3 x}{5} \leq 2-4$

$\Rightarrow 16 \times \dfrac{5}{3}<x \leq-2\left(\dfrac{5}{3}\right)$

$\Rightarrow-\dfrac{80}{3}<x \leq-\dfrac{10}{3}$

$\therefore x \in\left(-\dfrac{80}{3},-\dfrac{10}{3}\right]$

Solve the equations.
$-12 < 4 - \frac{3x} {-5} \leq 2$

Given:

$-12<4-\dfrac{3 x}{-5} \leq 2$

$\Rightarrow-12<4+\dfrac{3 x}{5} \leq 2$

$\Rightarrow-16<\dfrac{3 x}{5} \leq 2-4$

$\Rightarrow 16 \times \dfrac{5}{3}<x \leq-2\left(\dfrac{5}{3}\right)$

$\Rightarrow-\dfrac{80}{3}<x \leq-\dfrac{10}{3}$

$\therefore x \in\left(-\dfrac{80}{3},-\dfrac{10}{3}\right]$

Solve the equations.
$7 \leq \frac{3x + 11} {2} \leq 11$

Given

$7 \leq \dfrac{3 x+11}{2} \leq 11$

$\Rightarrow 14 \leq 3 x+11 \leq 22$

$\Rightarrow 14-11 \leq 3 x \leq 22-11$

$\Rightarrow \dfrac{3}{3} \leq x \leq \dfrac{11}{3}$

$\Rightarrow 1 \leq x \leq \dfrac{11}{3}$

$\therefore x \in\left[1, \dfrac{11}{3}\right]$

Find the integer values of $t$ which satisfy the inequalities.
$4t + 7 < 39 \nless 7t + 2$

1897735_1913365_ans_b4075b512b494eb6b426a35ba353eb1a.jpg

Solve the following inequalities, graphically :
$| x - y | \geq 1$

Given inequality

$|x - y| \geq 1$
Removing modulus, we get

$x - y \geq 1$
and

$x - y \leq -1$
Writing in equation form, we get

$x - y = 1$ ......(i)
or

$x - y = -1$ .....(ii)
In equation (i),
Putting

$x = 0, 0 - y = 1$

$y = -1$
Point

$(0, -1)$ will lie on y-axis.
Now, putting

$y = 0,$

$x - 0 = 1$

$x = 1$
Point

$(1, 0)$ will lie on x-axis.
from equation (ii),
Putting

$x = 0, 0 - y = -1$

$y = 1$
Point

$(0, 1)$ will lie on y-axis.
Now, putting

$y = 0$

$x - 0 = -1$

$x = -1$
Point

$(-1, 0)$ will lie on x-axis.
By joining point

$(0, -1), (1, 0)$ and

$(0,1), (-1,0)$ following graph is obtained.
Now, inequality

$x - y \geq 1$ does not satisfied by origin

$(0, 0).$
Its shaded portion will be region from line to opposite to origin.
Second inequality

$x - y \leq -1$ also not satisfied by origin

$(0, 0)$
i.e.,

$0 - 0 < -1$ is not true.
Therefore its shaded part will be the region from line to opposite to origin.
1955561_1875478_ans_437f3d58d28a4ebabb37777897949b9d.png

1955561_1875478_ans_437f3d58d28a4ebabb37777897949b9d.png

Solve the following inequalities, graphically :
$3x - 2y \leq x + y - 8$

Given inequality

$3x - 2y \leq x + y - 8$ writing this, in equation form, we get

$3x - 2y = x + y - 8$

$\Rightarrow 3x - x - 2y - y = -8$

$\Rightarrow 2x - 3y = -8$
Putting

$x = 0$ in equation, we get

$2 \times 0 - 3y = -8$

$-3y = -8$

$y = \dfrac{8}{3}$
point

$(0, \dfrac{8}{3})$ will cut y-axis.
Now, putting

$y = 0,$

$2x - 3 \times 0 = -8$

$x = -4$
Point

$(-4, 0)$ will cut x-axis.
Graph obtained from two points will be as follows:
Thus, inequality

$3x - 2y \leq x + y - 8$ does not satisfied by origin

$(0, 0).$

$\Rightarrow 3 \times 0 - 2 \times 0 \leq 0 + 0 - 8$ which is not true.
Therefore shaded area opposite to origin from line will be required solution set.
1955559_1875477_ans_bd720f9cf96b4966bbcd9bdee6c6673c.png

1955559_1875477_ans_bd720f9cf96b4966bbcd9bdee6c6673c.png

Solve the following linear programming problem by graphical method :
Minimize and maximize $Z = x + 2y$
Subject to the constraints,
$x + 2y \geq 100$
$2x y \leq 0$
$2x + y \leq 200$
and $x \geq 0,y \geq 0$

Converting the given inequations into equations

$x + 2y = 100$ …..(1)

$2x - y = 0$ …..(2)

$2x + y = 200$ …..(3)

$x = 0$ …..(4)

$y = 0$ …..(5)

Region represented by

$x + 2y \geq 100$ : The line

$x + 2y = 100$ meets the coordinate axis at

$A(10,0)$ and

$5(0, 5)$ .

Table for

$x + 2y = 100$

X 100 0

y 0 50

$A(100, 0); B(0, 50)$

Join the points A and B to obtain the line. We find that the point (0, 0) does not satisfy the inequation

$x + 2y \geq 100$ . So, the region opposite to the origin represents the solution set of the inequation.

Region represented by

$2x - y \leq 0$ : The line

$2x - y = 0$ meets the coordinate points at

$C(0, 0)$ and

$D( 100, 200)$ .

$2x - y = 0$

X 0 100

y 0 200

$C(0, 0), D(100, 200)$

Join the points

$C(0,0)$ to

$D(100, 200)$ to obtain the line. Clearly (0, 0) satisfies. The inequation

$2(0) - 0 = 0 \leq 0$ . So the region containing origin represents the solution set of the inequation.

Region represent by 2x + y ≤ 200 : The line 2x + y = 200 meets the coordinate axis at ,A(100,0) and B(0, 200).

2x + y = 200

X 100 0

y 0 200

$E(100, 0); F(0, 200)$

Join points E and F to obtain the line. Clearly

$(0, 0)$ satisfies the inequation

$2x + y \leq 200$ . So the region containing the origin represents the solution set of this inequation.

Region represented by

$x \geq 0$ and

$y \geq 0$ : Since every point in the first quadrant satisfies these inequations. So the first quadrant is the region represented by the inequations

$x \geq 0$ and

$y \geq 0$ .

The shaded region BEFD represents the common region of the above inequations. This region is the feasible region of the given LPP.

where point

$E(20,40)$ is the point of intersection of lines

$x + 2y = 100$ and

$2x - y = 0.$

Point

$F(50, 100)$ is the point of intersection of lines

$2x + y = 200$ and

$2x - y = 0$ .

The value of the objective function on these comer points of the feasible region are

$E(20, 40), B(0, 50), D(0, 200)$ and

$F(50, 100).$

Point x-coordinate y-coordinate Objective function Z = x + 2y

E 20 40

$Z_E = 20 + 2 x 40 = 100$

B 0 50

$Z_B = 0 + 2 x 50 = 100$

F 50 100

$Z_F = 50 + 2 x 100 = 250$

D 0 200

$Z_D = 0 + 2 x 200 = 400$

It is clear from the table that minimum value of objective function is

$= 100$ at points

$E(20, 40)$ and

$B(0, 50)$ which is minimum on every point on line AEB and maximum value of objective function is on the point

$D(0, 200)$ where

$Z = 400$ .

Hence Minimum value of

$Z = 100$

and Maximum value of

$Z = 400.$

1866565_1873287_ans_b61ff9b54d144d6bb780485bd733ff50.PNG

By graphical method, show the solution set of the following inequalities :
(i) $x \geq 2$
(ii) $2x + 3y \leq 6$

Given inequality

$x \geq 2$
Writing its equation from,

$x = 2$
It is clear that straight line is paralleled to y-axis and will pass through point

$(2, 0)$ of x-axis and obtained the following graph.
Graph of two meeting points is as follows:
Now, inequality

$2x + 3y \leq 6$ satisfied by origin

$(0, 0).$
Therefore region from line to origin, shaded part is required solution set.
1955556_1875474_ans_911fc1be379f46aa8b09b23d9e94dc85.png

Maximise $Z = x + 2y$
subject to the constraints
$x + 2y \geq 100$
$2x - y \leq 0$
$2x + y \leq 200$
$x, y \geq 0$
Solve the above LPP graphically.

Minimize & Maximize

$Z = x + 2y$

Subject to,

$x + 2y \ge 100$

$2x - y \le 0$

$2x + y \le 200$

$x, y \ge 0$

$x + 2y \ge 100$

$x$	$0$	$100$
$y$	$50$	$0$

$2x - y \le 0$

$x$	$0$	$50$
$y$	$0$	$100$

$2x + y \le 200$

$y$	$100$	$0$
$x$	$0$	$200$

$\therefore Z = 400$ is maximum at

$(0,200)$

Also,

$Z$ is minimum at two points

$(0,50) \ \& \ (20,40)$

$\therefore Z = 100$ is minimum at all points joining

$(0,50)$ &

$(20, 40)$

1964810_1944427_ans_34c5c729578d480fa209af4019af635b.png

A manufacturer has employed $5$ skilled men and $10$ semi-skilled men and makes two models $A$ and $B$ of an article. The making of one item of model $A$ requires 2 hours work by a skilled man and 2 hours work by a semi-skilled man. One item of model $B$ requires 1 hour by a skilled man and 3 hours by a semi-skilled man. No man is expected to work more than 8 hours per day. The manufacturers profit on an item of model A is Rs. 15 and on an item of model $B$ is Rs. How many of items of each model should be made per day in order to maximize daily profit? Formulate the above LPP and solve it graphically and find the maximum profit.

Let

$x$ article of delux model and

$y$ article of an ordinary model be made.

Numbers cannot be negative.

Therefore,

$x,y\ge 0$

According to the question, the profit on each model of deluxe and ordinary type model are

$Rs.15$ and

$Rs.10$ respectively,

So, profits on

$x$ deluxe model and

$y$ ordinary models are

$15x$ and

$10y$

Let

$Z$ be total profit, then,

$Z=15x+10y$

Since, the making of a deluxe and ordinary model requires

$2$ hrs. and

$1$ hr work by skilled men, so

$x$ deluxe and

$y$ ordinary models require

$2x$ and

$y$ hours of skilled men but time available by skilled men is

$5\times 8=40$ hours.

So,

$2x+y\le 40$ (First Constraint)

Since, the making of a deluxe and ordinary model requires

$2$ hrs. and

$3$ hr work by semi-skilled men, so

$x$ deluxe and

$y$ ordinary models require

$2x$ and

$y$ hours of skilled men but time available by skilled men is

$10\times 8=80$ hours.

$2x+3y\le 80$ (Second Constraint)

Hence the mathematical formulation of LPP is,

Max

$z=15x+10y$

subject to constraints,

$2x+y\le 40$

$2x+3y\le 80$

$x,y\ge 0$

Region

$2x+y\le 40$ ; line

$2x+4y=40$ meets axes at

$A_1(20,0),B_1(0,40)$ respectively. Region containing origin represents

$2x+3y\le 40$ as

$(0,0)$ satisfies

$2x+y\le 40$

Region

$2x+3y\le 80$ ; line

$2x+3y=80$ meets axes at

$A_2(40,0),B_2(0,\dfrac{80}{3})$ respectively. Region containing origin represents

$2x+3y\le 80.$

Refer fig. 1

The corner points are

$A_1(20,0),P(10,20),B_2(0,\dfrac{80}{3})$

The value of

$z=15x+10y$ at these corner points are

Refer fig . 2

The maximum value of

$Z$ is

$300$ which is attained at

$P(10,20)$

Thus, maximum profit is obtained when

$10$ units of deluxe model and $$20 init of ordinary model is produced.

1964935_1944301_ans_7682eca23267482b9a192eff3a33648b.png

How many liters of water will have to be added to 1125 liters of the $45 \%$ solution of acid so that the resulting mixture will contain more than $25 \%$ but less than $30 \%$ acid content?

Let

${ }^{\circ} x^{\prime}$ liters of water be added.

$\therefore 25 \%$ of

$(1125+x)<45 \%$ of

$1125<30 \%$ of

$(1125+x) \quad$

$\Rightarrow \dfrac{25}{100}(1125+x)<\dfrac{45}{100}(1125)<\dfrac{30}{100}(1125+x)$

$\Rightarrow 25(1125+x)<45(1125)<30(1125+x)$

$\Rightarrow 5(1125+x)<9(1125)<6(1125+x)$
Consider

$5(1125+x)<9(1125)$

$\Rightarrow 5 x<1125(9-5) \therefore x<\dfrac{1125(4)}{5}=900$

$x<900 \quad \cdots(1)$
Also,

$9(1125)<6(1125+x)$

$\Rightarrow 1125(9-6)<6 x \quad \therefore \dfrac{1125(3)}{6}<x$

$562.5<x \quad \cdots(2)$
From ( 1 ) and ( 2 ) we get

$562.5<x<900$

Solve the following inequalities, graphically :
$|x | \leq 3$

Removing modules from given

$|x | \leq 3$
inequality

$|x | \leq 3,$ we get

$\Rightarrow -3 \leq x \leq 3$

$\Rightarrow x \leq 3$ .........(i)
and

$x \geq -3$ ......(ii)
writing inequality (i) in equation form,

$x = 3$
line

$x = 3,$ will be right side of y-axis and parallel to y-axis writing inequality (ii) in equation form,

$r = -3.$
Line

$x = -3,$ will be left side of y-axis and II to y-axis.
Graph of both linear equations is as follows:
Thus, inequality

$|x| \leq 3$ satisfy by origin

$(0, 0)$
Therefore shaded part from line

$x = 3$ to

$x = -3$ will be required solution set.
1955558_1875475_ans_d88788cfc92f49ac9e9f603bc8300987.png

If $(xy)^{a-1}=z$
$(yz)^{b-1}=x$
$(zx)^{c-1}=y$
then $ab+bc+ca=?$

$(xy)^{a-1}=z$

$(yz)^{b-1}=x$

$(zx)^{c-1}=y$
Multiplying the equations we get,

$(xy)^{a-1}\times(yz)^{b-1}\times(zx)^{c-1}$ =

$x y z$
Comparing we get,

$a+c-2=1\Rightarrow a+c=3$

$a+b-2=1\Rightarrow a+b=3$

$b+c-2=1\Rightarrow b+c=3$
By solving these eqautions we get,

$a=b=c=\dfrac{3}{2}$
Therefore,

$ab+bc+ca=\dfrac { 3 }{ 2 } \times \dfrac { 3 }{ 2 } +\dfrac { 3 }{ 2 } \times \dfrac { 3 }{ 2 } +\dfrac { 3 }{ 2 } \times \dfrac { 3 }{ 2 } =\dfrac { 27 }{ 4 }$

Solve the following system of inequations graphically:
$2x+y> 1,x-2y< -1$

1947900_1708876_ans_1b0bef7a69084285b39d18f599e2db1d.jpg

Solve the following system of inequations graphically:
$x+y\leq 9,y> x,x\geq 0$

1947899_1708874_ans_5c21f55fbda44ff9bab361c45489d4b1.jpg

Solve the following systems of inequations graphically: $x+2y\leq 100,2x+y\leq 120,x+y\leq 70,x\geq 0,y\geq 0$

Inequality	Equation	x coordinate	y coordinate	Line passes through (x,y)	Sign of inequality	Region
$x+2y\le 100$	$x+2y=100$	100	0	$\left( 100,0 \right)$	$\le$	Origin side
		0	50	$\left( 0,50 \right)$
$2x+y\le 120$	$2x+y=120$	60	0	$\left( 60,0 \right)$	$\le$	Origin side
		0	120	$\left( 0,120 \right)$
$x+y\le 70$	$x+y=70$	70	0	$\left( 70,0 \right)$	$\le$	Origin side
		0	70	$\left( 0,70 \right)$

1858438_1708920_ans_ffc0511938c04994b3eb4856952936a4.png

Solve the following system of inequations graphically:
$3x+2y\leq 12,x\geq 1,y\geq 2$

1947897_1708854_ans_dbc37c2870f641ae9d09351440ce7b9b.jpg

Solve the following system of inequations graphically:
$4x+3y\leq 60,y\geq 2x,x\geq 3,x\geq 0,y\geq 0$

Inequality	Equation	x coordinate	y coordinate	Line passes through (x,y)	Sign of inequality	Region
$4x+3y\le 60$	$4x+3y=60$	15	0	$\left( 15,0 \right)$	$\le$	Origin side
		0	20	$\left( 0,20 \right)$
$2x-y\le 0$	$2x-y=0$	0	0	$\left( 0,0 \right)$	$\le$	Origin side
		1	2	$\left( 1,2 \right)$
$x\ge 3$	$x=3$	3	1	$\left( 3,1 \right)$	$\ge$	Non origin side
		3	2	$\left( 3,2 \right)$

1857279_1708910_ans_4b2904d5a3d9454cb0d6b79b3bdf9bbc.png

Solve the following system of inequations graphically:
$x+2y\leq 10,x+y\geq 1,x-y\leq 0,x\geq 0,y\geq 0$

Inequation	Equation	x coordinate	y coordinate	Line passes through (x,y)	Sign of inequality	Region
$x+2y\le 10$	$x+2y=10$	10	0	$\left( 10,0 \right)$	$\le$	Origin side
		0	5	$\left( 0,5 \right)$
$x+y\ge 1$	$x+y=1$	1	0	$\left( 1,0 \right)$	$\ge$	Non Origin side
		0	1	$\left( 0,1 \right)$
$x-y\le 0$	$x-y=0$	0	0	$\left( 0,0 \right)$	$\le$	Origin side
		1	1	$\left( 1,1 \right)$

Refer to the graph. Every point in the hatched region satisfies all the inequalities.

Thus, hatched region is the solution of given inequalities.

1857250_1708903_ans_677327f8eff4480eae5f60827589381e.png

Solve the following systems of inequations graphically: $x-2y\leq 2,x+y\geq 3,-2x+y\leq 4,x\geq 0,y\geq 0$

Inequality	Equation	x coordinate	y coordinate	Line passes through (x,y)	Sign of inequality	Region
$x-2y\le 2$	$x-2y=2$	2	0	$\left( 2,0 \right)$	$\le$	Origin side
		0	-1	$\left( 0,-1 \right)$
$x+y\ge 3$	$x+y=3$	3	0	$\left( 3,0 \right)$	$\ge$	Non origin side
		0	3	$\left( 0,3 \right)$
$-2x+y\le 4$	$-2x+y=4$	-2	0	$\left( -2,0 \right)$	$\le$	Origin side
		0	4	$\left( 0,4 \right)$

1858435_1708914_ans_f77fb71e99cb4f11a8690100519b7db8.png

Solve the following system of inequations graphically:
$2x+y\geq 6,3x+4y\leq 12$

1947898_1708868_ans_8435e125e9ea45679ae845f5a365f0ed.jpg

Solve the following systems of inequations graphically: $x+2y\leq 2000,x+y\leq 1500,y\leq 600,x\geq 0,y\geq 0$

Inequality	Equation	x coordinate	y coordinate	Line passes through (x,y)	Sign of inequality	Region
$x+2y\le 2000$	$x+2y=2000$	2000	0	$\left( 2000,0 \right)$	$\le$	Origin side
		0	1000	$\left( 0,1000 \right)$
$x+y\le 1500$	$x+y=1500$	1500	0	$\left( 1500,0 \right)$	$\le$	Origin side
		0	1500	$\left( 0,1500 \right)$
$0x+y\le 600$	$0x+y=600$	0	600	$\left( 0,600 \right)$	$\le$	Origin side
		100	600	$\left( 100,600 \right)$

Find the linear inequations for which the shaded area is the solution set in the figure given below.

1969570_1708975_ans_0dbd341122ba4e90bf4cee98848551a5.jpg

Find the linear inequations for which the shaded area in the figure given below, is the solution set.

1947814_1709007_ans_eaa979bd49af43368d8c4e2c19f13091.jpg

A firm manufactures headache pills in two sizes A and B. Size A contains $2$ grams of aspirin $5$ grams of bicarbonate and $1$ gram of caffeine; size B contains $1$ gram of aspirin of $8$ grams of bicarbonate and $6$ grains of caffeine. It has been found by users that it requires at least $12$ grams of aspirin $74$ grams of bicarbonate and $24$ grams of caffeine for providing immediate effects. Determine graphically the least number of pills a patient should have to get immediate relief.

Let the patient should have x pills of size A and y pills of size B.

∴ Maximum number of pills

$= x + y$

According to the question,

A type pills have a quantity of Aspirin

$= 2x$ grams and B type pills have a quantity of Aspirin

$= 1\,y$ grams

∴ Constraints for the quantity of Aspirin

$2x + y \geq 12$ grams

Similarly constraints for Bicarbonate

$5x + 8 \geq 74$ grams

and constraints for the quantity of caffeine

$x + 6.6y \geq 24$ grams

∵ x and y are the numbers of pills.

∴

$x \geq 0$ and

$y \geq 0$

So mathematical formulation of linear programming problem is the following :

Minimize

$Z = x + y$

Subject to the constraints

$2x + y \geq 12$

$5x + 8 \geq 74$

$x + 6y \geq 24$

$x \geq 0, y \geq 0$

Converting the constraints into the equations

$2x + y = 12$ …..(1)

$5x + 8y = 74$ …..(2)

$x + 6y = 24$ …..(3)

Region represented by

$2x + y \geq 12$ : The line

$2x + y = 12$ meets the coordinate axis at the points

$A(6,0)$ and

$B(0, 12).$

Table for

$2x + y = 12$

X 6 0

y 0 12

$A(6,0); B(0, 12)$

Join the point A to

$5$ to obtain the line. Since the origin

$(0, 0)$ does not satisfy the inequation

$2(0) + (0) = 0 \geq 12$ . So the region opposite to origin represents the solution set of the inequation.

Region represented by

$5x + 8y \geq 74$ : The line

$5x + 8y = 74$ meets the coordinate axis at the point

$C(74/5, 0)$ and

$D(0, 74/8)$ $.

Tabel for

$5x + 8y = 74$

X 74/5 0

y 0 74/8

Join the points C and D to obtain the line. Since the origin (0,0) does not satisfy the inequation

$5(0) + 8(0) \geq 74$ . So the region opposite to origin represents the solution set of inequation.

Region represented by

$x + 6y \geq 24$ : The line

$x + 6.6y = 24$ meets the coordinate axis at the point

$E(24,0)$ and

$F(0,4).$

The table for

$x + 6y = 24$

X 24 0

y 0 4

$A(24, 0); B(0, 4)$

Join the points E and F to obtain the line. Since the origin,

$(0, 0)$ does not satisfy the inequation. So the region opposite to origin represents the solution set of the inequation.

Region represented by

$x \geq 0, y \geq 0$ : Since every point in first quadrant satisfies the inequations so the first quadrant represents the solution set of these inequations.

The point of intersection of lines

$2x +y - 12$ and

$5x + 8y = 74$ is

$G(2, 8)$ .

The point of intersection of lines

$5x + 8y = 74$ and

$x + 6y = 24$ is

$H (,)$ .

From the above table the value of objective function is minimum at the point

$G(2, 8)$ where

$x = 2$ and

$y = 8$ and

$z = 2 + 8 = 10.$

Since the feasible region of the solution set is an open region. So the graph of

$x + y \leq 10$ passes through point

$G(2,8)$ which satisfies the inequation

$z + 8 = 10$ so the solution of this LPP is

$G(2, 8)$ where

$x = 2$ and

$y = 8$ .

Hence the patient should have a minimum of

$2$ pills of type A and 8 pills of type B to get immediate relax.

1866574_1873304_ans_8b068a44aae345259deeb38f4ae88a0e.PNG

Linear Inequalities - Class 11 Engineering Maths - Extra Questions

Solve the following inequalities graphically in two-dimensional plane:x>−3x >- 3

Solve 4x - 12 ≥\displaystyle \geq 0 graphically in two dimensional plane

Solve the given inequalities graphically:x≥3,y≥2x {\geq} 3, y {\geq} 2

Solve the inequalities and represent the solution graphically on number line.5x+1>−24,5x−1<245x+1 > -24, 5x-1 < 24

Solve 3x+2y>63x+2y>6 graphically.

State which of the following are equation (with variable). Given reasons for your an mention which variable is used from the equation with a variable.9n−9>59n-9>5

Solve: |2x−5|<1|2x-5|< 1.

Find the solution of the inequality 3(1−x)<2(x+4)3\left( 1-x \right) <2\left( x+4 \right) and draw on the number line.

Solve: −(x−2)+4>−3x+10-(x-2)+4 > - 3x+10.

Fill in the blank with >, <>,\ < or == sign(−25)−(−42)(-25)-(-42)________(−42)−(−25)(-42)-(-25)

Fill in the blank with >, <>,\ < or == sign45−(−11)45-(-11)________57+(−4)57+(-4)

Solve 3x−2<1\dfrac{3}{x-2}<1, when x∈Rx\in R

Solve :3x−2y=13x-2y=12x+y=32x+y=3

Solve the inequalities for real xx -3(2−x)≥2(1−x)3\left( 2-x \right) \ge 2\left( 1-x \right)

Find the pairs of consecutive even positive integers, both of which are larger than 55 such that their sum is less than 2323.

Solve the following inequation and represent the solution set on a number line−812<−12−4x≤712,x∈1-8\cfrac{1}{2}< -\cfrac{1}{2}-4x\le 7\cfrac{1}{2},x\in 1

Represent the following inequalities on real number line:−5<x≤−1 -5<x \leq-1

Enter 1 if it is true else enter 0.If xϵWx\, \epsilon\, W, then the solution set of x<2x<2 is x={0,1}x = \{0,1\}

Fill in the blanks with correct inequality sign (>,<,≥,≤)(>, <, \ge, \le ).p−q=−3⇒p.....qp-q=-3\Rightarrow p.....q

Solve 2(x–3)<1,x∈1,2,3,….102(x – 3) < 1, x ∈ {1, 2, 3, …. 10}

Solve the equation7≤3x+112≤11 7 \leq \dfrac{3 x+11}{2} \leq 11

If x∈Ix \in I, find the smallest value of xx which satisfies the inequation 2x+52>5x3+22x+\dfrac{5}{2}>\dfrac{5x}{3}+2

Solve the inequation −3≤3−2x<9,x∈R.- 3 \le 3 - 2x < 9, x \in R. Represent your solution on a number line.

Solve:Solve: 4x+3<5x+74x + 3 < 5x + 7

Solve the inequality and show the graph of the solution on number line:3(1−x)<2(x+4)3 (1 -x) < 2 (x + 4)

Solve & graph the solution set of 3x+6≥93x + 6 \geq 9 and −5x>−15-5x > -15, x∈Rx \in R.

Solve & graph the solution set of 3x+6≥9\displaystyle 3x+6\geq 9 and -5x > -15 xϵR\displaystyle x\: \epsilon \: R

Solve 24x<10024 x < 100, when xx is a natural number.

Solve the inequality and show the graph of the solution on number line:3x−2<2x+13x- 2 < 2x + 1

Solve the inequality and show the graph of the solution on number line:x2≥(5x−2)3−(7x−3)5\displaystyle {\frac{x}{2}\geq \frac{(5x-2)}{3}-\frac{(7x-3)}{5}}

Solve the inequality and show the graph of the solution on number line:5x−3≥3x−55x -3 \geq 3x- 5

Solve the following system of inequalities graphically x+2y≤8\displaystyle x+2y\leq 8 ......(1)2x+y≤8\displaystyle 2x+y\leq 8 ......(2)x≥0\displaystyle x\geq 0 ......(3)y≥0\displaystyle y\geq 0 ......(4)

Find x satisfying | x - 5 | ≤\displaystyle \leq 3

Solve the following inequalities graphically in two-dimensional plane:2x−3y>62x -3y > 6

Solve the following inequalities graphically in two-dimensional plane:y<−2y < -2

Solve the following inequalities graphically in two-dimensional plane:3x + 4y≤{ \leq} 12

Solve the following inequalities graphically in two-dimensional plane:x+y<5x + y < 5

Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23

Solve the following inequalities graphically in two-dimensional plane:−3x+2y≥−6-3x + 2y {\geq} -6

Solve the following inequalities graphically in two-dimensional plane:x−y≤2x -y {\leq} 2

The longest side of a triangle is 33 times the shortest side and the third side is 22 cm shorter than the longest side. If the perimeter of the triangle is at least 6161 cm, find the minimum length in cm. of the shortest side.

Solve the following inequalities graphically in two-dimensional plane: 2x+y≥62x + y {\geq} 6

Solve the following inequalities graphically in two-dimensional plane:y+8≥2xy + 8 {\geq}2x

Solve the given inequalities graphically:x+yx + y ≤{\leq} 6,x+y6, x + y ≥{\geq} 44

Solve the given inequalities graphically:x+yx + y ≥{\geq} 44 and 2x−y>0 2x -y > 0

Solve the following inequations graphically:2x−y>1,x−2y<−12x- y >1, x -2y < -1

Solve the system of inequalities graphically:2x+y2x + y ≥{\geq} 4,x+y4, x + y ≤{\leq}3,2x−3y 3, 2x- 3y ≤{\leq} 66

Solve the system of inequalities graphically:x−2y≤3,3x+4y≥12,x≥0,y≥1{x- 2y \leq 3, 3x + 4y \geq 12, x \geq 0 , y \geq 1}

Solve the given inequalities graphically:2x+y2x + y ≥{\geq} 8,x+2y8, x + 2y ≥{\geq}10 10

Solve the given inequalities graphically:2x+y2x + y ≥{\geq} 66 and 3x+4y3x + 4y ≤{\leq} 1212

Solve the given inequalities graphically:3x+2y3x + 2y ≤{\leq} 12,x12, x ≥{\geq} 1,y1, y ≥{\geq}2 2

Solve the system of inequalities graphically:5x+4y5x + 4y ≤{\leq} 20,x20, x ≥{\geq}1,y1, y ≥{\geq}2 2

Solve the given inequalities graphically:x+y≤9,y>x,x≥0x + y {\leq} 9, y > x, x {\geq} 0

Solve, and express the answer graphically.4x+4x−4≤0\displaystyle\frac{4x+4}{x-4} \le 0

Solve the inequalities and represent the solution graphically on number line.3x−7>2(x−6),6−x>11−2x3x-7 >2(x-6), 6-x> 11-2x

Solve the system of inequalities graphically3x+2y≤150,x+4y≤80,x≤15,y≥0,x≥0{3x + 2y \leq 150, x + 4y \leq 80, x \leq 15, y \geq 0, x \geq 0}

Solve the given inequalities graphically:x+2y≤10,x+y≥1,x−y≤0,x≥0,y≥0{x + 2y \leq 10, x + y \geq 1, x- y \leq 0, x \geq 0, y \geq 0}

Solve the following inequality−2x+5x+6>−2\displaystyle \frac{-2x+5}{x+6}>-2

Solve the inequality and represent the solution graphically on number line.2(x−1)<x+5,3(x+2)>2−x2(x-1)< x+5, 3(x+2)> 2-x

Solve the inequality and represent the solution graphically on number line.5(2x−7)−3(2x+3)≤0,2x+19≤6x+475(2x-7)-3(2x+3)\le 0, 2x+19\le 6x+47

A solution is to be kept between 68oF68^{o}F and 77oF77^{o}F . What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by F=95C+32?F=\dfrac{9}{5}C+32?

A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

How many litres of water will have to be added to 11251125 litres of the 45%45\% solution of acid so that the resulting mixture will contain more than 25%25\% but less than 30%30\% acid content?

Find the values of xx, which satisfy the inequation −256<12−2x3≤2,x∈W-2\dfrac {5}{6} < \dfrac {1}{2} - \dfrac {2x}{3} \leq 2, x \in W.

Solve the following inequation and write the solution set : 13x−5<15x+4<7x+12,x∈R13x - 5 < 15x + 4 < 7x + 12, x \in RRepresent the solution on a real number line.

Solve the following inequation and represent the solution set on the number line:4x−19<3x5−2≤−25+x,x∈R4x-19<\dfrac{3x}{5}-2\le-\dfrac{2}{5}+x,\,\,\,\,\,\, x\in R

Solve the following inequation, write the solution set and represent it on the number line.−3(x−7)≥15−7x>x+13,x∈R-3 (x - 7)\geq 15 - 7x > \dfrac {x + 1}{3}, x\in R

Solve the following inequation, write the solution set and represent it on the number line: −x3≤x2−113<16,x∈R-\displaystyle\frac{x}{3}\leq \frac{x}{2}-1\frac{1}{3} < \frac{1}{6}, x\in R.

Minimize and maximize z=5x+10yz=5x+10ysubject to the constraints x+2y≤120x+2y \le 120x+y≥60x+y\ge 60x−2y≥0x-2y\ge 0 and x≥0,y≥0x \ge 0,\, y \ge 0 by graphical method.

Solve by graphical method x+y≥5,x−y≤3x+y \geq 5, x-y \leq 3

Indicate the polygonal region represented by the system of inequationsx≥0,y≥0,x≤4,x≥yx\ge 0, y≥0, x\le 4, x\ge y.

Solve: 5x−7<3(x+3),1−3x2≥x−45x-7 < 3(x+3), 1-\displaystyle\frac{3x}{2} \geq x-4.

Solve the following inequalities.2−5xx+1>2\displaystyle\, \frac{2 \, - \, 5x \, }{x \, + \, 1}\, > \, 2

Solve by graphical method: y−2x≤1,x+y≤3,x≥0,y≥0y-2x\leq 1, x+y\leq 3, x\geq 0, y\geq 0.

Show triangle formed x + 2y = 5 and 4x - y = 5 and x-axis graphically.

Represent the solution of the following inequality on the real number line:2x+53>3x−3{\dfrac{2x + 5}{3}} > 3x - 3

Represent xx on number line or find xx3x−25>4x−32\dfrac{3x - 2} {5} > \dfrac{4x - 3}{2}

Solve the following equation:- 8 < - (3x - 5) < 13- 8 < - (3x - 5) < 13

Find the largest value of x for which 2(x - 1) \le 9 - x2(x - 1) \le 9 - x and x \in Wx \in W.

Solve the following inequalities graphically in two-dimensional plane:
$x >- 3$

Solve 4x - 12 $\displaystyle \geq$ 0 graphically in two dimensional plane

Solve the given inequalities graphically:
$x {\geq} 3, y {\geq} 2$

Solve the inequalities and represent the solution graphically on number line.
$5x+1 > -24, 5x-1 < 24$

Solve $3x+2y>6$ graphically.

State which of the following are equation (with variable). Given reasons for your an mention which variable is used from the equation with a variable.
$9n-9>5$

Solve: $|2x-5|< 1$ .

Find the solution of the inequality $3\left( 1-x \right) <2\left( x+4 \right)$ and draw on the number line.

Solve: $-(x-2)+4 > - 3x+10$ .

Fill in the blank with $>,\ <$ or $=$ sign
$(-25)-(-42)$ ________ $(-42)-(-25)$

Fill in the blank with $>,\ <$ or $=$ sign
$45-(-11)$ ________ $57+(-4)$

Solve $\dfrac{3}{x-2}<1$ , when $x\in R$

Solve :
$3x-2y=1$
$2x+y=3$

Solve the inequalities for real $x$ -
$3\left( 2-x \right) \ge 2\left( 1-x \right)$

Find the pairs of consecutive even positive integers, both of which are larger than $5$ such that their sum is less than $23$ .

Solve the following inequation and represent the solution set on a number line
$-8\cfrac{1}{2}< -\cfrac{1}{2}-4x\le 7\cfrac{1}{2},x\in 1$

Represent the following inequalities on real number line:
$-5<x \leq-1$

Enter 1 if it is true else enter 0.
If $x\, \epsilon\, W$ , then the solution set of $x<2$ is $x = \{0,1\}$

Fill in the blanks with correct inequality sign $(>, <, \ge, \le )$ .
$p-q=-3\Rightarrow p.....q$

Solve $2(x – 3) < 1, x ∈ {1, 2, 3, …. 10}$

Solve the equation
$7 \leq \dfrac{3 x+11}{2} \leq 11$

If $x \in I$ , find the smallest value of $x$ which satisfies the inequation $2x+\dfrac{5}{2}>\dfrac{5x}{3}+2$

Solve the inequation $- 3 \le 3 - 2x < 9, x \in R.$ Represent your solution on a number line.

$Solve:$ $4x + 3 < 5x + 7$

Solve the inequality and show the graph of the solution on number line:
$3 (1 -x) < 2 (x + 4)$

Solve & graph the solution set of $3x + 6 \geq 9$ and $-5x > -15$ , $x \in R$ .

Solve & graph the solution set of $\displaystyle 3x+6\geq 9$ and -5x > -15 $\displaystyle x\: \epsilon \: R$

Solve $24 x < 100$ , when $x$ is a natural number.

Solve the inequality and show the graph of the solution on number line:
$3x- 2 < 2x + 1$

Solve the inequality and show the graph of the solution on number line:
$\displaystyle {\frac{x}{2}\geq \frac{(5x-2)}{3}-\frac{(7x-3)}{5}}$

Solve the inequality and show the graph of the solution on number line:
$5x -3 \geq 3x- 5$

Solve the following system of inequalities graphically
$\displaystyle x+2y\leq 8$ ......(1)
$\displaystyle 2x+y\leq 8$ ......(2)
$\displaystyle x\geq 0$ ......(3)
$\displaystyle y\geq 0$ ......(4)

Find x satisfying | x - 5 | $\displaystyle \leq$ 3

$2x -3y > 6$

Solve the following inequalities graphically in two-dimensional plane:
$y < -2$

Solve the following inequalities graphically in two-dimensional plane:
3x + 4y ${ \leq}$ 12

$x + y < 5$

Solve the following inequalities graphically in two-dimensional plane:
$-3x + 2y {\geq} -6$

$x -y {\leq} 2$

The longest side of a triangle is $3$ times the shortest side and the third side is $2$ cm shorter than the longest side. If the perimeter of the triangle is at least $61$ cm, find the minimum length in cm. of the shortest side.

Solve the following inequalities graphically in two-dimensional plane: $2x + y {\geq} 6$

$y + 8 {\geq}2x$

Solve the given inequalities graphically:
$x + y$ ${\leq}$ $6, x + y$ ${\geq}$ $4$

Solve the given inequalities graphically:
$x + y$ ${\geq}$ $4$ and $2x -y > 0$

Solve the following inequations graphically:
$2x- y >1, x -2y < -1$

Solve the system of inequalities graphically:
$2x + y$ ${\geq}$ $4, x + y$ ${\leq}$ $3, 2x- 3y$ ${\leq}$ $6$

Solve the system of inequalities graphically:
${x- 2y \leq 3, 3x + 4y \geq 12, x \geq 0 , y \geq 1}$

Solve the given inequalities graphically:
$2x + y$ ${\geq}$ $8, x + 2y$ ${\geq}$ $10$

Solve the given inequalities graphically:
$2x + y$ ${\geq}$ $6$ and $3x + 4y$ ${\leq}$ $12$

Solve the given inequalities graphically:
$3x + 2y$ ${\leq}$ $12, x$ ${\geq}$ $1, y$ ${\geq}$ $2$

Solve the system of inequalities graphically:
$5x + 4y$ ${\leq}$ $20, x$ ${\geq}$ $1, y$ ${\geq}$ $2$

Solve the given inequalities graphically:
$x + y {\leq} 9, y > x, x {\geq} 0$

Solve, and express the answer graphically.
$\displaystyle\frac{4x+4}{x-4} \le 0$

Solve the inequalities and represent the solution graphically on number line.
$3x-7 >2(x-6), 6-x> 11-2x$

Solve the system of inequalities graphically
${3x + 2y \leq 150, x + 4y \leq 80, x \leq 15, y \geq 0, x \geq 0}$

Solve the given inequalities graphically:
${x + 2y \leq 10, x + y \geq 1, x- y \leq 0, x \geq 0, y \geq 0}$

Solve the following inequality
$\displaystyle \frac{-2x+5}{x+6}>-2$

Solve the inequality and represent the solution graphically on number line.
$2(x-1)< x+5, 3(x+2)> 2-x$

Solve the inequality and represent the solution graphically on number line.
$5(2x-7)-3(2x+3)\le 0, 2x+19\le 6x+47$

A solution is to be kept between $68^{o}F$ and $77^{o}F$ . What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by $F=\dfrac{9}{5}C+32?$

How many litres of water will have to be added to $1125$ litres of the $45\%$ solution of acid so that the resulting mixture will contain more than $25\%$ but less than $30\%$ acid content?

Find the values of $x$ , which satisfy the inequation $-2\dfrac {5}{6} < \dfrac {1}{2} - \dfrac {2x}{3} \leq 2, x \in W$ .

Solve the following inequation and write the solution set :
$13x - 5 < 15x + 4 < 7x + 12, x \in R$
Represent the solution on a real number line.

Solve the following inequation and represent the solution set on the number line:
$4x-19<\dfrac{3x}{5}-2\le-\dfrac{2}{5}+x,\,\,\,\,\,\, x\in R$

Solve the following inequation, write the solution set and represent it on the number line.
$-3 (x - 7)\geq 15 - 7x > \dfrac {x + 1}{3}, x\in R$

Solve the following inequation, write the solution set and represent it on the number line: $-\displaystyle\frac{x}{3}\leq \frac{x}{2}-1\frac{1}{3} < \frac{1}{6}, x\in R$ .

Minimize and maximize $z=5x+10y$
subject to the constraints
$x+2y \le 120$
$x+y\ge 60$
$x-2y\ge 0$ and $x \ge 0,\, y \ge 0$ by graphical method.

Solve by graphical method $x+y \geq 5, x-y \leq 3$

Indicate the polygonal region represented by the system of inequations
$x\ge 0, y≥0, x\le 4, x\ge y$ .

Solve: $5x-7 < 3(x+3), 1-\displaystyle\frac{3x}{2} \geq x-4$ .

Solve the following inequalities.
$\displaystyle\, \frac{2 \, - \, 5x \, }{x \, + \, 1}\, > \, 2$

Solve by graphical method: $y-2x\leq 1, x+y\leq 3, x\geq 0, y\geq 0$ .

Represent the solution of the following inequality on the real number line:
${\dfrac{2x + 5}{3}} > 3x - 3$

Represent $x$ on number line or find $x$
$\dfrac{3x - 2} {5} > \dfrac{4x - 3}{2}$

Solve the following equation:
$- 8 < - (3x - 5) < 13$

Find the largest value of x for which $2(x - 1) \le 9 - x$ and $x \in W$ .

Solve the following and find $x$
$3x + 9 \ge - x + 19$

$x \in$ {real numbers} and $- 1 < 3 - 2x \le 7$ , evaluate x and represent it on a number line.

Maximize $z=x+y$ subject to $x+y \le 10,3y-2x \le 15,x \le 6,x,y \ge 0$ . Find the maximum value.

Solve the inequalities for real $x$ .
$\dfrac {x}{3}>\dfrac {x}{2}+1$