# How do you solve using gaussian elimination or gauss-jordan elimination, 3x - 10y = -25, 4x + 40y = 20?

Jul 7, 2017

color(brown)(((1,0,|,-5),(0,1,|,1)))color(blue)(=>((" "x=-5),(" "y=+1))

#### Explanation:

Given:
$3 x - 10 y = - 25$
$4 x + 40 y = 20$

Using the requested method.

$\left[\left(\textcolor{w h i t e}{.} x , \text{ } y , | , \textcolor{w h i t e}{.} c \textcolor{w h i t e}{. .}\right)\right]$

$\textcolor{b r o w n}{\left(\begin{matrix}3 & - 10 & | & - 25 \\ 4 & 40 & | & 20\end{matrix}\right)}$
$\text{ } {R}_{1} \times 4 \mathmr{and} {R}_{2} \times 3$
$\text{ } \downarrow$

$\textcolor{b r o w n}{\left(\begin{matrix}12 & - 40 & | & - 100 \\ 12 & 120 & | & 60\end{matrix}\right)}$
$\text{ } {R}_{2} - {R}_{1}$
$\text{ } \downarrow$

$\textcolor{b r o w n}{\left(\begin{matrix}12 & - 40 & | & - 100 \\ 0 & 160 & | & 160\end{matrix}\right)}$
$\text{ } {R}_{2} \div 160$
$\text{ } \downarrow$

$\textcolor{b r o w n}{\left(\begin{matrix}12 & - 40 & | & - 100 \\ 0 & 1 & | & 1\end{matrix}\right)} \textcolor{b l u e}{\implies y = 1}$
$\text{ } {R}_{1} \div 12$
$\text{ } \downarrow$

$\textcolor{b r o w n}{\left(\begin{matrix}1 & - \frac{10}{3} & | & - \frac{25}{3} \\ 0 & 1 & | & 1\end{matrix}\right)}$
$\text{ } {R}_{1} + \frac{10}{3} {R}_{2}$
$\text{ } \downarrow$

$\textcolor{b r o w n}{\left(\begin{matrix}1 & 0 & | & - 5 \\ 0 & 1 & | & 1\end{matrix}\right)} \textcolor{b l u e}{\implies x = - 5}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:
$3 x - 10 y \text{ "->" "3(-5)-10(1)" "->" } - 15 - 10 = - 25$

Required solution is -25 so the values are correct