We have,

…eq.1

…eq.2

Let us first simplify eq.1 & eq.2, by taking LCM of denominators,

Eq.1 ⇒

⇒

⇒ 4x + 3y = 132 …eq.3

Eq.2 ⇒

⇒

⇒ 5x – 2y = - 42 …eq.4

To solve these equations, we need to make one of the variables (in both the equations) have same coefficient.

Lets multiply eq.3 by 2 and eq.4 by 3, so that variable y in both the equations have same coefficient.

Recalling equations 3 & 4,

4x + 3y = 132 [×2

5x – 2y = - 42 [×3

⇒ 8x + 6y = 264

15x – 6y = - 126

23x + 0 = 138

⇒ 23x = 138

⇒ x = 6

Substitute x = 6 in eq.3/eq.4, as per convenience of solving.

Thus, substituting in eq.4, we get

5(6) – 2y = - 42

⇒ 30 – 2y = - 42

⇒ 2y = 30 + 42

⇒ 2y = 72

⇒ y = 36

Hence, we have x = 6 and y = 36.