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
Solution:
(b) If x = 2, y = – 3 is a unique solution of any pair of equation, then these values must satisfy that pair of equations.
From option (b), LHS = 2x + 5y = 2(2) + 5(- 3) = 4 – 15 = – 11 = RHS
and LHS = 4x + 10y = 4(2) + 10(- 3)= 8 – 30 = – 22 = RHS
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