# How do you solve 1/2x+1/3y=-1/6 and 1/2x+2y=13/2 using substitution?

May 9, 2016

(x,y)=color(blue)(""(-3,4))

#### Explanation:

To simplify this process multiply each equation by appropriate constants to clear the fractions:
[1a]$\textcolor{w h i t e}{\text{XXX}} \frac{1}{2} x + \frac{1}{3} y = - \frac{1}{6}$
$\textcolor{w h i t e}{\text{XXXXXXXXXXXXX}} \rightarrow$[1b]$\textcolor{w h i t e}{\text{XXX}} 3 x + 2 y = - 1$

[2a]$\textcolor{w h i t e}{\text{XXX}} \frac{1}{2} x + 2 y = \frac{13}{2}$
$\textcolor{w h i t e}{\text{XXXXXXXXXXXXX}} \rightarrow$[2b]$\textcolor{w h i t e}{\text{XXX}} x + 4 y = 13$

[2b} appears to be the easiest one to convert into a form for substitution:
[2c]$\textcolor{w h i t e}{\text{XXXXXXXXXXX}} x = 13 - 4 y$

Substituting this back into [1b]
$\textcolor{w h i t e}{\text{XXXXX}} 3 \left(13 - 4 y\right) + 2 y = - 1$

$\textcolor{w h i t e}{\text{XXXXX}} 39 - 10 y = - 1$

$\textcolor{w h i t e}{\text{XXXXX}} 10 y = 40$

$\textcolor{w h i t e}{\text{XXXXX}} y = 4$

We can now substitute $4$ for $y$ back in [2c] to get
$\textcolor{w h i t e}{\text{XXXXX}} x = - 3$

[as always, you should substitute these values back into the original equations to verify the results... assume that I've done this.]