# How do solve the following linear system?:  6x-4y=-1 , 3 x-y=4 ?

Jun 1, 2016

Given
$\textcolor{w h i t e}{\text{XXX}} 6 x - 4 y = - 1$
$\textcolor{w h i t e}{\text{XXX}} 3 x - y = 4$

Subtract 2 times  from 
$\textcolor{w h i t e}{\text{XXXxXX}} 6 x - 4 y = - 1$
$2 \times$color(white)("XXX")underline(6x-2y=color(white)("X"x)8)
$\textcolor{w h i t e}{\text{XXXXXxXX}} - 2 y = - 9$

Divide both sides of  by $\left(- 2\right)$
$\textcolor{w h i t e}{\text{XXXXXXXXX}} y = \frac{9}{2}$

Substitute $\frac{9}{2}$ for $y$ in 
$\textcolor{w h i t e}{\text{XXXXXX}} 6 x - 4 \left(\frac{9}{2}\right) = - 1$

Simplify
$\textcolor{w h i t e}{\text{XXXXXX}} 6 x - 18 = - 1$

$\textcolor{w h i t e}{\text{XXXXXX}} 6 x = 17$

$\textcolor{w h i t e}{\text{XXXXXX}} x = \frac{17}{6}$

So the solution is $\left(x , y\right) = \left(\frac{17}{6} , \frac{9}{2}\right)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As always you should verify this result.

In this case substitute $\frac{17}{6}$ for $x$ and $\frac{9}{2}$ for $y$ in 

$\textcolor{w h i t e}{\text{XXX}} 3 x - y$

$\textcolor{w h i t e}{\text{XXXXXX}} = 3 \left(\frac{17}{6}\right) - \frac{9}{2}$

$\textcolor{w h i t e}{\text{XXXXXX}} = \frac{17}{2} - \frac{9}{2}$

$\textcolor{w h i t e}{\text{XXXXXX}} = \frac{8}{2}$

$\textcolor{w h i t e}{\text{XXXXXX}} = 4$ (which agrees with  so our work is valid)