Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 53

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Question 1
1 / -0

Solve the equations graphically: $$7x + 8y = 64$$ and $$x + 4y = 12$$

A
$$(8, 1)$$

B
$$(-8, -1)$$

C
$$(-8, 1)$$

D
$$(8, -1)$$

Solution

$$7x + 8y = 64$$ ----- (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 $$x = 8, y = 1$$
Question 2
1 / -0

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

A
$$(1, -2)$$

B
$$(-1, 2)$$

C
$$(1, 2)$$

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

Solution

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 $$(1, 2)$$.
Question 3
1 / -0

Check whether the system of equation is inconsistent:
$$4x + y = 23$$ and $$8x + 2y = 10$$

A
inconsistent

B
consistent

C
dependent

D
non-linear

Solution

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.
Question 4
1 / -0

Using graphical method check whether the given equation is consistent: $$2x + 5y = 13$$ and $$x + 6y = 10$$

A
$$(4, 1)$$

B
$$(-4, -1)$$

C
$$(-4, 1)$$

D
$$(4, -1)$$

Solution

$$2x + 5y = 13$$ ----- (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
Question 5
1 / -0

Using graphical method check whether the given equation is consistent: $$2x + 3y = 8$$ and $$3x + 6y = 15$$

A
$$(2, 1)$$

B
$$(-5, -1)$$

C
$$(-2, 1)$$

D
$$(2, -1)$$

Solution

$$2x + 3y = 8$$ ----- (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.
Question 6
1 / -0

Solve the following pair of equations by reducing them to a pair of linear equations:
$$\dfrac{2x-3y}{xy}=4$$ and $$\dfrac{15x+3y}{xy}=30$$

A
$$x=-1, y=0$$

B
$$x=1, y=\propto $$

C
$$x=0, y=2$$

D
$$x=\propto , y=\frac{1}{2}$$

Solution

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 $$(x=\propto, y=\dfrac{1}{2})$$
Question 7
1 / -0

Using graphical method check whether the given equation is consistent: $$6x + 7y = 49$$ and $$3x + 7y = 28$$

A
$$(7, 1)$$

B
$$(-7, -1)$$

C
$$(-7, 1)$$

D
$$(7, -1)$$

Solution

$$6x + 7y = 49$$ ----- (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.
Question 8
1 / -0

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

A
(1, 1)

B
(0, 1)

C
(1, 0)

D
(0, 0)

Solution

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)
Question 9
1 / -0

Which option of linear equation is inconsistent?

A
$$x + y = 1$$,
$$2x + y = 0$$

B
$$x - y = 2$$,
$$2x - 2y = 1$$

C
$$2x - 3y=0$$,
$$2y + x = 1$$

D
$$2x - 4y = 19$$,
$$-2x + y = 0$$

Solution

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.
Question 10
1 / -0

Two lines are represented by the equations $$2x-y=1$$ and $$x+2y=2$$. Represent this situation graphically and check if the system has....

A
No solution

B
Unique Solution

C
Infinite Solutions

D
None of the these

Solution

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.