Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 51

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Question 1
1 / -0

Find the value of a and b using substitution method:
$$a + b = 3$$ and $$-3a + 2b = 1$$

A
1 and 2

B
2 and 1

C
-2 and -1

D
2 and -1

Solution

$$a + b = 3$$ ------- (1)
$$-3a + 2b = 1$$ ------ (2)
From equation 1:
$$a = 3 - b$$
Substitute the value of a in equation 2 :
$$-3a + 2b = 1$$
$$-3(3- b) + 2b = 1$$
$$-9 + 3b + 2b = 1$$
$$5b = 10$$
$$b = \dfrac{10}{5} = 2$$
Now, Substitute $$b = 2$$ in equation 1:
$$a + b = 3$$
$$a + 2 = 3$$
$$a = 3- 2$$
$$a = 1$$
Therefore the solution is: $$a = 1$$ and $$b = 2$$
Question 2
1 / -0

Solve the equations using the method of substitution method :
$$3x+4y=-43$$ and $$-2x+3y=11$$

A
7 and 5

B
5 and -7

C
-5 and 7

D
5 and 7

Solution

$$3x+4y=-43$$ ------- (1)
$$-2x+3y=11$$ ------ (2)
From equation 1:
$$3x+4y=-43$$
$$4y=43-3x$$
$$y=\frac{43-3x}{4}$$
Substitute the value of $$y$$ in equation 2:
$$-2x+3(\frac{43-3x}{4})=11$$
$$-8x + 129-9x= 44$$
$$-17x = 44-129$$
$$x =\frac{-85}{-17}=5$$
The substitute the value of $$x$$ in equation 1:
$$3x+4y=43$$
$$3(5)+4y=43$$
$$15+4y=43$$
$$4y=43-15$$
$$4y=28$$
$$y=\frac{28}{4}$$
$$y=7$$
Therefore the solution is: $$x = 5$$ and $$y=7$$
Question 3
1 / -0

Solve the equations using graphical method: $$2x- y = 2$$ and $$-4x- y = -4$$

A
(1, -1)

B
(1, 0)

C
(2, -1)

D
(0, 1)

Solution

$$2x- y = 2$$ ----- (1)
$$-4x- y = -4$$ ----- (2)
From equation (1)
$$y = 2x - 2$$ ------ (3)
Assume the value of x = 1, 2 and put those values in equation (3)
If $$x = 1, y = 2(1)- 2 = 0$$
If $$x = 2, y = 2(2) - 2=2$$
Now plotting (1, 0), (2, 2) and joining them, we get a straight line.
From equation (2),
$$-4x- y = -4$$
$$y = -4x + 4$$ ------ (4)
Assume the value of $$x = 1, 2$$ and put those values in equation (3)
If $$x = 1, y =-4(1)+4= 0$$
If $$x = 2, y = -4(1)+4 = -2$$
Plotting (1, 0), (2, -2) and joining them, we get another straight line.
These lines intersect at the point (1, 0) and therefore the solution of the equation is $$x = 1, y = 0$$
Question 4
1 / -0

Solve the equations using graphical method: $$x + y = 1$$ and $$2x -3y = 7$$

A
(2, -1)

B
(-2, -1)

C
(-2, 1)

D
(2, 1)

Solution

$$x + y = 1$$ ------- (I)
$$2x - 3y = 7$$ ------ (II)
Question 5
1 / -0

Solve the following equations using substitution method:
$$2x+3y=13$$ and $$4x+5y=23$$

A
(-2, 3)

B
(2, 3)

C
(2, -3)

D
(-2, -3)

Solution

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 $$y=3$$
Question 6
1 / -0

Solve the equation using substitution method:
$$3x+y+1=0$$ and $$3x-4y=10$$

A
(0.4, 2.2)

B
(0.4, -2.2)

C
(0.4, 1.2)

D
(-0.4, 2.2)

Solution

$$3x+y+1=0$$ ------- (1)
$$3x-4y=10$$ --------(2)
From equation (1)
$$3x+y=-1$$
$$y=-1-3x$$
Substitute the value of $$y$$ in equation (2)
$$3x-4(-1-3x)=10$$
$$3x+4+12x=10$$
$$15x=10-4$$
$$x=\dfrac{6}{15}$$ or $$\dfrac{2}{5}\simeq 0.4$$
The substitute the value of x in equation (1)
$$3x+y=-1$$
$$3(\dfrac{2}{5})+y=-1$$
$$6+5y=-5$$
$$5y=-5-6$$
$$5y=-11$$
$$y=\dfrac{-11}{5}$$ or -2.2
therefore solution is $$x= 0.4$$ and $$y=-2.2$$
Question 7
1 / -0

Solve the equation using elimination method:
$$\displaystyle \frac{3}{x}+\frac{2}{y} = 7$$ and $$\displaystyle \frac{4}{x} -\frac{1}{y}=2$$

A
$$\left (1, \dfrac{1}{2}\right)$$

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

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

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

Solution

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 $$\left(1, \cfrac{1}{2}\right)$$.
Question 8
1 / -0

Solve the equation using substitution method:
$$x+y=-3$$ and $$x+2y=4$$

A
-10 and 7

B
10 and 7

C
10 and -7

D
-10 and -7

Solution

$$x+y=-3$$ ------(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 $$y=7$$
Question 9
1 / -0

Solve the equations using graphical method: $$x + y = 7$$ and $$2x - 3y = 9$$

A
$$(6, -1)$$

B
$$(-6, 1)$$

C
$$(6, 1)$$

D
$$(-6, -1)$$

Solution

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)
If $$x = 0, y =\dfrac{2x-9}{3} = \dfrac{2(0)-9}{3}=\dfrac{-9}{3} = - 3$$
If $$x = 6, y =\dfrac{2(6)-9}{3} = \dfrac{12-9}{3}=\dfrac{3}{3} =1$$

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 $$x = 6, y = 1$$ .
Question 10
1 / -0

Solve the equation using substitution method:
$$4x-6y=-4$$ and $$8x-2y=48$$

A
5.6 and 7.4

B
-7.4 and 5.6

C
7.4 and 5.6

D
8.6 and 5.6

Solution

$$4x-6y=-4$$ ------- (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 $$y=5.6$$

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

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

Pair of Linear Equations in Two Variables Test - 51

Result

Pair of Linear Equations in Two Variables Test - 51

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Rank

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SHARING IS CARING

Question 1

1 and 2

2 and 1

-2 and -1

2 and -1

Solution

Question 2

7 and 5

5 and -7

-5 and 7

5 and 7

Solution

Question 3

(1, -1)

(1, 0)

(2, -1)

(0, 1)

Solution

Question 4

(2, -1)

(-2, -1)

(-2, 1)

(2, 1)

Solution

Question 5

(-2, 3)

(2, 3)

(2, -3)

(-2, -3)

Solution

Question 6

(0.4, 2.2)

(0.4, -2.2)

(0.4, 1.2)

(-0.4, 2.2)

Solution

Question 7

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

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

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

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

Solution

Question 8

-10 and 7

10 and 7

10 and -7

-10 and -7

Solution

Question 9

$$(6, -1)$$

$$(-6, 1)$$

$$(6, 1)$$

$$(-6, -1)$$

Solution

Question 10

5.6 and 7.4

-7.4 and 5.6

7.4 and 5.6

8.6 and 5.6

Solution

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