Answer
$(x,y,z)=(\frac{23}{4},-\frac{1}{2},-5)$.
Work Step by Step
The augmented matrix is
$\begin{bmatrix}
2&-1 &1 &7\\
0 & -2 & 1 & -4\\
0& 0 & -3 & 15
\end{bmatrix}$.
For the linear equation.
First column $=$ coefficients of $x$.
Second column $=$ coefficients of $y$
Third column $=$ coefficients of $z$
and fourth column $=$ the right-hand side.
The given matrix is a upper triangular form.
Rewrite the third row as a linear equation.
$\Rightarrow 0x+0y-3z=15$
$\Rightarrow -3z=15$
Divide both sides by $-3$.
$\Rightarrow \frac{-3z}{-3}=\frac{15}{-3}$
Simplify.
$\Rightarrow z=-5$
Rewrite the second row as a linear equation.
$\Rightarrow 0x-2y+1z=-4$
$\Rightarrow -2y+z=-4$
Substitute $z=-5$.
$\Rightarrow -2y+(-5)=-4$
Clear the parentheses.
$\Rightarrow -2y-5=-4$
Add $5$ to both sides.
$\Rightarrow -2y-5+5=-4+5$
Simplify.
$\Rightarrow -2y=1$
Divide both sides by $-2$.
$\Rightarrow \frac{-2y}{*2}=\frac{1}{-2}$
Simplify.
$\Rightarrow y=-\frac{1}{2}$
Rewrite the first row as a linear equation.
$\Rightarrow 2x-1y+1z=7$
$\Rightarrow 2x-y+z=7$
Substitute $y=-\frac{1}{2}$ and $z=-5$.
$\Rightarrow 2x-(-\frac{1}{2})+(-5)=7$
Clear the parentheses.
$\Rightarrow 2x+\frac{1}{2}-5=7$
$\Rightarrow 2x+\frac{1-10}{2}=7$
$\Rightarrow 2x-\frac{9}{2}=7$
Add $\frac{9}{2}$ to both sides.
$\Rightarrow 2x-\frac{9}{2}+\frac{9}{2}=7+\frac{9}{2}$
Simplify.
$\Rightarrow 2x=\frac{14+9}{2}$
$\Rightarrow 2x=\frac{23}{2}$
Divide both sides by $2$.
$\Rightarrow \frac{2x}{2}=\frac{23}{2}\cdot \frac{1}{2}$
Simplify.
$\Rightarrow x=\frac{23}{4}$
The solution set is $(x,y,z)=(\frac{23}{4},-\frac{1}{2},-5)$.
