# How do you solve the following system?:  2x + 7y = -8, -x-8y=32

Apr 25, 2016

$x = \frac{160}{9}$ and $y = - \frac{56}{9}$.

#### Explanation:

The equations are $2 x + 7 y = - 8$ --(A) and $- x - 8 y = 32$ -- (B)

Now multiplying (B) by $2$ and adding to (A) cancels out $x$, and we get

$7 y - 16 y = - 8 + 64$ or $- 9 y = 56$ or $y = - \frac{56}{9}$. Now putting this in (B), we get

$- x - 8 \times - \frac{56}{9} = 32$ or $- x + \frac{448}{9} = 32$ or $- x = 32 - \frac{448}{9} = \frac{288}{9} - \frac{448}{9} = - \frac{160}{9}$ or $x = \frac{160}{9}$