Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 52

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Question 1
1 / -0

Solve the equations using graphical method: $$x- y = 3$$ and $$x + y = 11$$

A
$$(7, 4)$$

B
$$(-7, 3)$$

C
$$(7, -4)$$

D
$$(-7, 4)$$

Solution

$$x- y = 3$$ ----- (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 $$x = 7, y = 4$$ .
Question 2
1 / -0

Find the value x and y using substitution method:
$$2x-5y=9$$ and $$5x+6y=8$$

A
$$94$$ and $$199$$

B
$$\cfrac{94}{43}$$ and $$\cfrac{199}{215}$$

C
$$\cfrac{-94}{43}$$ and $$\cfrac{199}{215}$$

D
$$\cfrac{-94}{43}$$ and $$\cfrac{-199}{215}$$

Solution

$$2x-5y=9$$ -----(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 $$y=\cfrac{199}{215}$$
Question 3
1 / -0

Solve the equations graphically: $$2x- y = -8$$ and $$x + 2y = -14$$

A
$$(6, 4)$$

B
$$(-6, -4)$$

C
$$(-6, 4)$$

D
$$(6, -4)$$
Question 4
1 / -0

Solve the equations: $$5x -2 y = 6$$ and $$2x + 5y = 3$$ and then identify the system of equations.

A
Consistent

B
Inconsistent

C
Dependent

D
All the above

Solution

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.
Question 5
1 / -0

Find the value of x and y using graphical method: $$8x -y = 6$$ and $$6x - y = 4$$

A
$$(1, 2)$$

B
$$(-1, -2)$$

C
$$(-1, 2)$$

D
$$(1, -2)$$

Solution

$$8x -y = 6$$ ----- (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 $$ x = 1, y = 2$$.
Question 6
1 / -0

Solve the equations: $$5x + 8y = 9$$ and $$2x + 3y = 4$$ using graphical method.

A
(5, -2)

B
(-5, 2)

C
(5, 2)

D
(-5, -2)

Solution

$$5x + 8y = 9$$ ----- (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 $$x = 5, y = -2$$.
Question 7
1 / -0

Solve the equations: $$x -5y = 15$$ and $$2x + 5y = 0$$ using graphical method.

A
(5, 2)

B
(-5, 2)

C
(5, -2)

D
(-5, -2)

Solution

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 $$(5,-2)$$.
Question 8
1 / -0

Solve the equations: $$x- y = -1$$ and $$2y + 3x = 12$$ using graphical method.

A
$$(1, -2)$$

B
$$(2, 3)$$

C
$$(3, 3)$$

D
$$(-1, -2)$$

Solution

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

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

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 $$x = 2, y = 3$$ .
Question 9
1 / -0

Solve the equations: $$2x + 2y = 4$$ and $$-2x + 3y = 1$$ using graphical method.

A
(1, 1)

B
(-1, 1)

C
(1, -1)

D
(-1, -1)

Solution

$$2x + 2y = 4$$ ----- (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 $$x = 1, y = 1$$
Question 10
1 / -0

Solve the equations: $$x- y = -5$$ and $$x + y = 1$$ using graphical method

A
(-2, -3)

B
(-2, 3)

C
(3, 2)

D
(-3, -2)

Solution

$$x- y = -5$$ ----- (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 $$x = -2, y = 3$$