# How do you solve the system using the elimination method for 2x-y=-3 and x+3y =16?

Jun 27, 2015

I found:
$x = 1$
$y = 5$

#### Explanation:

Isolate, say, $x$ from the second equation as:
$x = 16 - 3 y$ (1)
Substitute for $x$ into the first equation:
$2 \left(16 - 3 y\right) - y = - 3$
$32 - 6 y - y = - 3$
$- 7 y = - 35$
$y = \frac{- 35}{-} 7 = 5$
substitute this value into equation (1):
$x = 16 - 3 \cdot 5 = 1$