Question

On solving $31x+47y=15\mathrm{and}47\mathrm{x}+31\mathrm{y}=63,$ we get:
(a) x = 2, y = −1
(b) x = −2, y = 1
(c) x = 1, y = 2
(d) x = 3, y = −2

Answer 1

The correct option is (a).

The given equations are as follows:
31x + 47y = 15 ....(i)
47x + 31y = 63 ....(ii)
On adding (i) and (ii), we get:
78x + 78y = 78
78(x + y) = 78
⇒ (x + y) = 1 ....(iii)
On subtracting (i) from (ii), we get:
16x − 16y = 48
16(x − y) = 48
⇒(x − y) = 3 ....(iv)
On adding (iii) and (iv), we get:
2x = 4 ⇒ x = 2
On substituting x = 2 in (iii), we get:
2 + y = 1 ⇒ y = −1
∴ x = 2 and y = −1