Question

A pair of linear equations which has a unique solution x = 2 and y = - 3 is 

A. x + y = 1 and 2x - 3y = - 5

B. 2x+ 5y = - 11 and 4x + 10y = - 22 

C. 2x - y = 1 and 3x + 2y = 0 

D. x-4y-14 = 0 and 5x - y - 13 = 0

Answer 1

B. 2x+ 5y= -11 and 4x + 10y = -22

D. x-4y-14 = 0 and 5x - y - 13 = 0

If x = 2 and y = - 3 is a unique solution of any pair of equation, then these values must satisfy that pair of equations.

Putting the values in the equations for every option and checking it -

From option (b).

LHS = 2x + 5y = 2(2+(- 3) = 4 + (- 15) = - 11 = RHS

and

LHS = 4x + 10y = 4(2) +10(- 3) = 8 + (- 30) = - 22 = RHS

It satisfies the pair of linear equation and hence is the unique solution for the equation.

From option (d)

LHS = x – 4y – 14 = 2 – 4 x –3 – 14
= 2 + 12 – 14
= 14 – 14
= 0
= RHS

And

LHS = 5x – y – 13 = 5 x 2 –(–3) – 13
= 10 + 3 – 13
= 13 – 13
= 0
= RHS

Hence, x = 2 & y = –3 satisfies equations which are present in option (d).
Therefore, (D) is also correct.

Answer 2

Correct option:

B. 2x+ 5y= -11 and 4x + 10y = -22

D. x-4y-14 = 0 and 5x - y - 13 = 0

Solution:

As x = 2, y = –3 is unique solution of system of equations so these values of must satisfy both equations.

(a) x + y = –1 and 2x – 3y = –5

Put x = 2 and y = - 3 in both the equations.

LHS = x + y ⇒ 2 + (– 3) = 2 - 3 = –1 (RHS)

LHS = 2x – 3y ⇒ 2(2) –3(–3) ⇒ 4 + 9 = 13 \(\ne\) RHS

(b) 2x + 5y = –11 and 4x + 10y = –22

Put x = 2 and y = –3 in both the equations.

LHS = 2x + 5y ⇒ 2 × 2 + 5 (–3) ⇒ 4 – 15 = –11 = RHS

LHS = 4x + 10y ⇒ 4(2) + 10(–3) ⇒ 8 – 30 = –22 = RHS

(c) 2x – y = 1 and 3x + 2y = 0

Put x = 2 and y = –3 in both the equations.

LHS = 2x – y ⇒ 2(2) - (-3) = 4 + 3 ⇒ 7 \(\ne\) RHS

LHS = 3x + 2y ⇒ 3(2) + 2(–3) ⇒ 6 – 6 = 0 = RHS

(d) x – 4y – 14 = 0 and 5x – y – 13 = 0

Put x = 2 and y = –3 in both the equations.

LHS = x – 4y - 14 ⇒ 2 – 4(–3) - 14 ⇒ 2 + 12 - 14 = 0 = RHS

LHS = 5x – y -13 ⇒ 5(2) – (–3) - 13 ⇒ 10 + 3 - 13 = 0 = RHS

Hence, the pair of equations is (B) and (D).