Question

On solving $3x+4y+7=0\mathrm{and}5\mathrm{x}-7\mathrm{y}-2=0,$ we get:
(a) x = 1, y = −1
(b) x = −1, y = 1
(c) x = −1, y = −1
(d) x = 0, y = $\frac{-2}{7}$

Answer 1

The correct option is (c).

The given equations are as follows:
3x + 4y = −7 ...(i)
5x − 7y = 2 ...(ii)
On multiplying (i) by 5 and (ii) by 3, we get:
15x + 20y = −35 ...(iii)
15x − 21y = 6 ...(iv)
On subtracting (iii) from (iv), we get:
−41y = 41 ⇒ y = −1
On substituting the value of y in (i), we get:
3x − 4 = − 7 ⇒ 3x = (−7 + 4)= −3 ⇒ x = −1
∴​ x = −1 and y = −1