# How do you solve the system using the elimination method for R – 9S = 2 and 3R – 3S = -10?

Jul 10, 2015

I found:
$R = - 4$
$S = - \frac{2}{3}$

#### Explanation:

I would multiply the first equation by $- 3$ to get:
{-3R+27S=-6
{3R-3S=-10 add:
$0 + 24 S = - 16$
so:
$S = - \frac{16}{24} = - \frac{2}{3}$
substitute back into the first equation:
$R - 9 \left(- \frac{2}{3}\right) = 2$
$R + 6 = 2$
$R = - 4$