We have

7(y + 3) – 2(x + 2) = 14

4(y – 2) + 3(x – 3) = 2

Lets simplify these equations. We can rewrite them,

7(y + 3) – 2(x + 2) = 14

⇒ 7y + 21 – 2x – 4 = 14

⇒ 7y – 2x + 17 = 14

⇒ 2x – 7y = 3 …(i)

4(y – 2) + 3(x – 3) = 2

⇒ 4y – 8 + 3x – 9 = 2

⇒ 3x + 4y – 17 = 2

⇒ 3x + 4y = 19 …(ii)

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

Lets multiply eq.(i) by 3 and eq.(ii) by 2, so that variable x in both the equations have same coefficient.

Recalling equations (i) & (ii),

2x – 7y = 3 [×3

3x + 4y = 19 [×2

⇒ - 29y = - 29

⇒ y = 1

Substitute y = 1 in eq.(i) or eq.(ii), as per convenience of solving.

Thus, substituting in equation (i), we get

2x – 7(1) = 3

⇒ 2x – 7 = 3

⇒ 2x = 7 + 3

⇒ 2x = 10

⇒ x = 5

Hence, we have x = 5 and y = 1