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.
