# How do you solve the following system: 5x-9y=-2, 7x-2y= -67 ?

Apr 10, 2018

$\left(x , y\right) = \left(- 11 \frac{16}{53} , - 6 \frac{3}{53}\right)$

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXX}} 5 x - 9 y = - 2$
[2]$\textcolor{w h i t e}{\text{XXX}} 7 x - 2 y = - 67$

Convert to equivalent equations with the same coefficient for $y$
by multiplying [1] by $2$
and multiplying [2] by $9$
[3]$\textcolor{w h i t e}{\text{XXX}} 10 x - 18 y = - 4$
[4]$\textcolor{w h i t e}{\text{XXX}} 63 x - 18 y = - 603$

Subtract [3] from [4] to eliminate the $y$ term
[5]$\textcolor{w h i t e}{\text{XXX}} 53 x = - 599$

Divide both sides of [5] by $53$
[6]$\textcolor{w h i t e}{\text{XXX}} x = - \frac{599}{53} = - 11 \frac{16}{53}$

We could substitute $- \frac{599}{53}$ for $x$ in one of the given equation (either one) and solve for $y$
or repeat a similar process to the one above to eliminate the $x$ factors:

Multiply [1] by $7$ and [2] by $5$
[7]$\textcolor{w h i t e}{\text{XXX}} 35 x - 63 y = - 14$
[8]$\textcolor{w h i t e}{\text{XXX}} 35 x - 10 y = - 335$

Subtract [7] from [8]
[9]$\textcolor{w h i t e}{\text{XXX}} 53 y = - 321$

Divide both sides of [9} by $53$
[10]$\textcolor{w h i t e}{\text{XXX}} y = - \frac{321}{53} = - 6 \frac{3}{53}$