MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 13

Solve the following equations using substitution method:

$2x+3y=13$ and

$4x+5y=23$

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

Explanation

Given:

$2x+3y=13$ ------(1)

$4x+5y=23$ ------(2)

From equation (1),

$2x+3y =13$

$y=\cfrac{13-2x}{3}$

Substitute the value of

$y$ in equation (2):

$4x+5y=23$

$4x+5(\cfrac{13-2x}{3}) = 23$

$12x+65-10x=69$

$2x=69-65$

$2x=4$

$x=2$

Then substitute the value of

$x$ in equation (1):

$2(2)+3y=13$

$4+3y=13$

$3y=13-4$

$3y=9$

$y=3$
Therefore solution is

$x=2$ ans

$y=3$

Solve the equation using substitution method:

$4x-6y=-4$ and

$8x-2y=48$

0%
5.6 and 7.4
0%
-7.4 and 5.6
0%
7.4 and 5.6
0%
8.6 and 5.6

Explanation

$4x-6y=-4$ ------- (1)

$8x-2y=48$ --------(2)
From equation (1)

$4x-6y=-4$

$4x+4=6y$

$y=\cfrac{4x+4}{6}$
Substitute the value of

$y$ in equation (2)

$8x-2(\cfrac{4x+4}{6})=48$

$42x-4x4=144$

$20x=144+4$

$x=7.4$
The substitute the value of

$x$ in equation (1)

$4(7.4)-6y=-4$

$29.6-6y=-4$

$29.6+4=6y$

$\cfrac{33.6}{6}=y$

$y=5.6$
Therefore solution is

$x= 7.4$ and

$y=5.6$

Solve the equations using graphical method:

$x- y = 3$ and

$x + y = 11$

0%
$(7, 4)$
0%
$(-7, 3)$
0%
$(7, -4)$
0%
$(-7, 4)$

Explanation

$x- y = 3$ ----- (1)

$x + y = 11$ ----- (2)
From equation (1)

$y = x - 3$ ------ (3)
Assume the value of

$x = 6, 7$ and put those values in equation (3)
If

$x = 6, y = 6 -3 \Rightarrow y= 3$
If

$x = 7, y = 7 -3 \Rightarrow y = 4$
On the above basis, the following points are:

$x$	$6$	$7$
$y$	$3$	$4$

Now plotting

$(6, 3), (7, 4)$ and joining them, we get a straight line.
From equation (2),

$x + y = 11$

$y = 11 - x$ ------ (4)
Assume the value of

$x = 6, 7$ and put those values in equation (3)
If

$x = 6, y =11- 6 \Rightarrow y= 5$
If

$x = 7, y = 11- 7 \Rightarrow y = 4$
Now table the following points:

$x$	$6$	$7$
$y$	$5$	$4$

Plotting

$(6, 5), (7, 4)$ and joining them, we get another straight line.
These lines intersect at the point

$(7, 4)$ and therefore the solution of the equation is

$x = 7, y = 4$ .

Solve the equation using elimination method:

$\displaystyle \frac{3}{x}+\frac{2}{y} = 7$ and

$\displaystyle \frac{4}{x} -\frac{1}{y}=2$

0%
$\left (1, \dfrac{1}{2}\right)$
0%
$\left (-1, \dfrac{1}{2}\right)$
0%
$\left (1, \dfrac{-1}{2}\right)$
0%
$\left (-1, \dfrac{-1}{2}\right)$

Explanation

Let us assume

$\cfrac{1}{x} = u; \cfrac{1}{y} = v$ , then the equation becomes,

$3u+2v=7$ .....(1)

$4y-v=2$ .....(2)

Multiply the equation (2) by

$2$ and then subtract the equation

$3u+2v=7$

$8u-2v=4$

__________

$11u =11$

$\Rightarrow u=\cfrac{11}{11}=1$

$\Rightarrow u=\cfrac{1}{x}; x=1$

Put the value of

$u$ in equation (2), we get

$4u-v=2$

$\Rightarrow 4(1)-v=2$

$\Rightarrow 4-2=v$

$\Rightarrow v=2$

Therefore, the solution is

$\left(1, \cfrac{1}{2}\right)$ .

Solve the equations using graphical method:

$x + y = 7$ and

$2x - 3y = 9$

0%
$(6, -1)$
0%
$(-6, 1)$
0%
$(6, 1)$
0%
$(-6, -1)$

Explanation

For line

$2x-3y=9$

$x$	$0$	$6$
$y$	$-3$	$1$

For line

$x+y=7$

$x$	$5$	$2$
$y$	$6$	$1$

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

$2x - 3y = 9$ ----- (2)
From equation (1)

$y = 7 - x$ ------ (3)
Assume the value of

$x = 5,$ 6 and put those values in equation (3)
If

$x = 5, y = 7-5 = 2$
If

$x = 6, y = 7 - 6 = 1$

Now plotting (5, 2), (6, 1) and joining them, we get a straight line.
From equation (2),

$2x - 3y = 9$

$y = \dfrac{2x-9}{3}$ ------ (4)
Assume the value of

$x = 0, 6$ and put those values in equation (4)

$x = 0, y =\dfrac{2x-9}{3} = \dfrac{2(0)-9}{3}=\dfrac{-9}{3} = - 3$

$x = 6, y =\dfrac{2(6)-9}{3} = \dfrac{12-9}{3}=\dfrac{3}{3} =1$

Plotting

$(0, -3), (6, 1)$ and joining them, we get another straight line.
These lines intersect at the point

$(6, 1)$ and therefore the solution of the equation is

$x = 6, y = 1$ .

Solve the equation using substitution method:

$x+y=-3$ and

$x+2y=4$

0%
-10 and 7
0%
10 and 7
0%
10 and -7
0%
-10 and -7

Explanation

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

$x+2y=4$ ------(2)
From equation (1)

$x+y=-3$

$y=-3-x$
Substitute the value of

$y$ in equation (2)

$x+2y=4$

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

$x-6-2x=4$

$-x=4+6$

$x=-10$
Substitute the value of

$x$ in equation (1)

$x+y=-3$

$-10+y=-3$

$y=-3+10$

$y=7$
Therefore solution is

$x=-10$ and

$y=7$

Find the value x and y using substitution method:

$2x-5y=9$ and

$5x+6y=8$

0%
$94$ and $199$
0%
$\cfrac{94}{43}$ and $\cfrac{199}{215}$
0%
$\cfrac{-94}{43}$ and $\cfrac{199}{215}$
0%
$\cfrac{-94}{43}$ and $\cfrac{-199}{215}$

Explanation

$2x-5y=9$ -----(1)

$5x+6y=8$ -----(2)
from equation (1)

$2x-5y=9$

$y=\cfrac{2x-9}{5}$
substitute the balue of

$y$ in equation (2)

$5x+6(\cfrac{2x-9}{5}) = 8$

$25x+18x-54=40$

$43x=40+54$

$43x=94$

$x=\cfrac{94}{43}$
Substitute the value of

$x$ in equation (1)

$2(\cfrac{94}{43})-5y =9$

$188-215y=387$

$188-387=215y$

$y=\cfrac{199}{215}$
Therefore solution is

$x=\cfrac{94}{43}$ and

$y=\cfrac{199}{215}$

Solve the equations using graphical method:

$x + y = 3$ and

$-3x + 2y = 1$

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

Explanation

For

$x + y = 3$ :

x	0	3	2
y	3	0	1

For

$-3x + 2y = 1$ :

x	-1	1	-3
y	-1	2	-4

As the graph shows, both lines intersect at

$(1, 2)$ .

Find the value of x and y using graphical method:

$8x -y = 6$ and

$6x - y = 4$

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

Explanation

$8x -y = 6$ ----- (1)

$6x - y = 4$ ----- (2)
From equation (1) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$8x -y - 6 = 0$ ------ (3)
Put

$x = 0, y = -6$ in equation (3)

$8x -y - 6 = 0$

$8(0)- (-6)- 6 = 0$

$6-6 = 0$

$0 = 0$
Again put

$x = 1, y = 2$ in equation (3)

$8(1) -2 - 6 = 0$

$6 - 6 = 0$

$0 = 0$
Now plotting

$(0, -6), (1, 2)$ and joining them, we get a straight line.
From equation (2) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$6x - y - 4 = 0$ ----- (4)
Put

$x = 0, y = -4$ in equation (4)

$6x - y - 4 = 0$

$6(0)- (-4)- 4 = 0$

$4- 4 = 0$

$0 = 0$
Put

$x = 1, y = 2$ in equation (4)

$6(1)-2-4=0$

$6-6=0$

$0=0$
Plotting

$(0, -4), (1, 2)$ and joining them, we get another straight line.
These lines intersect at the point

$(1, 2)$ and therefore the solution of the equation is

$x = 1, y = 2$ .

Solve the equations graphically:

$2x- y = -8$ and

$x + 2y = -14$

0%
$(6, 4)$
0%
$(-6, -4)$
0%
$(-6, 4)$
0%
$(6, -4)$

Solve the equations:

$5x -2 y = 6$ and

$2x + 5y = 3$ and then identify the system of equations.

0%
Consistent
0%
Inconsistent
0%
Dependent
0%
All the above

Explanation

The general form for a pair of linear equations in two variables x and y is

$a_1x + b_1y + c_1 = 0$ and

$a_2x + b_2y + c_2 = 0$

$5x - 2y - 6=0$ ----- (1)

$2x + 5y -3=0$ ----- (2)
Comparing equations 1 and 2 with the general form of equation to write their co-efficient,

$a_1= 5; a_2 = 2; b_1 = -2; b_2 = 5; c_1 = -6; c_2 = -3$
Compare the ratios of,

$\displaystyle \dfrac{a_1}{a_2} and \dfrac{b_1}{b_2}$

$\displaystyle \dfrac{5}{2}\neq\dfrac{-2}{5}$
If

$\displaystyle \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$ , then the pair of linear equations has exactly one solution.
So, the given system of equations is consistent.

Solve the equations:

$5x + 8y = 9$ and

$2x + 3y = 4$ using graphical method.

0%
(5, -2)
0%
(-5, 2)
0%
(5, 2)
0%
(-5, -2)

Explanation

$5x + 8y = 9$ ----- (1)

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

From equation (1) assume the values of

$x$ and

$y$ to satisfy the equation to zero.

For

$x=1$ , we have:

$5(1)+8y=9$

$\Rightarrow y=0.5$

Similarly, we can find the values of

$x$ and

$y$ that satisfy the equation (1).

Hence, we have:

$x$	$-3$	$1$	$5$
$y$	$3$	$0.5$	$-2$

In the same way, we can get the values of

$x$ and

$y$ that satisfy the equation (2).

Hence, we have:

$x$	$-4$	$2$	$5$
$y$	$3$	$0$	$-2$

Plotting these points in the graph, we will get two straight lines.

These lines intersect at the point

$(5, -2)$ , as seen from the graph.

Hence, the solution of the given equations is

$x = 5, y = -2$ .

Solve the equations:

$x -5y = 15$ and

$2x + 5y = 0$ using graphical method.

0%
(5, 2)
0%
(-5, 2)
0%
(5, -2)
0%
(-5, -2)

Explanation

Given equations are:

$x- 5y = 15$ ----- (1)

$2x + 5y = 0$ ----- (2)

From equation (1) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$x- 5y- 15 = 0$ ------ (3)
Put

$x = 0, y = -3$ in equation (3)

$0- 5(-3) -15 = 0$

$15- 15 = 0$

$0 = 0$

Again, put

$x = 5, y = -2$ in equation (3)

$5- 5(-2)- 15 = 0$

$5 + 10 -15 = 0$

$15- 15 = 0$

$0 = 0$
Now plotting

$(0, -3), (5, -2)$ and joining them, we get a straight line.

From equation (2) assume the value of

$x$ and

$y$ to satisfy the equation to zero.
Put

$x = 5, y = -2$ in equation (2)

$2x + 5y = 0$

$2(5) + 5(-2) = 0$

$10- 10 = 0$

$0 = 0$

Again, put

$x = 0, y = 0$ in equation (2)

$2(0) + 5(0) = 0$

$0 + 0 = 0$

$0 = 0$
Plotting

$(5, -2), (0, 0)$ and joining them, we get another straight line.

The graph of both the equations is as shown above.

As can be seen from the graph, these lines intersect at the point

$(5, -2)$ and therefore the solution of the equation is

$x = 5, y = -2$ .

Hence, the solution of the given equations is

$(5,-2)$ .

Using graphical method check whether the given equation is consistent:

$3x- y = 4$ and

$7x + 2y = 18$

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

Explanation

$3x- y = 4$ ----- (1)

$7x + 2y = 18$ ----- (2)
From equation (1) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$3x- y- 4 = 0$ ------ (3)
Put

$x = 0, y = -4$ in equation (3)

$3(0)- (-4)- 4 = 0$

$4- 4 = 0$

$0 = 0$
Again put

$x = 2, y = 2$ in equation (3)

$3(2)- (2) -4 = 0$

$6- 2 - 4 = 0$

$6- 6 = 0$

$0 = 0$
Now plotting

$(0, -4), (2, 2)$ and joining them, we get a straight line.
From equation (2) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$7x + 2y- 18 = 0$ ----- (4)
Put

$x = 0, y = 9$ in equation (4)

$7(0) + 2(9) -18 = 0$

$18- 18 = 0$

$0 = 0$
Again put

$x = 2, y = 2$ in equation (4)

$7(2) + 2(2)- 18 = 0$

$14 + 4- 18 = 0$

$18 -18 = 0$

$0 = 0$
Plotting

$(0, 9), (2, 2)$ and joining them, we get another straight line.
These lines intersect at the point

$(2, 2)$ and therefore the solution of the equation is

$x = 2, y = 2$ .

Solve the equations:

$x- y = -1$ and

$2y + 3x = 12$ using graphical method.

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

Explanation

$x- y = -1$ -----

$(1)$

$2y + 3x = 12$ -----

$(2)$

From equation

$(1)$ assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$x- y + 1 = 0$ -------

$(3)$

Put

$x = 2, y = 3$ in equation

$(3)$

$2- 3 + 1 = 0$

$-1 + 1 = 0$

$0 = 0$

Again put

$x = 4, y = 5$ in equation

$(3)$

$4- 5 + 1 = 0$

$-1 + 1 = 0$

$0 = 0$

Now plotting

$(2, 3), (4, 5)$ and joining them, we get a straight line.

From equation

$(2)$ assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$2y + 3x - 12 = 0$ -----

$(4)$

Put

$x = 2, y = 3$ in equation

$(4)$

$2(3) + 3(2) - 12 = 0$

$6 + 6- 12 = 0$

$12- 12 = 0$

$0 = 0$

Again put

$x=4, y=0$ in equation

$(4)$

$2(0) + 3(4) - 12 = 0$

$0 + 12- 12 = 0$

$12 -12 = 0$

$0 =0$

Plotting

$(2, 3), (4, 0)$ and joining them, we get another straight line.

These lines intersect at the point

$(2, 3)$ and therefore the solution of the equation is

$x = 2, y = 3$ .

Solve the equations:

$2x + 2y = 4$ and

$-2x + 3y = 1$ using graphical method.

0%
(1, 1)
0%
(-1, 1)
0%
(1, -1)
0%
(-1, -1)

Explanation

$2x + 2y = 4$ ----- (1)

$-2x + 3y = 1$ ----- (2)
From equation (1) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$2x + 2y - 4 = 0$ ------ (3)
Put

$x = 0, y = 2$ in equation (3)

$2(0) + 2(2) - 4 = 0$

$4- 4 = 0$

$0 = 0$
Again put

$x = 1, y = 1$ in equation (3)

$2(1) + 2(1)- 4 = 0$

$4 - 4 = 0$

$0 = 0$
Now plotting

$(0, 2), (1, 1)$ and joining them, we get a straight line.
From equation (2) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$-2x + 3y - 1 = 0$ ----- (4)
Put

$x = -0.5, y = 0$ in equation (4)

$-2 (-0.5) + 3(0) - 1 = 0$

$1 + 0 - 1 = 0$

$0 = 0$
Again put

$x = 1, y = 1$ in equation (4)

$-2(1) + 3(1)- 1 = 0$

$-2 + 3- 1 = 0$

$-3 + 3 = 0$

$0 = 0$
Plotting

$(-0.5, 0), (1, 1)$ and joining them, we get another straight line.
These lines intersect at the point (1, 1) and therefore the solution of the equation is

$x = 1, y = 1$

Solve the equations:

$x- y = -5$ and

$x + y = 1$ using graphical method

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

Explanation

$x- y = -5$ ----- (1)

$x + y = 1$ ----- (2)
From equation (1) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$x- y + 5 = 0$ ------ (3)
Put

$x = 0, y = 5$ in equation (3)

$0- 5 + 5 = 0$

$0 = 0$
Again put

$x = -2, y = 3$ in equation (3)

$-2 -3 + 5 = 0$

$-5 + 5 = 0$

$0 = 0$
Now plotting

$(0, 5), (-2, 3)$ and joining them, we get a straight line.
From equation (2) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$x + y- 1 = 0$ ----- (4)
Put

$x = 0, y = 1$ in equation (4)

$0 + 1 - 1 = 0$

$0 = 0$
Again put

$x = -2, y = 3$ in equation (4)

$-2 + 3- 1 = 0$

$-3 + 3 = 0$

$0 = 0$
Plotting

$(0, 1), (-2, 3)$ and joining them, we get another straight line.
These lines intersect at the point

$(-2, 3)$ and therefore the solution of the equation is

$x = -2, y = 3$

Solve the equations graphically:

$7x + 8y = 64$ and

$x + 4y = 12$

0%
$(8, 1)$
0%
$(-8, -1)$
0%
$(-8, 1)$
0%
$(8, -1)$

Explanation

$7x + 8y = 64$ ----- (1)

$x + 4y = 12$ ----- (2)
From equation (1) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$7x + 8y-64=0$ ------ (3)
Put

$x = 0, y = 8$ in equation (3)

$7(0) + 8(8)- 64 = 0$

$0 + 64 -64 = 0$

$0 = 0$
Again put

$x = 8, y = 1$ in equation (3)

$7(8) + 8(1)- 64 = 0$

$56 + 8 -64 = 0$

$0 = 0$
Now plotting

$(0, 8), (8, 1)$ and joining them, we get a straight line.
From equation (2) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$x + 4y -12=0$ ----- (4)
Put

$x = 0, y = 3$ in equation (4)

$0 + 4(3) -12 = 0$

$12 -12 = 0$

$0 = 0$
Again put

$x = 8, y = 1$ in equation (4)

$8+ 4(1) -12 = 0$

$8 + 4- 12 = 0$

$0 = 0$
Plotting

$(0, 3), (8, 1)$ and joining them, we get another straight line.
These lines intersect at the point

$(8, 1)$ and therefore the solution of the equation is

$x = 8, y = 1$

Using graphical method check whether the given equation is consistent:

$2x + 5y = 13$ and

$x + 6y = 10$

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

Explanation

$2x + 5y = 13$ ----- (1)

$x + 6y = 10$ ----- (2)
From equation (1) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$2x + 5y - 13 = 0$ ------ (3)
Put

$x = 6, y = 0.2$ in equation (3)

$2(6) + 5(0.2) - 13 = 0$

$13 - 13 = 0$

$0 = 0$
Again put

$x = 4, y = 1$ in equation (3)

$2(4) + 5(1)- 13 = 0$

$8 + 5 13 = 0$

$13- 13 = 0$

$0 = 0$
Now plotting

$(6, 0.2), (4, 1)$ and joining them, we get a straight line.
From equation (2) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$x + 6y - 10 = 0$ ----- (4)
Put

$x = -2, y = 2$ in equation (4)

$-2 + 6(2)- 10 = 0$

$-2 + 12- 10 = 0$

$10- 10 = 0$

$0 = 0$
Again put

$x = 4, y = 1$ in equation (4)

$4 + 6(1) - 10 = 0$

$4 + 6 - 10 = 0$

$10 -10 = 0$

$0 = 0$
Plotting

$(-2, 2), (4, 1)$ and joining them, we get another straight line.
These lines intersect at the point (4, 1) and therefore the solution of the equation is

$x = 4, y = 1$ .
In the above graph, the lines intersect each other at a point. In this case, the system will have exactly one solution. So, the equations are consistent

Check whether the system of equation is inconsistent:

$4x + y = 23$ and

$8x + 2y = 10$

0%
inconsistent
0%
consistent
0%
dependent
0%
non-linear

Explanation

Given,

$4x + y = 23$ and

$8x + 2y = 10$
The general form for a pair of linear equations in two variables

$x$ and

$y$ is:

$a_{1}x + b_{1}y + c_{1}= 0$ and

$4x + y = 23$ ---(1)

$8x + 2y = 10$ ---(2)
Comparing equations (1) and (2) with the general form of equation to consider their co-efficient,

$a_{1}=4$ ,

$a_{2} =8$ ,

$b_{1}= 1$ ,

$b_{2}= 2$ ,

$c_{1}= 23$ ,

$c_{2}= 10$
Now, ratio of coefficient =>

$\dfrac{4}{8}=\dfrac{1}{2}\neq\dfrac{23}{10}$
We know that when

$\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}\neq\dfrac{c_{1}}{c_{2}}$ then equations are said to be inconsistent.
Hence, the equations are inconsistent.

Using graphical method check whether the given equation is consistent:

$2x + 3y = 8$ and

$3x + 6y = 15$

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

Explanation

$2x + 3y = 8$ ----- (1)

$3x + 6y = 15$ ----- (2)
From equation (1) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$2x + 3y - 8 = 0$ ------ (3)
Put

$x = 4, y = 0$ in equation (3)

$2(4) + 3(0) - 8 = 0$

$8 - 8 = 0$

$0 = 0$
Again put

$x = 1, y = 2$ in equation (3)

$2(1) + 3(2) -8 = 0$

$2 + 6 -8 = 0$

$8- 8 = 0$

$0 = 0$
Now plotting

$(4, 0), (1, 2)$ and joining them, we get a straight line.
From equation (2) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$3x + 6y - 15 = 0$ ----- (4)
Put

$x = 5, y = 0$ in equation (4)

$3(5) + 6(0) -15 = 0$

$15 + 0- 15 = 0$

$15- 15 = 0$

$0 = 0$
Again put

$x = 1, y = 2$ in equation (4)

$3(1) + 6(2) -15 = 0$

$3 + 12- 15 = 0$

$15- 15 = 0$
Plotting

$(5, 0), (1, 2)$ and joining them, we get another straight line.
These lines intersect at the point

$(2, 1)$ and therefore the solution of the equation is

$x = 2, y = 1$ .
In the above graph, the lines intersect each other at a point. In this case, the system will have exactly one solution. So, the equations are consistent.

Solve the following pair of equations by reducing them to a pair of linear equations:

$\dfrac{2x-3y}{xy}=4$ and

$\dfrac{15x+3y}{xy}=30$

0%
$x=-1, y=0$
0%
$x=1, y=\propto$
0%
$x=0, y=2$
0%
$x=\propto , y=\frac{1}{2}$

Explanation

Simplify the equations,

$\dfrac{2x}{xy} - \dfrac{3y}{xy}=4$ ;

$\dfrac{15x}{xy}+\dfrac{3y}{xy} = 30$

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

$\dfrac{15}{y}+\dfrac{3}{x} = 30$
Assume

$\dfrac{1}{x} = a; \dfrac{1}{y}=b$

$2b-3a=4$ ----- (1)

$15b+3a=30$ ---- (2)
___________

$17b = 34$

$b=\dfrac{34}{17}=2$
Put the value of b in eqn (1)

$2(2) -3a=4$

$4-4=3a$

$0=3a$

$a=0$

$\therefore \dfrac{1}{x}=a; \dfrac{1}{y}=b$

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

$\therefore$ the solution is

$(x=\propto, y=\dfrac{1}{2})$

Using graphical method check whether the given equation is consistent:

$6x + 7y = 49$ and

$3x + 7y = 28$

0%
$(7, 1)$
0%
$(-7, -1)$
0%
$(-7, 1)$
0%
$(7, -1)$

Explanation

$6x + 7y = 49$ ----- (1)

$3x + 7y = 28$ ----- (2)
From equation (1) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$6x + 7y - 49 = 0$ ------ (3)
Put

$x = 0, y = 7$ in equation (3)

$6(0) + 7(7) - 49 = 0$

$49- 49 = 0$

$0 = 0$
Again put

$x = 7, y = 1$ in equation (3)

$6(7) + 7(1) - 49 = 0$

$42 + 7 - 49 = 0$

$49- 49 = 0$

$0 = 0$
Now plotting

$(0, 7), (7, 1)$ and joining them, we get a straight line.
From equation (2) assume the value of

$x$ and

$y$ to satisfy the equation to zero.

$3x + 7y - 28 = 0$ ----- (4)
Put

$x = 0, y = 4$ in equation (4)

$3(0) + 7(4) - 28 = 0$

$28- 28 = 0$

$0 = 0$
Again put

$x = 7, y = 1$ in equation (4)

$3(7) + 7(1) - 28 = 0$

$21 + 7 - 28 = 0$

$28- 28 = 0$

$0 = 0$
Plotting

$(0, 4), (7, 1)$ and joining them, we get another straight line.
These lines intersect at the point

$(7, 1)$ and therefore the solution of the equation is

$x = 7, y = 1$ .
In the above graph, the lines intersect each other at a point. In this case, the system will have exactly one solution. So, the equations are consistent.

Solve the following pairs of equations by reducing them to a pair of linear equations.

$\displaystyle \frac{3}{x+1}-\frac{1}{y+1}=2$ and

$\dfrac{6}{x+1}-\dfrac{1}{y+1}=5$

0%
(1, 1)
0%
(0, 1)
0%
(1, 0)
0%
(0, 0)

Explanation

Assume

$\dfrac{1}{x+1}=p; \dfrac{1}{y+1}=q$

$3p-q=2$ ----- (1)

$6p-q=5$ -----(2)
- + -
________

$-3p = -3$

$p=1$
Put the value of p in equation (1)

$3(1)-q=2$

$3-2=q$

$q=1$

$\dfrac{1}{x+1}=p$

$1=x+1$

$x=1-1$

$x=0$

$\dfrac{1}{y+1}=q$

$1=y+1$

$y=1-1$

$y=0$

$\therefore$ the solution is (0,0)

Which option of linear equation is inconsistent?

0%
$x + y = 1$ ,
$2x + y = 0$
0%
$x - y = 2$ ,
$2x - 2y = 1$
0%
$2x - 3y=0$ ,
$2y + x = 1$
0%
$2x - 4y = 19$ ,
$-2x + y = 0$

Explanation

The general form for a pair of linear equations in two variables

$x$ and

$y$ is

Option A:-

$x+y-1=0\Rightarrow a_1=1,\,b_1=1,\,c_1=-1$

$2x+y=0\Rightarrow a_2=2,\,b_2=1,\,c_2=0$

$\dfrac{a_1}{a_2}=\dfrac12$

$\dfrac{b_1}{b_2}=\dfrac11$

$\therefore \dfrac{a_1}{a_2}\ne\dfrac{b_1}{b_2}$
system of equations in option A is consistent

option B:-

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

$2x - 2y = 1$ ---(2)
Comparing equations (1) and (2) with the general form of equation to consider their co-efficient,

$a_{1}=1$ ,

$a_{2} =2$ ,

$b_{1}= -1$ ,

$b_{2}= -2$ ,

$c_{1}= 2$ ,

$c_{2}= 1$
So,

$\dfrac{1}{2}=\dfrac{-1}{-2}\neq\dfrac{2}{1}$
We know that

$\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}\neq\dfrac{c_{1}}{c_{2}}$ is said to be inconsistent.
Hence, the equations are inconsistent.

The equation

$-4x - 4y = 2$ and

$-2x - 2y = 1$ have _________.

0%
2 solutions
0%
unique solution
0%
no solutions
0%
infinitely many solutions

Explanation

The general form for a pair of linear equations in two variables

$x$ and

$y$ is

$a_{1}x + b_{1}y + c_{1}= 0$ and

$a_{2}x + b_{2}y + c_{2}= 0$

$-4x -4y = 2$ ---(1)

$-2x - 2y = 1$ ---(2)
Comparing equations (1) and (2) with the general form of equation to consider their co-efficient,

$a_{1}=-4, a_{2} =-2,b_{1}= -4, b_{2}= -2, c_{1}=2, c_{2}=1$
So,

$\dfrac{-4}{-2}=\dfrac{-4}{-2}=\dfrac{2}{1}$
We know that

$\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}$ is said to be dependent.
Hence, the equations are satisfying the above condition.
It is dependent and hence has infinitely many solutions

Solve the following pairs of linear equations by elimination method

$3x+4y=25$

$5x-6y=-9$

0%
$x=3$ and $y=3$
0%
$x=4$ and $y=4$
0%
$x=3$ and $y=4$
0%
None of these

Explanation

$3x+4y=25\_\_\_\_\_\_(i)\times 3$

$5x-6y=-9\_\_\_\_\_\_(ii)\times 2$

$9x+12y=75$

$10x-12y=-18$

(+)

$\quad$ (+)

$\quad\quad\quad$ (+)

$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

$19x\quad=57$

$\Rightarrow x=3$

$\Rightarrow 3 \times 3+4y=25$

$\Rightarrow 4y=16\Rightarrow y=4$

Two lines are represented by the equations

$2x-y=1$ and

$x+2y=2$ . Represent this situation graphically and check if the system has....

0%
No solution
0%
Unique Solution
0%
Infinite Solutions
0%
None of the these

Explanation

For

$2x - y = 1$ :

x	0	1	2
y	-1	1	3

For

$x + 2y = 2$ :

x	0	2	-2
y	1	0	2

As the lines are intersecting at one point. It has unique solution.

The equations

$4x - y = 1$ and

$2x - 2y = 2$ are __________.

0%
Non-linear
0%
Inconsistent
0%
Consistent
0%
Dependent

Explanation

The general form for a pair of linear equations in two variables

$x$ and

$y$ is

$a_{1}x + b_{1}y + c_{1}= 0$ and

$a_{2}x + b_{2}y + c_{2}= 0$

$4x - y - 1=0$ ---

$(1)$

$2x - 2y -2=0$ ---

$(2)$
Comparing equations

$(1)$ and

$(2)$ with the general form of equations to consider their co-efficient,

$a_{1}=4, a_{2} =2,b_{1}= -1, b_{2}= -2, c_{1}=-1, c_{2}=-2$
We have,

$\dfrac{4}{2}\neq\dfrac{-1}{-2}$

$\Rightarrow \dfrac{a_{1}}{a_{2}}\neq\dfrac{b_{1}}{b_{2}}$

Hence, the equations are consistent.

The sum of unit place and tens place digits of a two digit number is

$12$ . If the new number formed by reversing those digits is less than the original number by

$18$ . Find the original number.

0%
$70$
0%
$75$
0%
$80$
0%
$85$

Explanation

Let the unit place digit $= x$
Ten's place digit $= 12 - x$
Original number $=$ $10\times(12 - x) + x$
$= 120 - 10x + x$
$=120 - 9x$
Reversing the digit:
Unit digit $= 12 - x$
Ten's digit $= x$
New number formed $=$ $10\times x + 12 - x$
$= 9x + 12$
As new number is less than the original number by $18$
New number $=$ original number $- 18$
$9x + 12 = 120 - 9x - 18$
$18x = 120 - 30$
$18x =90$
$x = 5$
Original number $= 120 - 9x$
$= 120 - 9 \times 5$
$= 120 -45$
$= 75$
Original number $= 75.$

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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