# How do you solve -3x-y=0, y-4z=15, x-2y+z=-10?

Jan 1, 2017

$x = - 1 \text{; "y=3"; } z = - 3$

#### Explanation:

$\left(\begin{matrix}x & y & z & | & - \\ - 3 & - 1 & 0 & | & 0 \\ 0 & 1 & - 4 & | & 15 \\ 1 & - 2 & 1 & | & - 10\end{matrix}\right)$
$\textcolor{b r o w n}{\text{ } {R}_{3} + 2 {R}_{2}}$
$\textcolor{b r o w n}{\text{ } \downarrow}$

$\left(\begin{matrix}- 3 & - 1 & 0 & | & 0 \\ 0 & 1 & - 4 & | & 15 \\ 1 & 0 & - 7 & | & + 20\end{matrix}\right)$
$\textcolor{b r o w n}{\text{ } {R}_{1} + 3 {R}_{3}}$
$\textcolor{b r o w n}{\text{ } \downarrow}$

$\left(\begin{matrix}0 & - 1 & - 21 & | & 60 \\ 0 & 1 & - 4 & | & 15 \\ 1 & 0 & - 7 & | & + 20\end{matrix}\right)$
$\textcolor{b r o w n}{\text{ } {R}_{1} + {R}_{2}}$
$\textcolor{b r o w n}{\text{ } \downarrow}$

$\left(\begin{matrix}0 & 0 & - 25 & | & 75 \\ 0 & 1 & - 4 & | & 15 \\ 1 & 0 & - 7 & | & + 20\end{matrix}\right)$
$\textcolor{b r o w n}{\text{ } {R}_{1} \div \left(- 25\right)}$
$\textcolor{b r o w n}{\text{ } \downarrow}$

$\left(\begin{matrix}0 & 0 & 1 & | & - 3 \\ 0 & 1 & - 4 & | & 15 \\ 1 & 0 & - 7 & | & + 20\end{matrix}\right)$
color(brown)(" "R_2+4R_1
$\textcolor{b r o w n}{\text{ } \downarrow}$

$\left(\begin{matrix}0 & 0 & 1 & | & - 3 \\ 0 & 1 & 0 & | & 3 \\ 1 & 0 & - 7 & | & + 20\end{matrix}\right)$
color(brown)(" "R_3+7R_1
$\textcolor{b r o w n}{\text{ } \downarrow}$

$\left(\begin{matrix}0 & 0 & 1 & | & - 3 \\ 0 & 1 & 0 & | & 3 \\ 1 & 0 & 0 & | & - 1\end{matrix}\right)$
color(brown)(" "R_3+7R_1

$x = - 1 \text{; "y=3"; } z = - 3$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{check}}$

-3x-y=0" " ->" " -3(-1)-3= 0 " "color(red)(larr"True")

y-4z=15" " ->" "3-4(-3)=15" "color(red)(larr"True")

x-2y+z=-10" "->""-1-2(3)-3= -10" "color(red)(larr"True")