# How do you solve using gaussian elimination or gauss-jordan elimination, -2x -3y = -7, 5x - 16 = -6y?

Jan 14, 2016

#### Answer:

So $x = \frac{10}{3}$ and $y = \frac{1}{9}$

#### Explanation:

You have
$\left[\begin{matrix}- 2 & - 3 & | - \textcolor{w h i t e}{0} 7 \\ 5 & - 6 & | \textcolor{w h i t e}{-} 16\end{matrix}\right]$

After a quick switch of the last equation to a $a x + b y = c$ format, now we want

$\left[\begin{matrix}1 & 0 & | h \\ 0 & 1 & | q\end{matrix}\right]$

So we know $x = h$ and $y = q$ as solutions.

Adding $\frac{5}{2} {L}_{1}$ to ${L}_{2}$ we have

$\left[\begin{matrix}- 2 & - 3 & | - \textcolor{w h i t e}{0} 7 \textcolor{w h i t e}{+ \frac{5}{2} \left(- 7\right)} \\ 5 + \frac{5}{2} \left(- 2\right) & - 6 + \frac{5}{2} \left(- 3\right) & | \textcolor{w h i t e}{-} 16 + \frac{5}{2} \left(- 7\right)\end{matrix}\right]$

$\left[\begin{matrix}- 2 & - 3 & | - \textcolor{w h i t e}{0} 7 \\ 0 & - \frac{27}{2} & | \textcolor{w h i t e}{} - \frac{3}{2}\end{matrix}\right]$

Multiplying ${L}_{2}$ by $2$

$\left[\begin{matrix}- 2 & - 3 & | \textcolor{w h i t e}{0} - 7 \\ 0 & - 27 & | \textcolor{w h i t e}{0} - 3\end{matrix}\right]$

Add $- \frac{1}{9} {L}_{2}$ to ${L}_{1}$

$\left[\begin{matrix}- 2 - 0 & - 3 + 3 & | \textcolor{w h i t e}{0} - 7 + \frac{1}{3} \\ 0 & - 27 & | \textcolor{w h i t e}{0} - 3\end{matrix}\right]$

$\left[\begin{matrix}- 2 & 0 & | \textcolor{w h i t e}{0} - \frac{20}{3} \\ 0 & - 27 & | \textcolor{w h i t e}{0} - 3\end{matrix}\right]$

Multiply ${L}_{1}$ by $- \frac{1}{2}$

$\left[\begin{matrix}1 & 0 & | \textcolor{w h i t e}{0} \frac{10}{3} \\ 0 & - 27 & | \textcolor{w h i t e}{0} - 3\end{matrix}\right]$

Multiply ${L}_{2}$ by $- \frac{1}{27}$

$\left[\begin{matrix}1 & 0 & | \textcolor{w h i t e}{0} \frac{10}{3} \\ 0 & 1 & | \textcolor{w h i t e}{0} \frac{1}{9}\end{matrix}\right]$

So $x = \frac{10}{3}$ and $y = \frac{1}{9}$