How do you solve the following system?: 15x +57y =62, 3x -y = -2

Nov 16, 2015

$\left(x , y\right) = \left(\frac{26}{93} , \frac{36}{31}\right)$

Explanation:

Given:
[1]$\textcolor{w h i t e}{\text{XXX}} 15 x + 57 y = 62$
[2]$\textcolor{w h i t e}{\text{XXX}} 3 x - y = - 2$

Multiply [2] by 5
[3]$\textcolor{w h i t e}{\text{XXX}} 15 x - 5 y = - 10$

Subtract [3] from [1]
[4]$\textcolor{w h i t e}{\text{XXX}} 62 y = 72$

Divide [4] by 62
[5]$\textcolor{w h i t e}{\text{XXX}} y = \frac{36}{31}$

Substitute $\frac{36}{31}$ for $y$ in [2]
[6]$\textcolor{w h i t e}{\text{XXX}} 3 x - \frac{36}{31} = - 2$

Add $\frac{36}{31}$ to both sides of [6]
[7]$\textcolor{w h i t e}{\text{XXX}} 3 x = - \frac{26}{31}$

Divide [7] by 3
[8]$\textcolor{w h i t e}{\text{XXX}} - \frac{26}{93}$