Question

Question 10
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

If x = 2, y = - 3 is a unique solution of any pair of equation, then these values must satisfy both the equations in that pair of equations.
Substituting these values in the equations given in the options,

From option (A)
LHS = x + y = (2) + (-3) = 2 – 3 = - 1 $\ne$ RHS
$\therefore$ A cannot be the answer

From option (B)
LHS = 2x + 5y = 2(2) + 5(-3) = 4 – 15 = - 11 = RHS
LHS = 4x + 10y = 4(2) + 10( - 3) = 8 – 30 = -22 = RHS

From option (C)
LHS = 2x - y = 2(2) - (-3) = 4 + 3 = 7 $\ne$ RHS
$\therefore$ C cannot be the answer

From option (D)
LHS = x - 4y + 14 = (2) - 4(-3) + 14 = 2 + 12 + 14 = 28 $\ne$ RHS
$\therefore$ D cannot be the answer

$\therefore$ , Required answer is (B)