Question

On solving $3x-4y+11=0\mathrm{and}7\mathrm{x}+11\mathrm{y}-15=0,$ we get:
(a) x = 1, y = −2
(b) x = −1, y = 2
(c) x = 0, y = $\frac{15}{11}$
(d) x = $\frac{-11}{3}$, y = 0

Answer 1

The correct option is (b).

The given equations are as follows:
3x − 4y = −11 ...(i)
7x + 11y = 15 ...(ii)
On multiplying (i) by 7 and (ii) by 3, we get:
21x − 28y = −77 ...(iii)
21x + 33y = 45 ...(iv)
On subtracting (iii) from (iv), we get:
61y = 122 ⇒ y = 2
On substituting y = 2 in (i), we get:
3x − 8 = −11 ⇒ 3x = (−11 + 8)= −3 ⇒ x = −1
∴​ x = −1 and y = 2