Linear Inequalities - Class 11 Engineering Maths - Extra Questions
Solve the following inequalities graphically in two-dimensional plane:$$x >- 3$$
Yellow line is $$x=-3$$
$$x>-3$$
$$x\in(-3,\infty)$$
$$x>-3$$
$$x\in(-3,\infty)$$
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
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
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 $$x\geq3\;\; and\;\; y\geq2$$
$$\Rightarrow x\in [3,\infty), y\in [2,\infty)$$
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.
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.
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.
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: $$-(x-2)+4 > - 3x+10$$.
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 :
$$3x-2y=1$$$$2x+y=3$$
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 $$x<0, y,0.$$
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
$$-8\cfrac{1}{2}< -\cfrac{1}{2}-4x\le 7\cfrac{1}{2},x\in 1$$
$$-8\dfrac{1}{2} < \dfrac{-1}{2} -4x \leq 7\dfrac{1}{2}$$
$$\dfrac{-17}{2}+\dfrac{1}{2} < -4x \leq \dfrac{15}{2}+\dfrac{1}{2}$$
$$-8 < -4x \leq 8$$
$$8 > 4x \geq -8$$
$$2 > x \geq -2$$
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.
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.
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:
$$Solve:$$ $$4x + 3 < 5x + 7$$
Solve the inequality and show the graph of the solution on number line:
$$3 (1 -x) < 2 (x + 4)$$
Given, $$3(1-x)< 2(x+4)$$
$${\Rightarrow 3-3x < 2x +8}$$
$${\Rightarrow 3-8 <2x +3x}$$
$${\Rightarrow -5 < 5x}$$
$${\Rightarrow 3-3x < 2x +8}$$
$${\Rightarrow 3-8 <2x +3x}$$
$${\Rightarrow -5 < 5x}$$
$$\Rightarrow x>-1 \text{or} \;\; x\in (-1, \infty)$$
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 $$x>-1\;\; \Rightarrow\;\; 2x+8>3-3x $$
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 $$\displaystyle x\in [1,3)$$
$$\displaystyle \Rightarrow 3x\geq 3$$ 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 $$\displaystyle x\in [1,3)$$
$$\displaystyle \Rightarrow 3x\geq 3\Rightarrow -x>-3$$
$$\displaystyle \Rightarrow x\geq 1\: \: \: \: \Rightarrow \: \: x<3$$
Combining the solution
So the solutions is $$\displaystyle x\in [1,3)$$
Solve $$24 x < 100$$, when $$x$$ is a natural number.
Solve the inequality and show the graph of the solution on number line:
$$3x- 2 < 2x + 1$$
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 $$x=3$$
$${\Rightarrow 3x -2x <1 +2}$$
$${\Rightarrow x<3} \;\;\text{or} \;\;x\in (-\infty, 3)$$
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, $$\displaystyle{\frac{x}{2}\geq \frac{(5x-2)}{3}-\frac{(7x-3)}{5}}$$
$$\displaystyle{\Rightarrow \frac{x}{2}\geq \frac{5(5x-2)-3(7x-3)}{15}}$$
$${\Rightarrow\displaystyle \frac{x}{2}\geq \frac{25x-10-21x +9}{15}}$$
$${\Rightarrow \displaystyle\frac{x}{2}\geq \frac{4x-1}{15}}$$
$${\Rightarrow\displaystyle 15 x \geq 2(4x-1)}$$
$${\Rightarrow\displaystyle 15 x \geq 8x-2}$$
$${\Rightarrow \displaystyle 7x \geq -2}$$
$$\displaystyle{\Rightarrow \frac{x}{2}\geq \frac{5(5x-2)-3(7x-3)}{15}}$$
$${\Rightarrow\displaystyle \frac{x}{2}\geq \frac{25x-10-21x +9}{15}}$$
$${\Rightarrow \displaystyle\frac{x}{2}\geq \frac{4x-1}{15}}$$
$${\Rightarrow\displaystyle 15 x \geq 2(4x-1)}$$
$${\Rightarrow\displaystyle 15 x \geq 8x-2}$$
$${\Rightarrow \displaystyle 7x \geq -2}$$
$${\Rightarrow \displaystyle x\geq -\frac{2}{7}}$$
Solve the inequality and show the graph of the solution on number line:
$$5x -3 \geq 3x- 5$$
Given, $${ 5x -3 \geq 3x -5}$$
$${\Rightarrow 5x -3x \geq -5 +3}$$
$${\Rightarrow 2x \geq -2}$$
$${\Rightarrow x\geq -1} \;\; \text{or} \;\; x\in [-1, \infty)$$
$${\Rightarrow 5x -3x \geq -5 +3}$$
$${\Rightarrow 2x \geq -2}$$
$${\Rightarrow x\geq -1} \;\; \text{or} \;\; x\in [-1, \infty)$$
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
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
Find x satisfying | x - 5 | $$\displaystyle \leq $$ 3
as $$\displaystyle |x-a|\leq r\Leftrightarrow a-r\leq x\leq a+r$$ 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. $$\displaystyle x\in [2,8]$$
$$\displaystyle \therefore |x-5|\leq 3\Leftrightarrow 5-3\leq x\leq 5+3$$ i.e. $$\displaystyle 2\leq x\leq 8$$ i.e. $$\displaystyle x\in [2,8]$$
Solve the following inequalities graphically in two-dimensional plane:
$$2x -3y > 6$$
$$2x-3y>6$$
$$3y<2x-6$$
$$y<\dfrac{2x-6}{3}$$
So we have the region below the line.away from origin.
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.
Solve the following inequalities graphically in two-dimensional plane:
$$x + y < 5$$
$$x+y\lt5$$ 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
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.
Solve the following inequalities graphically in two-dimensional plane:
$$x -y {\leq} 2$$
$$x-y\le2$$ 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.
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.
Solve the following inequalities graphically in two-dimensional plane:
$$y + 8 {\geq}2x $$
$$y\geq2x-8$$
So all the points lies on line or lying origin side will be your answer
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.
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 $$x+y=3$$
$$x+y<3$$ 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.
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 $$2x+y=8$$
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
$$x, where\ y = 2$$
$$x$$ $$0$$ $$-1$$ $$4$$
$$y$$ $$2$$ $$2$$ $$2$$
Points to be plotted are $$(0,2), (-1, 2), (4, 2)$$
Also, $$x \ge 1$$
So, for all values of $$y, where \ x = 1$$
$$x$$ $$1$$ $$1$$ $$1$$
$$y$$ $$0$$ $$-1$$ $$3$$
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.
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
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:
$$\displaystyle\frac{4x+4}{x-4} = 0$$
$$4x + 4 = 0$$
$$x = -1$$
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:
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 $$x>5$$
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.
Solve the following inequality$$\displaystyle \frac{-2x+5}{x+6}>-2$$
Solve the inequality and represent the solution graphically on number line.
$$2(x-1)< x+5, 3(x+2)> 2-x$$
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 $$x\in(-1,7)$$
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
$$5(2x-7)-3(2x+3)\le 0$$
$$\Rightarrow 10x-35-6x-9\le 0$$
$$\Rightarrow 4x-44\le 0\Rightarrow x\le 11$$
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 $$F=\dfrac{9}{5}C+32?$$
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 $$1125$$ 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.
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 : $$\left\{x : \dfrac{-9}{2} < x < 1, x \in R\right\}$$
$$\Rightarrow$$ $$15x+4 > 13x-5$$ 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:
$$4x-19<\dfrac{3x}{5}-2\le-\dfrac{2}{5}+x,\,\,\,\,\,\, x\in R$$
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 $$45 - 21x > x + 1$$
$$-3x + 21\geq 15 - 7x > \dfrac {x + 1}{3}$$
$$\Rightarrow -3x + 21 \geq 15 - 7x$$ 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 $$= \left \{x : x \in R, \dfrac {-3}{2} \leq x < 2\right \}$$
$$\therefore$$ 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 $$\{x: 1.6\leq x\leq 3, x\epsilon R\}$$
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 $$x \ge 0,\, y \ge 0$$
$$x+y = 60$$
$$x-2y = 0$$
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.
$$x+2y = 120$$
| $$x$$ | $$0$$ | $$60$$ | $$120$$ |
| $$y$$ | $$60$$ | $$30$$ | $$0$$ |
| $$x$$ | $$0$$ | $$40$$ | $$60$$ |
| $$y$$ | $$60$$ | $$20$$ | $$0$$ |
| $$x$$ | $$0$$ | $$40$$ | $$60$$ |
| $$y$$ | $$0$$ | $$20$$ | $$30$$ |
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
At $$B(120,0),$$ $$Z=5\times 120+10\times 0=600$$
At $$D(60,0),$$ $$Z=5\times 60+10\times 0=300$$
At $$E(40,20),$$ $$Z=5\times 40+10\times 20=400$$
At $$F(60,30),$$ $$Z=5\times 60 + 10\times 30=600$$
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 $$x+y \geq 5, x-y \leq 3$$
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: $$y-2x\leq 1, x+y\leq 3, x\geq 0, y\geq 0$$.
$$x+y\leq 3$$ 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.
Represent the solution of the following inequality on the real number line:
$${\dfrac{2x + 5}{3}} > 3x - 3$$
Given condition:
$$\cfrac { 2x+5 }{ 3 } >3x-3$$
$$ 2x+5>9x-9$$
$$ 7x<14$$
$$ x<2$$
Represent $$x$$ on number line or find $$x$$$$\dfrac{3x - 2} {5} > \dfrac{4x - 3}{2}$$
$$\cfrac { 3x-2 }{ 5 } >\cfrac { 4x-3 }{ 2 } $$
$$6x-4>20x-15$$
$$ 11>14x$$
$$ x<\cfrac { 11 }{ 14 } $$
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.
$$(i)3-2x \le 7$$
$$ \quad -2x\le 4$$
$$ \quad x\ge -2$$
$$ (ii)-1<3-2x$$
$$ \quad 2x<4$$
$$ \quad x<2$$
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 $$B(6, 4), C(3, 7)$$
Solve the inequalities for real $$x$$.
$$\dfrac {x}{3}>\dfrac {x}{2}+1$$
Find $$x$$$$\dfrac{5x}{2} + \dfrac{3x}{4} \ge \dfrac{39} {4}$$
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.
Minimize $$z = 6x + 2y,$$ Subject to $$x + 2y \ge 3, x + 4y \ge 4 , 3x + y \ge 3 , x \ge 0, y \ge 0$$.
Solve the inequalities for real $$x$$.
$$\dfrac { 3\left( x-2 \right) }{ 5 } \le \dfrac { 5\left( 2-x \right) }{ 3 } $$
Find $$x$$$$ - \left( {x - 3} \right) + 4 < 5 - 2x$$
Find $$x$$$$\dfrac{3} {x - 2} < 1$$
Solve the inequalities for real $$x$$.
$$2(2x+3)-10 < 6(x-2)$$
Solve the inequalities for real $$x$$.
$$37-(3x+5) \ge 9x-8(x-3)$$
Solve the inequalities for real $$x$$.
$$\dfrac { x }{ 4 } <\dfrac { \left( 5x-2 \right) }{ 3 } -\dfrac { \left( 7x-3 \right) }{ 5 } $$
Solve the system of in equations
$$ \dfrac { x }{ 2x+1 } \ge \dfrac { 1 }{ 4 } ;\dfrac { 6x }{ 4x-1 } <\dfrac { 1 }{ 2 }$$
Solve:$$10 \le - 5(x - 2) < 20$$
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 $$(-\infty,-9)\cup (5,8)$$
Solve the inequalities for real $$x$$ -
$$\dfrac { 3\left( x-2 \right) }{ 5 } \le \dfrac { 5\left( 2-x \right) }{ 3 } $$
Solve the inequalities for real $$x$$.
$$\dfrac { \left( 2x-1 \right) }{ 3 } \ge \dfrac { \left( 3x-1 \right) }{ 4 } -\dfrac { \left( 2-x \right) }{ 5 } $$
Solve $$\dfrac{\left| x-3 \right|}{x^{2}-5x+6}\ge 2$$
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 $$y\le 4$$
Solve $$\frac{1}{2}\left( {\frac{{3x}}{5} + 4} \right) \ge \frac{1}{3}\left( {x - 6} \right)$$.
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 $$x-y=3$$
Solve:$$2\left(2x+3\right)-10<6\left(x-2\right)$$
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 $$x+2 y=1$$
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 $$x+y=4$$
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.
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.
Solve: $$3x+4y\ge 12,\:x\ge 0,\:y\ge 1$$ and $$4x+7y\le 28$$
Intersection of $$3x+4y=12$$ and $$4x+7y=28$$,
$$3x+4y=12\times 4\Rightarrow 12x+16y=48$$
$$4x+7y=28\times 3\Rightarrow 12x+21y=84$$
$$\Rightarrow 5y=36\Rightarrow y=\dfrac{36}{5}$$
$$x=\dfrac{12}{3}-\dfrac{4}{3}\times \dfrac{36}{5}$$
$$\Rightarrow 4-\dfrac{48}{5}=-\dfrac{28}{5}$$.
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.
Draw the graph of the solution set of the following inequations: $$ 3x+2y> 6 $$
Draw the graph of the solution set of the following inequations:
$$ 3x+2y> 6 $$
Given inequality,
$$3x+2y>6$$
at $$x=0,y=3$$ and $$y=0,x=2$$
At $$(0,0)$$
$$\implies 0>6$$ (False)
So solution region lies away from the origin.
Graph the solution sets of the following inequations :
$$2x - 3y < 4$$.
$$ Converting\text{ the inequality to equation} $$
$$ 2x=3y+4 $$
$$ at\,y=0 $$
$$ x=2 $$
$$ at\,y=2 $$
$$ x=5 $$
$$ at\,y=-1 $$
$$ x=1/2 $$
$$ plotting\,all\,the\,po\operatorname{int}son\,graph $$
$$ 2x=3y+4 $$
$$ at\,y=0 $$
$$ x=2 $$
$$ at\,y=2 $$
$$ x=5 $$
$$ at\,y=-1 $$
$$ x=1/2 $$
$$ plotting\,all\,the\,po\operatorname{int}son\,graph $$
Draw the graph of the solution set of the following inequations: $$ x-y\leq 3 $$
Draw the graph of the solution set of the following inequations:
$$ x-y\leq 3 $$
Given inequality,
$$x-y\le 3$$
at $$x=0,y=-3$$ and $$y=0,x=3$$
At $$(0,0)$$
$$0\le 3$$(true)
So solution region lies towards the origin.
Draw the graph of the solution set of the following inequations: $$ x\geq y-2 $$
Draw the graph of the solution set of the following inequations:
$$ x\geq y-2 $$
Given inequality,
$$x\ge y-2$$
At $$x=0,y=2$$ and $$y=0,x=-2$$
At $$(0,0)$$
$$0\ge -2$$ (true)
SO solution region lies towards the origin.
Draw the graph of the solution set of the following inequations: $$ 3x+5y< 15 $$
Given inequality,
$$3x+5y<15$$
at $$x=0,y=3$$ and $$y=0,x=5$$
At $$(0,0)$$
$$\implies 0<15$$ (true)
So solution region lies towards from the origin.
Graph the solution sets of the following inequations :
$$x \geq y - 2$$.
$$ Converting\text{ the inequality to equation} $$
$$ x=y-2 $$
$$ at\,y=0 $$
$$ x=-2 $$
$$ at\,y=2 $$
$$ x=0 $$
$$ at\,y=-1 $$
$$ x=-3 $$
$$ plotting\,all\,the\,po\operatorname{int}son\,graph $$
$$ x=y-2 $$
$$ at\,y=0 $$
$$ x=-2 $$
$$ at\,y=2 $$
$$ x=0 $$
$$ at\,y=-1 $$
$$ x=-3 $$
$$ plotting\,all\,the\,po\operatorname{int}son\,graph $$
Draw the graph of the solution set of the following inequations: $$ x+y\geq 4 $$
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.
Draw the graph of the solution set of the following inequations: $$ y-2\leq 3x$$
$$ y-2\leq 3x$$
Given inequality,
$$y-2\le 3x$$
At $$x=0,y=2$$ and $$ y=0,x=-\dfrac{2}{3}$$
At $$(0,0)$$
$$-2\le 0$$(true)
So solution region lies towards the origin.
Solve the following system of inequations graphically:$$ x\geq 2,y\geq 3$$
Graph the solution sets of the following inequations :
$$y - 2 \leq 3x$$.
$$ Converting\text{ the inequality to equation} $$
$$ y=3x+2 $$
$$ at\,x=0 $$
$$ y=2 $$
$$ at\,x=1 $$
$$ y=5 $$
$$ at\,x=-1 $$
$$ y=-1 $$
$$ plotting\,all\,the\,po\operatorname{int}son\,graph $$
$$ y=3x+2 $$
$$ at\,x=0 $$
$$ y=2 $$
$$ at\,x=1 $$
$$ y=5 $$
$$ at\,x=-1 $$
$$ y=-1 $$
$$ plotting\,all\,the\,po\operatorname{int}son\,graph $$
Find the solution set of the inequation $$\dfrac{x+1}{x+2}>1$$
Show that each of the following systems of linear inequations has no solution: $$ 3x+2y\geq 24,3x+y\leq 15,x\geq 4 $$
Solve the following system of inequations graphically: $$ 2x+y\geq 4,x+y\leq 3,2x-3y\leq 6 $$
Solve the following system of inequations graphically: $$ 3x+4y\leq 60,x+3y\leq 30,x\geq 0,y\geq 0$$
Find the solution set of the inequation $$\dfrac{[x-2]}{(x-2)}>0,x\neq 2$$
Solve the following system of inequations graphically: $$ 5x+4y\leq 20,x\geq 1,y\geq 2 $$
Show that each of the following systems of linear inequations has no solution: $$ 2x-y\leq -2,x-2y\geq 0,x\geq 0,y\geq 0 $$
Solve the following system of inequations graphically: $$ x+y\leq 6,x+y\geq 4 $$
Show that the solution set of the following system of inequations is an unbounded set: $$ 3x+y\geq 12,x+y\geq 9,x\geq 0,y\geq 0 $$
Find the solution set of the inequation $$\dfrac{1}{x-2}<0$$
Fill in the blanks with correct inequality sign $$(>, <, \ge, \le )$$.
$$-6x\le -18\Rightarrow x.....3$$
Solve $$|x|<4$$, where $$x\in R$$
Solve $$|x|>4$$, where $$x\in R$$
Solve $$\dfrac{x}{x-5}>\dfrac{1}{2}$$, when $$x\in R$$
Fill in the blanks with correct inequality sign $$(>, <, \ge, \le )$$.
$$-3x>9\Rightarrow x.....-3$$
Fill in the blanks with correct inequality sign $$(>, <, \ge, \le )$$.
$$4x>-16\Rightarrow x.....-4$$
Solve the system of inequation $$x-2\ge 0,2x-5\le 3$$
Solve $$-4x>16$$, when $$x\in R$$
Solve $$x+5>4x-10$$, when $$x\in R$$
Fill in the blanks with correct inequality sign $$(>, <, \ge, \le )$$.
$$5x<20\Rightarrow x.....4$$
Fill in the blanks with correct inequality sign $$(>, <, \ge, \le )$$.
$$a<b$$ and $$c<0\Rightarrow \dfrac{a}{c}.....\dfrac{b}{c}$$
Fill in the blanks with correct inequality sign $$(>, <, \ge, \le )$$.
$$u-v=2\Rightarrow u.....v$$
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 $$(-\infty, -2.5)$$.
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 $$=(5,\infty)$$
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 $$=(-\infty, 2]$$.
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 $$=(-\infty, -8)$$.
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 $$=(-\infty, 5.5]$$.
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 $$=(-\infty, \dfrac{5}{2}\left.\right]$$
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 $$(-2, \infty)$$.
Fill in the blanks with correct inequality sign $$(>, <, \ge, \le )$$.
$$x>-3\Rightarrow -2x.....6$$
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 $$=(-\infty, 2]$$.
Solve:
$$\dfrac{x-7}{x-2}\ge 0, x\in R$$
Solve:
$$\dfrac{1}{x-1}\le 2, x\in R$$
Solve:
$$\dfrac{x-3}{x+4}>0, x\in R$$
Solve:
$$\dfrac{x-3}{x+1}<0, x\in R$$
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 $$=(0, \infty)$$.
Solve:
$$\dfrac{3}{x-2}> 2, x\in R$$
Solve:
$$\dfrac{5x+8}{4-x}< 2, x\in R$$
Solve:
$$\dfrac{2x-3}{3x-7}>0, x\in R$$
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 $$=(4, \infty)$$.
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.
Solve:
$$|5-2x| \le 3, x\in R$$.
Solve:
$$|4x-5| \le \dfrac{1}{3}, x\in R$$.
List the solution set of $$30 4 (2x 1) < 30,$$ given that $$ x$$ is a positive integer.
Solve: $$ \dfrac{(2x – 3)}4 ≥ \dfrac12 , x \in \{0, 1, 2,…,8\}$$
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.
Solve:
$$|3x-7| > 4, x\in R$$.
If $$x$$ is a negative integer, find the solution set of $$\dfrac23 + \dfrac13 (x + 1) > 0$$.
Solve: $$2 (x – 2) < 3x – 2, x \in \{– 3, – 2, – 1, 0, 1, 2, 3\}.$$
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 $$-5\le x$$
$$\Rightarrow -5\le x<-1$$
As $$x \in Z$$
The solution set is $$\{-5,-4,-3,-2\}$$.
Representation on number line is as shown in the image.
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
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
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.
Solve $$ x 3 (2 + x) > 2 (3x 1), x { 3, 2, 1, 0, 1, 2, 3}. $$ Also represent its solution on the number line.
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.
Given $$x \in \{1, 2, 3, 4, 5, 6, 7, 9\}$$ solve $$x - 3 < 2x - 1$$.
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.
Given $$x \in \{1, 2, 3, 4, 5, 6, 7, \},$$ find the values of $$x$$ for which $$-3 < 2x – 1 < x + 4.$$
List the solution set of the inequation
$$\dfrac12 + 8x > 5x -\dfrac32, x \in Z$$
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:
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:
$$A = \{x \ ;\ 11x - 5 > 7x + 3, x \in R\}$$ 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:
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) $$Q – P$$
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:
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
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:
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:
Solve 3x/5 (2x 1)/3 > 1, x R and represent the solution set on the 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.
$$2x-5>11$$ is an equation
Solve the inequation:
$$\dfrac{5x+1}{7}-4\left(\dfrac{x}{7}+\dfrac{2}{5}\right)\leq 1\dfrac{3}{5}+\dfrac{3x-1}{7},xR$$
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:
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:
Solve the inequation: 6x 5 < 3x + 4, x I
Given 20 – 5 x < 5 (x + 8), find the smallest value of x, when
$$(i) x ∈ I$$
$$(ii) x ∈ W$$
$$(iii) x ∈ N.$$
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.
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.
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:
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, $$P \cap Q = \{4\}$$
Solve the inequality $$2\le 3x-4\le 5$$
Solve the inequality $$-15 < \dfrac {3(x-2)}{5}\le 0$$
Solve the inequalities
$$2(x-1)< x+5, 3(x+2) > 2-x$$
Solve the inequality $$-12 < 4-\dfrac {3x}{-5}\le 2$$
Solve the inequalities
$$3x-7 > 2(x-6), 6-x > 11-2x$$
Solve the inequality $$7\le \dfrac {(3x+11)}{2}\le 11$$
Solve the inequality $$6\le -3(2x-4) < 12$$
Solve the inequality $$-3\le 4-\dfrac {7x}{2} \le 18$$
Solve the inequalities
$$5(2x-7)-3(2x+3)\le 0, 2x+19\le 6x+47$$
Solve the inequality
$$5x+1 > -24, 5x-1 < 24$$
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.
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.
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.
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.
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.
$$x \ge 1$$ ......(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.
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.
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.
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.
$$ y \ge 2x$$ .....(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.
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.
$$3x+4y \ge 12$$ .......(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.
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.
$$2x-y > 0$$ .....(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.
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.
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.
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.
$$x-2y < -1$$ .....(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.
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.
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.
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.
$$ x+y \ge 1$$ .....(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.
Solve $$24x < 100$$, when
$$x$$ is an integer.
Solve $$5x-3 < 7$$, when
$$x$$ is a real number
Solve $$-12x > 30$$, when
$$x$$ is an integer
Solve $$-12x > 30$$, when
$$x$$ is a natural number
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.
$$ x+4y =80$$ .....(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.
Solve $$5x-3 < 7$$, when
$$x$$ is an integer
Solve $$3x+8 > 2$$, when
$$x$$ is an integer
Solve $$24x < 100$$, when
$$x$$ is a natural number
Solve $$3x+8 > 2$$, when
$$x$$ is a real number
Solve the given inequality for real $$x:3x-7> 5x-1$$
Solve the given inequality for real $$x:3(2-x)\ge (1-x)$$
Solve the given inequality for real $$x:\dfrac{1}{2}\left(\dfrac{3x}{5}+4\right)\ge \dfrac{1}{3}(x-6)$$
Solve the given inequality for real $$x:37-(3x+5)\ge 9x-8(x-3)$$
Solve the given inequality for real $$x:\dfrac{3(x-2)}{5}\le \dfrac{5(2-x)}{3}$$
Solve the given inequality $$x:2(2x+3)-10 < 6(x-2)$$
Solve the given inequality for real $$x:\dfrac{x}{3} > \dfrac{x}{2}+1$$
Solve the given inequality for real $$x:3(x-1) \le 2(x-3)$$
Solve the given inequality for real $$x:4x+3 < 5x+7$$
Solve the given inequality for real $$x:x+\dfrac{x}{2}+\dfrac{x}{3} < 11$$
Solve the given inequality for real $$x:\dfrac{(2x-1)}{3} \ge \dfrac{(3x-2)}{4}-\dfrac{(2-x)}{5}$$
Solve the given inequality and show the graph of the solution on number line:
$$5x-3\ge 3x-5$$
$$\Rightarrow 5x-3x\ge -5+3$$
$$\Rightarrow 2x\ge -2$$
$$\Rightarrow \dfrac{2x}{2}\ge \dfrac{-2}{2}$$
$$\Rightarrow x\ge -1$$
The graphical representation of the solution of the given inequality is as follows:
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.
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 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.
$$\Rightarrow \dfrac{x}{2}\ge \dfrac{5(5x-2)-3(7x-3)}{15}$$
$$\Rightarrow \dfrac{x}{2}\ge \dfrac{25x-10-21x+9}{15}$$
$$\Rightarrow \dfrac{x}{2}\ge \dfrac{4x-1}{15}$$
$$\Rightarrow 15x\ge 2(4x-1)$$
$$\Rightarrow 15\ge 8x-2$$
$$\Rightarrow 15-8x\ge 8x-2-8x$$
$$\Rightarrow 7x\ge -2$$
$$\Rightarrow x\ge -\dfrac{2}{7}$$
The graphical representation of the solution of the given inequality is as follows.
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:
$$\Rightarrow 3-3x < 2x+8$$
$$\Rightarrow 3-8 < 2x+3x$$
$$\Rightarrow -5x < 5x$$
$$\Rightarrow \dfrac{-5}{5} < \dfrac{5x}{5}$$
$$\Rightarrow -1 < x$$
The graphical representation of the solutions of the given inequality is as follows:
Find all pairs of consecutive odd positive integers both which are smaller than $$10$$ such that their sum is more than $$11$$.
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.
Solve the given inequality for real $$x:\dfrac{x}{4} < \dfrac{(5x-2)}{3}-\dfrac{(7x-3)}{5}$$
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:
$$\Rightarrow 3x-2x < 1+2$$
$$\Rightarrow x < 3$$
The graphical representation of the solutions of the given inequality is as follows:
Solve $$-7< 4x + 1 \leq 23, x \, \epsilon\, l$$
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.
Solve $$3 - 4x < x - 12, x\, \epsilon\, {-1, 0, 1, 2, 3, 4, 5, 6, 7}$$
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) $$ -1 < \dfrac{2x}{3} + 1 \leq 4, x\, \epsilon\, l $$
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.
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?
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) $$-3 \leq x \leq 2, x \epsilon l$$
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.
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,
$$ -2 \times 2 < \dfrac{x}{2} \times 2 \leq - 2 \times 2 $$
$$ -4 < x \leq - 4 $$
As $$ x \, \varepsilon\, I $$ ,
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.
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.
Solve the following inequation :
$$ \dfrac{2}{3}p + 5 < 9 , p\, \varepsilon \, W $$
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,
$$ \dfrac{-4}{4} \leq \dfrac{4x}{4} < \dfrac{14}{4} $$
$$ -1 \leq x \leq \dfrac{7}{2} $$
As $$ x \, \varepsilon \, N $$,
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.
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 ,
$$ x < \dfrac{3}{2} $$
As $$ x \, \varepsilon\, W $$,
Hence solution set $$ = \left \{ 0 , 1 \right \} $$
The graphical representation for this solution set is shown indicating with thick dots on the number line.
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.
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.
If $$ 25-4 x \leq 16, $$ find the smallest value of $$ \mathbf{x}, $$ when $$ \mathbf{x} $$ is a real number.
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.
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.
Represent the following inequality on real number line:
$$ -4<x<4 $$
$$ -4<x<4 $$
Solution on the number line is shown in figure.
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.
Solve the inequation:
$$ 3-2 \mathbf{x} \geq \mathbf{x}-12 $$ given that $$ \mathbf{x} \in \mathbf{N} $$
For given graph alongside, write an inequation taking $$ \mathbf{x} $$ as the variable:
If $$ 25-4 x \leq 16, $$ find the smallest value of $$ x $$, when $$ x $$ is an integer.
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.
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.
For the given graph alongside, write an inequation taking $$ \mathbf{x} $$ as the variable.
For the given graph alongside, write an inequation taking $$ \mathbf{x} $$ as the variable.
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.
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.
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.
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.
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.
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.
For the given graph alongside, write an inequation taking $$ \mathbf{x} $$ as the variable.
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.
Solve and graph the solution set of:
$$ x+5 \geq 4(x-1) $$ and $$ 3-2 x<-7, x \in R $$
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.
$$\mathrm{x} \in\{\text { real numbers }\} $$ 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.
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.
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.
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.
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.
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.
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.
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.
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,y0y=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.
To find the solution set, we have to check any point other than origin.
Let us check the point (1 ,1)
When y=1,y0y=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.
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.
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,x0x=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.
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,x0x=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.
Solve: $$3+x < 2, x\in N$$
Solve the inequation $$8-2x > x-5; x\in N$$
Solve the inequality $$18-3(2x-5) > 12; x\in W$$
Solve: $$\dfrac{2x+1}{3}+15 < 17; x\in W$$
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.
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.
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 $$2x - 5y \geq 10.$$
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 $$2x - 5y \geq 10.$$
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.
Since x≥0,y≥0.x≥0,y≥0. the solution set is in the first quadrant which is shaded in the graph.
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.
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.
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.
Since x≤0,y≤0,x≤0,y≤0, the solution set is in the third quadrant which is shaded in the graph.
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 $$x + 2y \leq 6.$$
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 $$x + 2y \leq 6.$$
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
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
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.
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.
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 $$5x - 3y \leq 0.$$
$$\therefore $$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 $$5x - 3y \leq 0.$$
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 $$x + 2y > 0.$$
$$\therefore $$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 $$x + 2y > 0.$$
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.
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.
The feasible solution is CEPQRC which is shaded in the graph.
| 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 |
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.
The solution set of the given system of inequalities is shaded in the graph.
| 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.
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.
The feasible solution is $$\Delta ABC$$ which is shaded in the graph.
| 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 |
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.
The solution set of the given system of inequalities is shaded in the graph.
| 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 |
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.
The feasible solution is shaded in the graph.
| 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 |
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.
The feasible solution is PQDB which is shaded in the graph.
| 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.
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.
The Solution set of the given system of inequalities shaded in the graph.
| 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 |
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.
The feasible solution is OCPQBO which is shaded in the graph.
| 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 |
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.
The solution set of the given system of inequalities is shaded in the graph.
| 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.
$$5y -12 \geq 0$$
Consider the line
whose equation is $$5y -12 \geq 0$$ i.e. $$y = \dfrac{12}{5}$$
This represents a line parallel to X-axis passing 3 through the
point (0,$$ \dfrac{12}{5})$$
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.
x is a natural number.
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.
x is an integer.
$$4x - 18 \geq 0$$
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.
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.
$$y \leq -3.5$$
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.
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.
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:
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.
The feasible solution id OCPBO which is shaded in the graph.
| Article A (x) | Article B (y) | Availability | |
| Gold | 1 | 2 | 4 |
| Silver | 3 | 2 | 6 |
$$x + 2y \leq 4$$ 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 |
$$- 11x - 55 \leq 0$$
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.
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.
$$x$$ is a real number
$$x$$ $$\epsilon$$ $$N$$
$$x$$ is an integer
x is an integer
$$x$$ is a real number.
Solve: $$4x + 3 < 6x + 7$$
$$x$$ $$\epsilon$$ $$Z$$
Solve: $$3x - 7 > 5x - 1$$
x is a real number.
Solve: $$37 - (3x + 5) \geq 9x - 8(x - 3)$$
Solve: $$3(2 - x) > 2(1 - x)$$
Solve : $$x+\cfrac{x}{2}+ \cfrac{x}{3}<11$$
Solve:
$$\cfrac{2 x-1}{3} \geq \cfrac{3 x-2}{4}- \cfrac{2-x}{5}$$
Solve:
$$\cfrac{x}{3}>\cfrac{x}{2} + 1$$
$$3x - 2 < 2x + 1$$
Solve: $$\frac{3(x - 2)}{5} \leq \frac{5(2 - x)}{3}$$
Solve: $$3(x - 1) < 2 (x - 3)$$
Solve: $$\cfrac{5 - 2 x}{3} \leq \cfrac{x}{6} - 5$$
Solve:
$$\cfrac{x}{4} <\cfrac{5 x - 2}{3} - \cfrac{7 x-3}{5}$$
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.
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.
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.
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.
$$5x - 3 \geq 3x - 5$$
$$5x - 3 \geq 3x - 5$$
$$\Rightarrow 2x \geq - 2$$
$$\Rightarrow x \geq -1$$
$$\therefore$$ Solution set [-1, $$\infty$$)
$$\Rightarrow 2x \geq - 2$$
$$\Rightarrow x \geq -1$$
$$\therefore$$ Solution set [-1, $$\infty$$)
$$7x + 3 < 5x + 9$$
$$7x + 3 < 5x + 9$$
$$\Rightarrow 2x < 6$$
$$\Rightarrow x < 3$$
$$\therefore$$ Solution set $$= (- \infty, 3)$$
$$\Rightarrow 2x < 6$$
$$\Rightarrow x < 3$$
$$\therefore$$ Solution set $$= (- \infty, 3)$$
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.
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.
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.
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.
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.
$$\cfrac{3 x-4}{2} \geq \cfrac{x+1}{4} - 1$$
$$\cfrac{3 x-4}{2} \geq \cfrac{x+1}{4} - 1$$
$$\Rightarrow 2(3x - 4) \geq x + 1 - 4$$
$$\Rightarrow 6x - x \geq -3 + 8$$
$$\Rightarrow x \geq 1$$
$$\therefore$$ Solution set = [1, $$\infty$$)
$$\Rightarrow 2(3x - 4) \geq x + 1 - 4$$
$$\Rightarrow 6x - x \geq -3 + 8$$
$$\Rightarrow x \geq 1$$
$$\therefore$$ Solution set = [1, $$\infty$$)
$$3(1 - x) < 2(x + 4)$$
$$3 - 3x < 2x + 8$$
$$\Rightarrow -5 < 5x$$
$$\Rightarrow x > - \frac{5}{8} = -1$$
$$\therefore$$ Solution set $$= (-1, \infty)$$
$$\Rightarrow -5 < 5x$$
$$\Rightarrow x > - \frac{5}{8} = -1$$
$$\therefore$$ Solution set $$= (-1, \infty)$$
$$\cfrac{x}{2} \geq \cfrac{5 x-2}{3} - \cfrac{7 x-3}{5}$$
$$\cfrac{x}{2} \geq \cfrac{5 x-2}{3} - \cfrac{7 x-3}{5}$$
$$\Rightarrow \frac{25x - 10 - 21x + 9}{15}$$
$$\Rightarrow 15x \geq 2(4x - 1) = 8x - 2$$
$$\Rightarrow 7x \geq - 2$$
$$\Rightarrow x \geq - \frac{2}{7}$$
$$\therefore$$ Solution set = [ - $$\frac{2}{7}, \infty$$ ).
$$\Rightarrow \frac{25x - 10 - 21x + 9}{15}$$
$$\Rightarrow 15x \geq 2(4x - 1) = 8x - 2$$
$$\Rightarrow 7x \geq - 2$$
$$\Rightarrow x \geq - \frac{2}{7}$$
$$\therefore$$ Solution set = [ - $$\frac{2}{7}, \infty$$ ).
Solve the following system of inequalities graphically.
$$2x + y \geq 8, x + 2y \geq 10$$
2x + y > 6
3x + 4y < 12
Solve the following system of inequalities graphically.
$$x + y \geq 4, 2x - y > 0$$
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.
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.
Solve the following system of inequalities graphically.
$$x + y \leq 6, x + y \geq 4$$
Solve the following system of inequalities graphically.
$$2x - y > 1, x - 2y < -1$$
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.
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.
Solve the following system of inequalities graphically.
$$8x + 3y \leq 100, x \geq 0, y \geq 0$$
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.
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.
Solve the following system of inequalities graphically.
$$5x + 4y \leq 20, x \geq 1, y \geq 2$$
Solve the following system of inequalities graphically.
$$x + 2y \leq 8, 2x + y \leq 8, x \geq 0, y \geq 0$$
Solve the following system of inequalities graphically.
$$3x + 4y \leq 60, x + 3y \leq 30, x \geq 0, y \geq 0$$
Solve the following system of inequalities graphically.
$$3x + 2y \leq 12, x \geq 1, y \geq 2$$
Solve the following system of inequalities graphically.
$$5x + 4y \leq 40, x \geq 0, y \geq 0$$
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.
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.
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.
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.
Solve the following system of inequalities graphically.
$$x + y \leq 9, y > x, x \geq 0$$
y + 8 > 2x
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.
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.
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.
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.
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.
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.
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.
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.
Solve the equations.
$$-5 \leq \frac{5 - 3x} {2} \leq 8$$
-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.
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.
Solve the equations.
$$6 < -3 \left(2JC - 4 \right) < 12$$
Solve the equations.
$$-15 \leq \frac{3 \left(x - 2 \right)} {5} \leq 0$$
Solve the equations.
$$-3 \leq 4 - \frac{7x} {2} \leq 18$$
Solve the equations.
$$2 \leq 3x - 4 \leq 5$$
Solve the equations.
$$-8 \leq 5x - 3 < 7$$
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?
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 $$ -7 \leq x \leq 11 $$.
$$ \therefore x \in[-7,11] $$
$$ \Rightarrow 10 x-35-6 x-9 \leq 0 $$
$$ \Rightarrow 4 x-44 \leq 0 $$
$$ \Rightarrow x \leq \dfrac{44}{4} $$
$$ \therefore x \leq 11 \ldots(1)$$
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 $$ -7 \leq x \leq 11 $$.
$$ \therefore x \in[-7,11] $$
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., $$ x \in(5, \infty) $$
$$ \Rightarrow 3 x-2 x>-12+7 $$
$$ \Rightarrow \quad x>-5 \ldots(1)$$
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., $$ x \in(5, \infty) $$
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 $$ -5<x<5 $$
$$ \therefore x \in(-5,5) $$
$$ \Rightarrow 5 x>-25 $$
$$ \therefore x>-5 \ldots(1)$$
Consider $$ 5 x-1<24 $$
$$ \Rightarrow 5 x<25 $$
$$ \therefore x<5 \ldots(2)$$
From ( 1 ) and (2) we get $$ -5<x<5 $$
$$ \therefore x \in(-5,5) $$
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 $$ 18 \%? $$
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 $$ -1<x<7 $$
$$ \therefore x \in(-1,7) $$
$$ \Rightarrow 2 x-x>2-6 $$
$$ x<7 \ldots(1)$$
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 $$ -1<x<7 $$
$$ \therefore x \in(-1,7) $$
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 $$ 2 \leq x<6 $$
$$ \therefore x \in[2,6) $$
$$ \Rightarrow 3 x-x<5+7 $$
$$ \Rightarrow x<\dfrac{12}{2}=6 $$
$$ x<6 \ldots(1)$$
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 $$ 2 \leq x<6 $$
$$ \therefore x \in[2,6) $$
Solve the equations.
$$ -12<4-\dfrac{3 x}{-5} \leq 2 $$
Solve the equations.
$$-12 < 4 - \frac{3x} {-5} \leq 2$$
Solve the equations.
$$7 \leq \frac{3x + 11} {2} \leq 11$$
Find the integer values of $$t$$ which satisfy the inequalities.
$$4t + 7 < 39 \nless 7t + 2 $$
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.
$$|x - y| \geq 1$$
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.
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.
$$3x - 2y = x + y - 8$$
$$\Rightarrow 3x - x - 2y - y = -8$$
$$\Rightarrow 2x - 3y = -8$$
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.
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 $$Z = 400.$$
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.
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.
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 $$Z = x + 2y$$
$$2x - y \le 0$$
$$2x + y \le 200$$
Subject to,
$$x + 2y \ge 100$$
$$2x - y \le 0$$
$$2x + y \le 200$$
$$x, y \ge 0$$
$$x + 2y \ge 100$$
| $$x$$ | $$0$$ | $$100$$ |
| $$ y$$ | $$50$$ | $$0$$ |
| $$x$$ | $$0$$ | $$50$$ |
| $$y$$ | $$0$$ | $$100$$ |
| $$y$$ | $$100$$ | $$0$$ |
| $$x$$ | $$0$$ | $$200$$ |
$$\therefore Z = 400$$ 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 $$(0,50)$$ & $$(20, 40)$$
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 $$20 init of ordinary model is produced.
How many liters of water will have to be added to 1125 liters of the $$ 45 \% $$ solution of acid so that the resulting mixture will contain more than $$ 25 \% $$ but less than $$ 30 \% $$ acid content?
Solve the following inequalities, graphically :
$$|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$$ will be required solution set.
inequality $$|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$$ will be required solution set.
If $$(xy)^{a-1}=z$$
$$(yz)^{b-1}=x$$
$$(zx)^{c-1}=y$$
then $$ab+bc+ca=?$$
Solve the following system of inequations graphically: $$ 2x+y> 1,x-2y< -1 $$
Solve the following system of inequations graphically: $$ x+y\leq 9,y> x,x\geq 0 $$
Solve the following systems of inequations graphically: $$ 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) $$ |
Solve the following system of inequations graphically: $$ 3x+2y\leq 12,x\geq 1,y\geq 2 $$
Solve the following system of inequations graphically: $$ 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) $$ |
Solve the following system of inequations graphically: $$ 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) $$ |
Thus, hatched region is the solution of given inequalities.
Solve the following systems of inequations graphically: $$ 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) $$ |
Solve the following system of inequations graphically: $$ 2x+y\geq 6,3x+4y\leq 12 $$
Solve the following systems of inequations graphically: $$ x+2y\leq 2000,x+y\leq 1500,y\leq 600,x\geq 0,y\geq 0 $$
Find the linear inequations for which the shaded area is the solution set in the figure given below.
Find the linear inequations for which the shaded area in the figure given below, is the solution set.
A firm manufactures headache pills in two sizes A and B. Size A contains $$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.
Class 11 Engineering Maths Extra Questions
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