We have,

…eq.1

5x = 2y + 7 or 5x – 2y = 7 …eq.2

Let us first simplify eq.1 by taking LCM of denominator,

Eq.2 ⇒

⇒ 8x – 3y = 12 …eq.3

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

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

Recalling equations 2 & 3,

5x – 2y = 7 [×3]

8x – 3y = 12 [×2]

⇒ 15x – 6y = 21

16x – 6y = 24

On solving the above equations we get,

- x – 0 = - 3

⇒ - x = - 3

⇒ x = 3

Substitute x = 3 in eq.2/eq.3, as per convenience of solving.

Thus, substituting in eq.2, we get

5(3) – 2y = 7

⇒ 15 – 2y = 7

⇒ 2y = 15 – 7

⇒ 2y = 8

⇒ y = 4

Hence, we have x = 3 and y = 4