# How do you solve 3r-5s=-35 and 2r-5s=-30?

Sep 5, 2015

The solution is:
color(blue)(r=-5, s=4

#### Explanation:

$3 r - \textcolor{b l u e}{5 s} = - 35$ ....equation $1$
$2 r - \textcolor{b l u e}{5 s} = - 30$......equation $2$

We can solve the system of equations by elimination as subtracting the two would eliminate color(blue)(5s

Subtracting equation $2$ from $1$
$3 r - \cancel{\textcolor{b l u e}{5 s}} = - 35$
$- 2 r + \cancel{\textcolor{b l u e}{5 s}} = + 30$

color(blue)(r=-5

Finding $s$ by substituting the value of $r$ in equation $1$
$3 r - 5 s = - 35$

$3 r + 35 = 5 s$

$3 r + 35 = 5 s$

$3 \cdot \left(\textcolor{b l u e}{- 5}\right) + 35 = 5 s$

$- 15 + 35 = 5 s$

$20 = 5 s$

$\frac{20}{5} = s$

color(blue)(s=4