How do you solve the following system?: 16x +3y =7, 9x +5y = -12

Jan 4, 2016

$\left(x , y\right) = \left(\frac{71}{53} , - \frac{255}{53}\right)$

Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} 16 x + 3 y = 7$
$\textcolor{w h i t e}{\text{XXX}} 9 x + 5 y = - 12$

Multiply  by $5$ and  by $3$ to get equations with equal coefficients for $y$
$\textcolor{w h i t e}{\text{XXX}} 80 x + 15 y = 35$
$\textcolor{w h i t e}{\text{XXX}} 27 x + 15 y = - 36$

Subtract  from 
$\textcolor{w h i t e}{\text{XXX}} 53 x = 71$

Divide by $23$
$\textcolor{w h i t e}{\text{XXX}} x = \frac{71}{53}$
$\textcolor{w h i t e}{\text{XXX}}$Yes; it's going to be ugly; but that doesn't mean it's going to be wrong.

Substitute $\frac{71}{53}$ for $x$ in 
$\textcolor{w h i t e}{\text{XXX}} \frac{16 \times 71}{53} + 3 y = 7$

Simplify
$\textcolor{w h i t e}{\text{XXX}} \frac{1136}{53} + \frac{159}{53} y = \frac{371}{53}$

$\textcolor{w h i t e}{\text{XXX}} 159 y = - 765$

$\textcolor{w h i t e}{\text{XXX}} y = - \frac{765}{159} = - \frac{255}{23}$

I would (strongly) recommend using a calculator or spreadsheet to verify these values in the original equations.