## Algebra: A Combined Approach (4th Edition)

2x + 3y = 13 ; x = y + 4 Substitute x = y + 4 to the first equation. 2 $\times$ (y + 4) + 3y = 13 2y + 8 + 3y = 13 5y + 8 = 13 5y + 8 - 8 = 13 - 8 5y = 5 y = $\frac{5}{5}$ y = 1 Substitute y = 1 to the second equation x = 1 + 4 x = 5