# How do you solve the following system?:  -5x - 5y = -6 , 3x – y = -1

##### 1 Answer
Feb 7, 2016

$\left(x , y\right) = \left(\frac{1}{20} , \frac{23}{20}\right)$

#### Explanation:

Given:
[1]$\textcolor{w h i t e}{\text{XXX}} - 5 x - 5 y = - 6$
[2]$\textcolor{w h i t e}{\text{XXX}} 3 x - y = - 1$

Rearrange [2] to isolate the $y$ term on the left side:
[3]$\textcolor{w h i t e}{\text{XXX}} - y = - 1 - 3 x$

(the next two steps are not necessary, but I find it easier to work with positives)
Multiply both [1] and [3] by $\left(- 1\right)$ to reverse the signs
[4]$\textcolor{w h i t e}{\text{XXX}} 5 x + 5 y = 6$
[5]$\textcolor{w h i t e}{\text{XXX}} y = 1 + 3 x$

Substitute $\left(1 + 3 x\right)$ from [5] for $y$ in [4]
[6]$\textcolor{w h i t e}{\text{XXX}} 5 x + 5 \left(1 + 3 x\right) = 6$

Simplify [6]
[7]$\textcolor{w h i t e}{\text{XXX}} 20 x + 5 = 6$

Subtract $5$ from both sides of [7]
[8]$\textcolor{w h i t e}{\text{XXX}} 20 x = 1$

Divide both sides of [8] by $20$
[9]$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{x = \frac{1}{20}}$

Substitute $\frac{1}{20}$ from [9] for $x$ in [2]
[10]$\textcolor{w h i t e}{\text{XXX}} 3 \times \frac{1}{20} - y = - 1$

Subtract $\left(\frac{3}{20}\right)$ from both sides of [10]
[11]$\textcolor{w h i t e}{\text{XXX}} - y = - \frac{23}{20}$

Multiply both sides of [11] by $\left(- 1\right)$
[12]$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{y = \frac{23}{20}}$