MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 5

Solve the following pairs of equations, graphically:

$3x - 2y = 0$ and

$y + 3 = 0$

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

Explanation

The solution set of equation

$3x-2y=0$ is as follows:

$x=0$ then

$3\times 0-2y=0\\ \Rightarrow 0-2y=0\\ \Rightarrow -2y=0\\ \Rightarrow y=0$

$x=-2$ then

$3\times (-2)-2y=0\\ \Rightarrow -6-2y=0\\ \Rightarrow -2y=6\\ \Rightarrow y=-3$

The required points are

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

Another equation of line given is

$y+3=0$ or

$y=-3$ .

Plot all the above points in the graph. Since the point of intersection in the graph is

$(-2,-3)$ .

Hence the solution is

$x=-2,y=-3$ .

Solve the following pairs of equations, graphically:

$x + y = 0$ and

$y = 5$

0%
$(5, -5)$
0%
$(-5, -5)$
0%
$(-5, 5)$
0%
$(5, 5)$

Explanation

The solution set of equation

$x+y=0$ is as follows:

$x=0$ then,

$0+y=0\\ \Rightarrow y=0$

$x=-5$ then

$-5+y=0\\ \Rightarrow y=5$

The required points to plot are

$(0,0),(-5,5)$ .

Another equation of a line given is

$y=5$ .

Plot all the above points of the line

$x+y=0$ and the line parallel to

$x$ axis that is

$y=5$ in the graph. Since the point of intersection in the graph is

$(-5,5)$ .

Hence the solution is

$x=-5,y=5$ .

Find a fraction which reduces to

$\displaystyle \frac{2}{3}$ if the numerator and the denominator are each increased by

$1$ , and reduces to

$\displaystyle \frac{3}{5}$ if the numerator and the denominator are each decreased by

$2$ .

0%
$\displaystyle \frac{13}{12}$
0%
$\displaystyle \frac{11}{17}$
0%
$\displaystyle \frac{9}{4}$
0%
$\displaystyle \frac{13}{3}$

Explanation

Let the fraction be

$\dfrac {x}{y}.$

Given, if the numerator and the denominator are each increased by

$1$ , then fraction

$= \dfrac {x + 1}{y + 1} = \dfrac {2}{3}$

$\Longrightarrow 3x + 3 = 2y + 2$

$\Longrightarrow 3x - 2y = -1$ .......................

$(1)$

Also, if the numerator and the denominator are each decreased by

$2$ , then fraction

$= \dfrac {x - 2}{y - 2} = \dfrac {3}{5}$

$\Longrightarrow 5x - 10 = 3y - 6$

$\Longrightarrow 5x - 3y = 4$ .......................

$(2)$

Multiplying equation

$(1)$ with

$5$ we get,

$15x - 10y = -5$ ...

$(3)$

Multiplying equation $(2)$ with $3$ we get, $15x - 9y = 12$ .... $(4)$

Subtracting equations $(3)$ from $(4)$ , we get,

$-9y-(-10y) = 12-(-5)$

$y=17$

Substituting $y = 17$ in the equation $(2)$ , we get,

$5x - 3(17) = 4$

$5x = 55$

$\Longrightarrow x = 11$

Hence, the fraction is $\dfrac {11}{17}.$

$A$ is

$25$ years older than

$B$ . In

$15$ years,

$A$ will be twice of

$B$ . Find the present ages of

$A$ and

$B$ .

0%
Present age of $A$ is $40$ years and

Present age of $B$ is $15$ years
0%
Present age of $A$ is $37$ years and

Present age of B is $12$ years
0%
Present age of $A$ is $35$ years and

Present age of $B$ is $10$ years
0%
Present age of $A$ is $45$ years and

Present age of $B$ is $20$ years

Explanation

Let the present age of

$A$ be

$x$ and the present age of

$B$ be

$y$ .

It is given that

$A$ is

$25$ years older than

$B$ that is:

$x=y+25$

$x-y=25..........(1)$

Also, In

$15$ years,

$A$ will be twice of

$B$ that is:

$x+15=2(y+15)$

$x+15=2y+30$

$x-2y=30-15$

$x-2y=15..........(2)$

Now subtract equation 1 from equation 2 to eliminate

$x$ , because the coefficients of

$x$ are same. So, we get

$(x-x)+(-2y+y)=15-25$

i.e.

$-y=-10$

i.e.

$y=10$

Substituting this value of

$y$ in (1), we get

$x-10=25$

i.e.

$x=25+10=35$

Hence, the present age of

$A$ is

$35$ years and the present age of

$B$ is

$10$ years.

Find the values of

$(x+y)$ and

$(x-y)$ without actually solving for

$x$ and

$y$ .

$13x+15y=114 \quad and \quad15x+13y=110$

0%
$x+y= 9, x-y= -2$
0%
$x+y= 8, x-y= -2$
0%
$x+y= 3, x-y= -2$
0%
$x+y= 1, x-y= -2$

Explanation

$13x+15y=114$ ...(i)

$15x+13y=110$ ...(ii)

On adding (i) and (ii) , we get

$28x+28y=224$

$\Rightarrow x+y=\frac {224}{28} =8$

On subtracting (i) from (ii) , we get

$2x-2y=-4$

$\Rightarrow x-y=-\frac 42 =-2$ .

A man is 24 years older than his son. 12 years ago, he was five times as old as his son. Find the present ages of both.

0%
Present age of father is $44$ years and Present age of son is $20$ years
0%
Present age of father is $42$ years and Present age of son is $18$ years
0%
Present age of father is $60$ years and Present age of son is $36$ years
0%
Present age of father is $48$ years and Present age of son is $24$ years

Explanation

Let the present age of the father be

$x$ years and the present age of the son be

$y$ years.

It is given that the man is

$24$ years older than his son that is:

$x=y+24$

$\Rightarrow x-y=24\quad ..........(1)$

Also,

$12$ years ago, he was five times as old as his son that is:

$(x-12)=5(y-12)$

$\Rightarrow x-12=5y-60$

$\Rightarrow x-5y=-60+12$

$\Rightarrow x-5y=-48\quad .......(2)$

On subtracting

$(1)$ from

$(2)$ , we get:

$(x-x)+(-5y+y)=-24-48$

$\Rightarrow -4y=-72$

$\Rightarrow y=18$

Substituting this value of

$y$ in

$(1)$ , we get:

$x-18=24$

$\Rightarrow x=24+18=42$

Hence, the present age of the father is

$42$ years and the present age of the son is

$18$ years.

Solve the following pairs of equations, graphically:

$\displaystyle\,\frac{x}{2}\,-\,\frac{y}{3}\,=\,3$ and

$x \,+ \,y \,=\, 1$

0%
$(-4, 3)$
0%
$(4, -3)$
0%
$(-4, -3)$
0%
$(4, 3)$

Explanation

The solution set of equation

$\frac { x }{ 2 } -\frac { y }{ 3 } =3$ or

$3x-2y=18$ is as follows:

$x=0$ then

$3\times 0-2y=18\\ \Rightarrow -2y=18\\ \Rightarrow y=-9$

$x=4$ then

$3\times 4-2y=18\\ \Rightarrow 12-2y=18\\ \Rightarrow y=-3$

The required points are

$(0,-9),(4,-3)$ .

Also, the solution set of equation

$x+y=1$ is as follows:

$x=0$ then

$0+y=1\\ \Rightarrow y=1$

$x=4$ then

$4+y=1\\ \Rightarrow y=1-4=-3$

The required points are

$(0,1),(4,-3)$ .

Plot all the above points in the graph. Since the point of intersection in the graph is

$(4,-3)$ .

Hence the solution is

$x=4,y=-3$ .

Find the fraction such that it becomes

$\displaystyle \frac{1}{2}$ if 1 is added to the numerator, and

$\displaystyle \frac{1}{3}$ if 1 is added to the denominator.

0%
$\dfrac{3}{8}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{8}{7}$
0%
$\dfrac{7}{3}$

Explanation

Let the fraction be

$\dfrac { x }{ y }$

$\therefore \dfrac { x+1 }{ y } = \dfrac { 1 }{ 2 }$ ...(1)

$\dfrac { x }{ y+1 } = \dfrac { 1 }{ 3 }$ ...(2)

So from

$1$ we get,

$2x+2 = y$ ...(3)

From

$2$ we get,

$3x = y+1$ ...(4)

Substitute

$eq(3)$ in

$eq(4)$ , we get

$3x = 2x+2+1$

$x = 3\\ y = 2x+2 = 8$

The fraction is

$\dfrac { 3 }{ 8 }$

Solve the following pair of equations:

$7x+6y= 71$

$5x-8y= -23$

0%
$x= -5;y= 4$
0%
$x= 3,y= -2$
0%
$x= 5,y= 6$
0%
$x= 1,y= 3$

Explanation

$\\ 7x+6y=71;\\ y=\dfrac{71-7x}{6};\\$

Substituting:

$5x-8\left(\dfrac{71-7x}{6}\right)=-23$

$\Rightarrow 30x-568+56x=-138$

$\Rightarrow 86x=430$

$\Rightarrow x=5$

Resubstituting the value of

$x$ , we get

$y=\dfrac{71-35}{6}=6$

Therefore

$x=5, y=6$

Solve the following pair of equations:

$\displaystyle \frac{6}{x}+\displaystyle \frac{4}{y}= 20, \displaystyle \frac{9}{x}-\displaystyle \frac{7}{y}= 10.5$

0%
$x= \displaystyle \frac{3}{7};y= \displaystyle \frac{2}{3}$
0%
$x= \displaystyle \frac{2}{5};y= \displaystyle \frac{2}{3}$
0%
$x= \displaystyle \frac{5}{7};y= \displaystyle \frac{1}{3}$
0%
$x= \displaystyle \frac{2}{5};y= \displaystyle \frac{1}{7}$

Explanation

Given equations are

$\dfrac { 6 }{ x } +\dfrac { 4 }{ y } =20\quad ...(1)\\ \dfrac { 9 }{ x } -\dfrac { 7 }{ y } =10.5\quad ...(2)$

Let

$\dfrac{1}{x} = a$ and

$\dfrac{1}{y} = b$

So, the original equations will reduce to

$6a + 4b = 20$ ...

$(1)$

$9a - 7b = 10.5$ ...

$(2)$

Multiplying

$(1)$ by

$7$ and

$(2)$ by

$4$ , we get

$42a + 28b = 140$ ...

$(3)$

$36a - 28b = 42$ ...

$(4)$

Adding

$(3)$ and

$(4)$ ,

$78a = 182$

$a = \dfrac{182}{78} = \dfrac{14}{6} = \dfrac{7}{3}$

Substituting

$a = \dfrac{7}{3}$ in

$(1)$ , we get

$b$ as

$14 + 4b = 20$

$4b = 6$

$b = \dfrac{6}{4} = \dfrac{3}{2}$

Now,

$a = \dfrac{1}{x}$ , so

$x = \dfrac{1}{a} = \dfrac{3}{7}$

$b = \dfrac{1}{y}$ , so

$y = \dfrac{1}{b} = \dfrac{2}{3}$

$x = \dfrac{3}{7}, y = \dfrac{2}{3}$

Option

$A$

Solve the following pair of equations using substitution method:

$\displaystyle \frac{5y}{2}-\displaystyle \frac{x}{3}= 8$

$\displaystyle \frac{y}{2}+\displaystyle \frac{5x}{3}= 12$

0%
$x= 2;y= 7$
0%
$x= 6;y= 4$
0%
$x= 7;y= 1$
0%
$x= 8;y= 3$

Explanation

$\dfrac { 5y }{ 2 } -\dfrac { x }{ 3 } =8\quad ...(1)\\ \dfrac { y }{ 2 } +\dfrac { 5x }{ 3 } =12\quad ...(2)$

Let

$\dfrac{y}{2} = a$ and

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

So, the original equations reduce to

$5a - b = 8$ ...

$(1)$

$a + 5b = 12$ ...

$(2)$

From

$(1), b = 5a - 8$

Substituting

$b = 5a - 8$ in

$(2)$ , we get

$a + 5(5a - 8) = 12$

$a + 25a - 40 = 12$

$26a = 52$

$a = 2$

So,

$b = 5a - 8$

$b = 10 - 8 = 2$

But

$\dfrac{y}{2} = a$ or

$y = 2a = 4$

$\dfrac{x}{3} = b$ or

$x = 3b = 6$

$x = 6, y = 4$

Option

$B$ .

Solve the following equations by substitution method.

$3y-2x=9; \, \, 2x+5y=15$

0%
$x = 0, y= 2$
0%
$x = 5, y= 2$
0%
$x = 1, y= 3$
0%
$x = 0, y= 3$

Explanation

$3y -2x =9; \, \, 2x +5y =15$

$y = \dfrac {9+2x}{3}$ Put it in second equation.

$2x + 5 \times \dfrac {9+2x}{3} =15$

$6x +45 +10x = 45$

$16x =0$

$\boxed{x =0}$

$y = \dfrac {9+2x}{3} = \dfrac {9+2 (0)}{3} = 3 \\ \boxed{y=3}$

Solve the following pair of equations:

$4x+\displaystyle \frac{6}{y}= 15$ and

$6x-\displaystyle \frac{8}{y}=14$

0%
$x= 5;y= 6$
0%
$x= 3;y= 2$
0%
$x= 7;y= 4$
0%
$x= 0;y= 6$

Explanation

The equation

$4x +\frac { 6 }{ y } =15$ can be solved as:

$4x+\frac { 6 }{ y } =15\\ \Rightarrow \frac { 4xy+6 }{ y } =15\\ \Rightarrow 4xy+6=15y\quad .........(1)$

The equation

$6x -\frac { 8 }{ y } =14$ can be solved as:

$6x-\frac { 8 }{ y } =14\\ \Rightarrow \frac { 6xy-8 }{ y } =14\\ \Rightarrow 6xy-8=14y\quad .........(2)$

Multiply the equation 1 by

$3$ and equation 2 by

$2$ . Then we get the equations:

$12xy+18=45y.........(3)$

$12xy-16=28y.........(4)$

Subtract Equation 4 from equation 3 to eliminate

$xy$ , because the coefficients of

$xy$ are same. So, we get

$(12xy-12xy)+(18+16)=45y-28y$

i.e.

$17y=34$

i.e.

$y=2$

Substituting this value of

$y$ in the equation 1, we get

$8x+6=30$

i.e.

$8x=30-6=24$

i.e.

$x=3$

Hence, the solution of the equations is

$x=3 ,y=2$ .

Solve the following pair of equations :

$x-y= 0.9$ and

$\displaystyle \frac{11}{2\left ( x+y \right )}= 1$

0%
$x= 1.5;y= 4$
0%
$x= 3;y= 2.5$
0%
$x= 5.2;y= 0.3$
0%
$x= 3.2;y= 2.3$

Explanation

$x=y+0.9$

Substituting in the 2nd equation:

$11=2(y+0.9+y)$

$\dfrac{11}{2}=2y+0.9\\5.5-0.9=2y\\4.6=2y\\ \therefore y=2.3$

and

$x=2.3+0.9=3.2$

Therefore,

$x=3.2$ and

$y=2.3$

Solve the following pair of linear (simultaneous) equations by the method of elimination :

$x+y= 7$

$5x+12y= 7$

0%
$x= 11$ and $y=-4$
0%
$x= 1$ and $y=7$
0%
$x= 13$ and $y=6$
0%
$x= 1$ and $y=-3$

Explanation

$x+y=7$ .... (1)

$5x+12y=7$ ....(2)

Multiply eq. (1) by 5, we get

$5x+5y=35$ ....(3)

Subtracting eq(2) from eq.(3)

$5x+5y=35$

$5x+12y=7$

$-$

---------------------

$-7y=28$

$y=-4$

Substituting value of

$y$ in eq,(1), we get,

$x-4=7$

$x=11$

Solve the following pairs of equations, graphically :

$4x + 3y = 1$ and

$2x - y = 3$

0%
$(1, 0)$
0%
$(1, -1)$
0%
$(2, -1)$
0%
$(2, -3)$

Explanation

Given,

$4x + 3y = 1$

$\Rightarrow y = \dfrac {1-4x}{3}$

Substituting different values of

$x$ to get corresponding values of

$y$ , we get

$( 0, \dfrac {1}{3}), (1, -1), (-1, \dfrac {5}{3}), (2, -\dfrac {7}{3}), (-2,3)$

Plotting these points on the graph, we get the graph of the equation,

$4x + 3y = 1$

Similarly,
Given second eqn,

$2x-y= 3$

$\Rightarrow y = 2x - 3$

Substituting different values of

$x$ to get corresponding values of

$y$ , we get

$( 1, -1), (0,-3), (2,1), (-1,-5), (3,3)$

Plotting these points on the graph, we get the graph of the equation,

$2x-y= 3$

The point of intersection of the graphs of the lines is the solution of the equations which is

$(1,-1)$

Solve the following pair of equations:

$\displaystyle \frac{9}{x}-\displaystyle \frac{4}{y}= 8$ ,

$\displaystyle \frac{13}{x}+\displaystyle \frac{7}{y}=101$

0%
$x= \displaystyle \frac{2}{3};y= \displaystyle \frac{4}{3}$
0%
$x= \displaystyle \frac{1}{4};y= \displaystyle \frac{1}{7}$
0%
$x= \displaystyle \frac{5}{4};y= \displaystyle \frac{2}{5}$
0%
$x= \displaystyle \frac{3}{2};y= \displaystyle \frac{6}{5}$

Explanation

The equation

$\dfrac { 9 }{ x } -\dfrac { 4 }{ y } =8$ can be solved as:

$\dfrac { 9 }{ x } -\dfrac { 4 }{ y } =8$

$\Rightarrow \dfrac { 9y-4x }{ xy } =8$

$\Rightarrow 9y-4x=8xy\quad .........(1)$

The equation

$\dfrac { 13 }{ x } +\dfrac { 7 }{ y } =101$ can be solved as:

$\dfrac { 13 }{ x } +\dfrac { 7 }{ y } =101$

$\Rightarrow \dfrac { 13y+7x }{ xy } =101$

$\Rightarrow 13y+7x=101xy\quad .........(2)$

Multiply the equation 1 by

$7$ and equation 2 by

$4$ . Then we get the equations:

$63y-28x=56xy.........(3)$

$52y+28x=404xy.........(4)$

Add equations 3 and 4 to eliminate

$x$ , because the coefficients of

$x$ are the opposite. So, we get

$(63y+52y)+(-28x+28x)=56xy+404xy$

i.e.

$115y=460xy$

i.e.

$115=460x$

i.e.

$x=\dfrac { 115 }{ 460 } =\dfrac { 1 }{ 4 }$

Substituting this value of

$x$ in the equation 1, we get

$9y-4\left( \dfrac { 1 }{ 4 } \right) =8\left( \dfrac { 1 }{ 4 } \right) y\\ \Rightarrow 9y-1=2y\\ \Rightarrow 9y-2y=1\\ \Rightarrow 7y=1\\ \Rightarrow y=\dfrac { 1 }{ 7 }$

Hence, the solution of the equations is

$x=\dfrac { 1 }{ 4 } ,y=\dfrac { 1 }{ 7 }$ .

Solve the following pair of equations:

$2x-3y-3= 0$ ,

$\displaystyle \frac{2x}{3}+4y+\displaystyle \frac{1}{2}= 0$

0%
$x= \displaystyle \frac{13}{20},y= -\displaystyle \frac{7}{10}$
0%
$x= \displaystyle \frac{17}{20},y= -\displaystyle \frac{1}{20}$
0%
$x= \displaystyle \frac{1}{10},y= -\displaystyle \frac{7}{10}$
0%
$x= \displaystyle \frac{21}{20},y= -\displaystyle \frac{3}{10}$

Explanation

$2x-3y-3= 0$ ...... (1)

$\displaystyle \frac{2x}{3}+4y=-\displaystyle \frac{1}{2}$ ...... (2)

$2x-3y=3;$

$x=\dfrac{3+3y}{2}$ .....(3)

Substituting in the 2nd equation

$\dfrac{-1}{2}=\dfrac{2}{3} \times \dfrac{3+3y}{2} +4y$

$5y=-\dfrac{3}{2} \implies y=\dfrac{-3}{10}$

Put the value of

$y$ in (3)

$x=\dfrac{3+3(\dfrac{-3}{10})}{2} \implies x=\dfrac{21}{20}$

$\therefore x=\dfrac{21}{20} \quad and \quad y=\dfrac{-3}{10}$

Solve the following pairs of linear equations, graphically :

$\displaystyle\,2x \,-\, 3y\, =\, -6\,$ and

$x\, -\,\dfrac{y}{2}\,=\,1$

0%
$(3, \,2)$
0%
$(3, \,3)$
0%
$(4, \,4)$
0%
$(3, \,4)$

Explanation

Given,

$2x - 3y = - 6$

$\Rightarrow y = \dfrac {2x}{3} + 2$
Substituting different values of

$x$ to get corresponding values of

$y$ , we get

$( 0, 2), (3,4), (6,6), (-3, 0)$
Plotting these points on the graph, we get the graph of the equation,

$2x - 3y = - 6$
Similarly, given second equation,

$x - \dfrac {y}{2} = 1$

$\Rightarrow y = 2(x-1)$
Substituting different values of

$x$ to get corresponding values of

$y$ , we get

$( 1, 0), (0,-2), (2,2), (-1,-4), (3,4)$
Plotting these points on the graph, we get the graph of the equation,

$y = 2(x-1)$ .
The point of intersection of the graphs of the lines is the solution of the equations which is

$(3,4)$ .

Solve the following equations by substitution method.

$x=2y-1; \, \, y=2x-7$

0%
$x = 8, y= 3$
0%
$x = 2, y= 3$
0%
$x = 5, y= 3$
0%
$x = 1, y= 3$

Explanation

$x=2y-1; \, \, y=2x - 7$

$x = 2y-1$ Put it in second equation.

$y = 2 \times (2y -1) -7$

$y =4y -9$

$\boxed{y = 3}$

and

$x =2y -1 = 2 \times 3 -1 =5$

$\boxed{x=5}$

$\boxed{\,x = 5 \ and \ y = 3 \,}$

Solve:

$\displaystyle \frac{3}{x}-\displaystyle \frac{2}{y}= 0$ and

$\displaystyle \frac{2}{x}+\displaystyle \frac{5}{y}= 19$ . Hence, find

$a$ if

$y= ax+3$ .

0%
$x=\displaystyle \frac{2}{3};y= \displaystyle \frac{5}{3}$ and $a= 6\displaystyle \frac{6}{5}$
0%
$x=\displaystyle \frac{3}{4};y= \displaystyle \frac{1}{3}$ and $a= -4\displaystyle \frac{7}{3}$
0%
$x=\displaystyle \frac{7}{2};y= \displaystyle \frac{8}{3}$ and $a= -7\displaystyle \frac{2}{9}$
0%
$x=\displaystyle \frac{1}{2};y= \displaystyle \frac{1}{3}$ and $a= -5\displaystyle \frac{1}{3}$

Explanation

Let

$\dfrac{1}{x} = u$ and

$\dfrac{1}{y} = v$

So, the equations will reduce to

$3u - 2v = 0$ ...

$(1)$

$2u + 5v = 19$ ...

$(2)$

Multiplying

$(1)$ by

$5$ and

$(2)$ by

$2$

$15u - 10v = 0$ ...

$(3)$

$4u + 10v = 38$ ...

$(4)$

Adding

$(3)$ and

$(4)$

$19u = 28$

$u = 2$

Substituting

$u = 2$ in

$(1)$ , we get

$6 - 2v = 0$

$v = 3$

$u = \dfrac{1}{x}$ So,

$x = \dfrac{1}{u} = \dfrac{1}{2}$

$v = \dfrac{1}{y}$ So,

$y = \dfrac{1}{v} = \dfrac{1}{3}$

$x = \dfrac{1}{2}, y = \dfrac{1}{3}$

$y = ax + 3$

$\dfrac { 1 }{ 3 } =\dfrac { a }{ 2 } +3\\ \dfrac { a }{ 2 } = -\dfrac { 8 }{ 3 } \\ a = \dfrac { -16 }{ 3 }$

Option

$D$

Solve the following equations by substitution method.

$2x-3y=14; \, \, 5x+2y=16$

0%
$x = 7, y= 1$
0%
$x = 4, y= -2$
0%
$x = 6, y= -2$
0%
$x = 3, y= -1$

Explanation

$2x-3y=14; \, \, 5x+2y=16$

$x = \dfrac {14+3y}{2}$ Put it in second equation.

$5 \times \dfrac {14+3y}{2} +2y =16$

$70 +15y +4y = 32$

$19y = -38$

$y = -2$

$x = \dfrac {14+3y}{2} = \dfrac {14+3(-2)}{2} = 4$

$x = 4$ and

$y = -2$

Solve the following equations by substitution method.

$x-2y+2=0; \, \, x+2y=10$

0%
$x = 4, y= 3$
0%
$x = 1, y= 3$
0%
$x = 9, y= 3$
0%
$x = 5, y= 3$

Explanation

$x-2y+2=0$ ....(i)

$x+2y=10$ ....(ii)
We pick either of the equations and write one variable in terms of the other.
Let us consider the Equation (i) and write it as

$x=2y-2$ .........(iii)
Substitute the value of

$x=2y-2$ in Equation (ii). We get

$2y-2+2y=10$

$\Rightarrow 4y=12$

$\Rightarrow y=3$
Substituting this value of

$y=3$ in Equation (iii), we get

$x=2(3)-2=4$

$\therefore x=4 , y=3$

Solve:

$\displaystyle \frac{34}{3x+4y}+\displaystyle \frac{15}{3x-2y}= 5$ and

$\displaystyle \frac{25}{3x-2y}-\displaystyle \frac{8.50}{3x+4y}= 4.5$

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$x= 1;y= 4$
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$x= 3;y= 2$
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$x= 5;y= 8$
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$x= 7;y= 6$

Explanation

On putting

$\dfrac1{3x+4y}=X$ and

$\dfrac1{3x-2y}=Y$ in given equations, we get

$34X+15Y=5$ ...(i)

$25Y-\dfrac{17}2X=4.5\Rightarrow -17X+50Y=9$ ...(ii)

On multiplying (i) by 1 and (ii) by 2 and adding, we get

$34X+15Y=5$

$\underline {-34X+100Y=18}$

$115Y=23$

$\therefore Y=\dfrac{23}{115}=\dfrac15$

On putting

$Y=\dfrac15$ in (i), we get

$34X+15\times\dfrac15=5\Rightarrow 34X=2\Rightarrow X=\dfrac1{17}$

$\therefore 3x+4y=17$ ...(iii)

$3x-2y=5$ ...(iv)

On subtracting (iv) from (iii), we get

$6y=12\Rightarrow y=2$

On putting

$y=2$ in (iv), we get

$3x-4=5\Rightarrow x=\dfrac93=3$

Therefore,

$x=3,y=2$

Solve:

$\displaystyle \frac{20}{x+y}+\displaystyle \frac{3}{x-y}= 7$ and

$\displaystyle \frac{8}{x-y}-\displaystyle \frac{15}{x+y}= 5$

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$x= 4;y= -7$
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$x= 4;y= -4$
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$x= 3;y= 2$
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$x= 1;y= 6$

Explanation

Let

$\dfrac {20}{x+y} + \dfrac {3}{x-y} = 7$ --- (1)

and

$\dfrac {8}{x-y} - \dfrac {15}{x+y} = 5$ --- (2)

Multiplying equation $(1)$ with $3$ we get, $\dfrac {60}{x+y} + \dfrac {9}{x-y} = 21$ ----- equation $(3)$

Multiplying equation $(2)$ with $4$ we get, $\cfrac {32}{x-y} - \dfrac {60}{x+y} = 20$ ----- equation $(4)$

Adding equations $3$ and $4$ , we get $\dfrac {41}{x-y} = 41 => x - y = 1$ ---- (5)

Substituting $x-y = 1$ in the equation $(1)$ , we get $\dfrac {20}{x+y} + \dfrac {3}{1} = 7 => \dfrac {20}{x+y} = 4$

=> $x +y= 5$ --- (6)

Adding equations $5$ and $6$ , we get $2x = 6 => x = 3$ ---- (5)

Substituting $x = 3$ in the equation $(5)$ , we get $3-y = 1 => y = 2$

Solve the following equations by substitution method:

$5x-2y=13; \, \, 4x+3y=15$

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$x=1,y=-3$
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$x=3,y=1$
0%
$x=-1,y=3$
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$x=-1,y=-3$

Explanation

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$5x-2y=13$ and write it as

$x=\frac { 13+2y }{ 5 }$

Substitute the value of

$x$ in equation

$4x+3y=15$ . We get

$4\left( \frac { 13+2y }{ 5 } \right) +3y=15\\ \Rightarrow 52+8y+15y=75\\ \Rightarrow 23y=23\\ \Rightarrow y=1$

Therefore,

$y = 1$

Substituting this value of

$y$ in the equation

$5x-2y=13$ , we get

$5x-2=13$

$5x=13+2=15$

$x=3$

Hence, the solution is

$x = 3, y = 1$ .

The sides of an equilateral triangle are given by

$x+3y$ ;

$3x+2y-2$ and

$4x+\displaystyle \frac{1}{2}y+1$ respectively. Find the lengths of the sides of the triangle.

0%
$15$ units each
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$17$ units each
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$12$ units each
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$11$ units each

Explanation

Since it is an equilateral triangle, the lengths of the sides are equal

So,

$x+3y = 3x+2y-2 = 4x+\dfrac { 1 }{ 2 } y+1$

$\therefore x + 3y = 3x+2y -2\\ \therefore y = 2x - 2$

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

Also,

$4x+\dfrac { 1 }{ 2 } y+1 = x+3y$

$\therefore 8x + y + 2 = 2x +6y\\ 6x + 2 = 5y$

$6x - 5y = -2$ ...(2)

Multiplying eq(1) by 5 we get,

$10x - 5y = 10$ ...(3)

Subtracting eq(2) from eq(3) we get,

$4x = 12 \Rightarrow x = 3$

$\therefore y = 2x -2 = 4$

The length of one side of equilateral triangle =

$x +3y$ =

$3 + 3(4) = 15$ units

Solve:

$4x+\displaystyle \frac{6}{y}= 15$ and

$6x-\displaystyle \frac{8}{y}= 14$ . Hence, find

$a$ if

$y= ax-2$

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$x= 2;y= 26$ and $a= 2\displaystyle \frac{5}{3}$
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$x= 3;y= 2$ and $a= 1\displaystyle \frac{1}{3}$
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$x= 1;y= 5$ and $a= 2\displaystyle \frac{7}{3}$
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$x= 5;y= 1$ and $a= 5\displaystyle \frac{1}{8}$

Explanation

The equation

$4x +\frac { 6 }{ y } =15$ can be solved as:

$4x+\frac { 6 }{ y } =15\\ \Rightarrow \frac { 4xy+6 }{ y } =15\\ \Rightarrow 4xy+6=15y\quad .........(1)$

The equation

$6x -\frac { 8 }{ y } =14$ can be solved as:

$6x-\frac { 8 }{ y } =14\\ \Rightarrow \frac { 6xy-8}{ y } =14\\ \Rightarrow 6xy-8=14y\quad .........(2)$

Multiply the equation 1 by

$3$ and equation 2 by

$2$ . Then we get:

$12xy+18=45y.........(3)$

$12xy-16=28y.........(4)$

Subtract equation 4 from equation 3 to eliminate

$xy$ , because the coefficients of

$xy$ are same. So, we get

$(12xy-12xy)+(18+16)=45y-28y$

i.e.

$17y=34$

i.e.

$y=2$

Substituting this value of

$y$ in the equation 1, we get

$8x+6=30 \\ \Rightarrow 8x=30-6 \\ \Rightarrow 8x=24 \\ \Rightarrow x=3$

Therefore, the solution of the equations is

$x=3,y=2$ .

Now substitute the values of

$x$ and

$y$ in

$y=ax-2$ as shown below:

$2=3a-2$

$3a=2+2$

$3a=4$

$a=\frac { 4 }{ 3 } =1\frac { 1 }{ 3 }$

Hence,

$a=1\frac { 1 }{ 3 }$ .

Solve the following equations by substitution method.

$2x-3y-3=7; \, \, 4x-5y-5=10$

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$x = \displaystyle \frac{-7}{2}, y= -5$
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$x = \displaystyle \frac{-3}{2}, y= -5$
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$x = \displaystyle \frac{-1}{2}, y= -5$
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$x = \displaystyle \frac{-5}{2}, y= -5$

Explanation

Rewriting the given equations, we get

$2x-3y=10$ ...(i)

$4x-5y=15$ ...(ii)
We pick either of the equations and write one variable in terms of the other. Consider the Equation (i) and write it as

$x=\displaystyle \frac {10+3y}{2}$ ...(iii)
Substitute the value of

$x=\displaystyle \frac {10+3y}{2}$ in Equation (ii). we get,

$4\left(\displaystyle \dfrac {10+3y}{2}\right)-5y=15$

$\Rightarrow 20+6y-5y=15$

$\Rightarrow y=-5$
Substituting this value of

$y=-5$ in Equation (iii). we get,

$x=\displaystyle \frac {10+3(-5)}{2}= \frac {-5}{2}$

$\therefore x=\dfrac {-5}{2} , y=-5$

Pooja and Ritu can do a piece of work in

$\displaystyle 17\frac{1}{7}$ days. If one day work of Pooja is three fourth of one day work of Ritu then find in how many days each alone will do the same work.

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Pooja in 40 days and Ritu in 30 days
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Pooja in 60 days and Ritu in 40 days
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Pooja in 30 days and Ritu in 10 days
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Pooja in 45 days and Ritu in 20 days

Explanation

Let the work done by pooja in one day be

$1/x$ and work done by Ritu in one day be

$1/y.$

Therefore, according to the question total work done by both in one day,

$\\ 1/x+1/y=\dfrac{1}{17\dfrac{1}{7}} = 7/120$ .....(1)

Also, one day work of Pooja is three fourth of one day work of Ritu

$\\ 1/x=3/4y$

Substituting above equation in equation (1) , we get

$y=40$

Substituting

$y=40$ in equation (1) we get

$x=30$
Hence pooja will complete the work in

$40$ days whereas ritu in

$30$ days

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 5 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 5 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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