# Solve for x and y: , 5x = 2y + 7

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

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