Question

On solving $x+\frac{6}{y}=11,3x+\frac{8}{y}=28,$ we get:
(a) x = 5, y = 1
(b) x = 8, y = 2
(c) x = 4, y = $\frac{1}{2}$
(d) x = 9, y = 3

Answer 1

The correct option is (b).

The given equations are as follows:
$x+\frac{6}{y}=11$
$3x+\frac{8}{y}=28$
Putting $\frac{1}{y}=u$ in the given equations, we get:
x + 6u = 11 ....(i)
3x + 8u = 28 ....(ii)
On multiplying (i) by 3, we get:
3x + 18u = 33 ....(iii)
On subtracting (ii) from (iii), we get:
10u = 5 ⇒ u = $\frac{1}{2}$ ⇒ $\frac{1}{y}=\frac{1}{2}\Rightarrow y=2$
Putting u = $\frac{1}{2}$ in (ii), we get:
3x + 4 = 28 ⇒ 3x = 24 ⇒ x = 8
∴​ x = 8 and y = 2