Question

$$\begin{gathered} \frac{x}{3}+\frac{y}{4}=11, \\ \frac{5x}{6}-\frac{y}{3}+7=0. \end{gathered}$$

Answer 1

The given equations are:
$\frac{x}{3}+\frac{y}{4}=11$
⇒ 4x + 3y = 132 ........(i)

and $\frac{5x}{6}-\frac{y}{3}+7=0$
⇒ 5x − 2y = −42..........(ii)

On multiplying (i) by 2 and (ii) by 3, we get:
8x + 6y = 264 ........(iii)
15x − 6y = −126 ...........(iv)

On adding (iii) and (iv), we get:
23x = 138
⇒ x = 6

On substituting x = 6 in (i), we get:
24 + 3y = 132
⇒ 3y = (132 − 24) = 108
⇒ y = 36

Hence, the solution is x = 6 and y = 36.