# How do you solve using elimination of x-15y=3 and 3x+y=21?

Dec 20, 2015

$\left(x , y\right) = \left(\frac{159}{23} , \frac{6}{23}\right)$

#### Explanation:

Given
[1]$\textcolor{w h i t e}{\text{XXX}} x - 15 y = 3$
[2]$\textcolor{w h i t e}{\text{XXX}} 3 x + y = 21$

Eliminate the $x$ terms by subtracting $3 \times$[1] from [2]

{: ([2],color(white)("XXX"),,"(",3x,+y,"=",21,")"), (3xx[1],color(white)("XXX"),-,"(",3x,-45y,"=",9,")") :}

[4]$\textcolor{w h i t e}{\text{XXX}} 46 y = 12$
[5]$\textcolor{w h i t e}{\text{XXX}} y = \frac{6}{23}$

Eliminate the $y$ terms by adding $15 \times \left[2\right]$ to [1]

{: ([1],color(white)("XXX"),,"(",x,-15y,"=",3,")"), (15xx[2],color(white)("XXX"),+,"(",45x,+15y,"=",315,")") :}

[6]$\textcolor{w h i t e}{\text{XXX}} 46 x = 318$
[7]$\textcolor{w h i t e}{\text{XXX}} x = \frac{159}{23}$

Dec 20, 2015

I found:
$x = \frac{159}{23}$
$y = \frac{6}{23}$

#### Explanation:

We can multiply the first equation by $- 3$ and add (in column) to the second:
$\left\{\begin{matrix}- 3 \times \left[x - 15 y = 3\right] \\ 3 x + y = 21\end{matrix}\right.$
$\left\{\begin{matrix}- 3 x + 45 y = - 9 \\ 3 x + y = 21\end{matrix}\right.$ add:
$0 + 46 y = 12$
So:
$y = \frac{12}{46} = \frac{6}{23}$
use this into the first equation:
$x - 15 \cdot \textcolor{red}{\frac{6}{23}} = 3$
$x = \frac{90}{23} + 3 = \frac{159}{23}$