Test: Pair of Linear Equations in Two Variables (Easy) - Class 10 MCQ

13x + 12y = 21....(2)

Adding (1) and (2)

25x + 25y = 50

x + y = 2

2x − 3y = 7 ....(1)

and 5x + y = 9 ....(2)

Multiply equation (2) by 3, we get

15x + 3y = 27 ....(3)

Add equations (1) and (3),

17x = 34

⇒ x = 2

Put this value in equation (2), we get 5(2) + y = 9 ⇒ 10 + y = 9

⇒ y = −1

Therefore, the solution is x = 2,y = −1.

x + y = 0 ...(2)

From equation 1 : y = x - 3

Substitute the value of y in equation 2: x + y = 0x + x - 3 = 0

2x = 3X= 3 / 2

Now, Substitute x = 3 / 2 in equation 1: x - y = 3x

3 / 2 - y = 3y = 3 / 2 - 3y = -3 / 2.

Therefore the solution is: x = 3 / 2 and y = -3 / 2

From (1), we get x = .....(3)

From (2), we get x = 5 - 2y …...(4)

= 5 - 2y or 8x - 3y = 10 - 4y or 3y = 10 - 4y or y = 2

Using y = 2 in (4), we get x = 5 - 4 = 1.

Therefore, x = 1, y = 2

Given, 25x + 20y = 430

5x + 4y = 86 ....(1)

Also x + y = 18 ....(2)

Multiply equation (1) by 4, we get 4x + 4y = 72 ....(3)

Subtract equations (1) and (3), we get x = 14

Hence, answer is 14.

6x − 14y = 70.........(1)

6x + 15y = 12.........(2)

Subtract Equation (2) from Equation (1) to eliminate x, because the coefficients of x are the same.

So, we get (6x − 6x) + (−14y − 15y) = 70 − 12

i.e. −29y = 58 i.e. y = −2

Substituting this value of y in (2), we get

6x − 30 = 12

i.e. 6x = 42 i.e. x = 7

Hence, the solution of the equations is x = 7, y = −2.

Then, 30 / x - 30 / 2x = 3

⇒ 6x = 30 ⇒ x = 5km/hr.

(i) If we interchange the digits of Original number then we get new number, i.e 10y + x

According to the given condition 10y + x + 18 = 10x + y ⇒ 9x - 9y = 18 ⇒ x — y = 2

(ii) Now multiplying equation (ii) by 4. we get, 4x - 4y = 8 ....(iii) On subtracting (iii)

from (i). we get, 4x - 5y - (4x - 4y) = 3 - 8 A y = 5 On substituting y = 5 in (ii). we get, x -

5 = 2 ⇒ x = 7 A number is 10x + y = 10(7) + 5 = 75

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 put x = 0, y = 4

in equation...(4)

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

28 - 28 = 0

0 = 0 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.

42x − 31y = 95 ---------(ii)

On adding (i) and (ii), we get 73x − 73y = 146

⇒ x − y = 2 ...(iii)

On subtracting (i) from (ii), we get 11x + 11y = 44

⇒ x + y = 4 ....(iv)

On adding (iii) and (iv), we get

2x - 6 ⇒ x = 3

By putting x = 3

3 + y = 4

⇒ y = 1

∴ x = 3, y = 1

a₁ / a₂ = 6 / 2 = 3 / 1

b₁ / b₂ = -3 / -1 = 3 / 1

c₁ / c₂ = 10 / 9

This implies

a₁ / a₂ = b₁ / b₂ ≠ c₁ / c₂

Here,

a₁ / a₂ = 1 / -3

b₁ / b₂ = 2 / -6 = 1 / -3

c₁ / c₂ = -5 / 15 = -1 / 3

This implies

a₁ / a₂ = b₁ / b₂ = c₁ / c₂