Answer
$x_1=8$
$x_2=10$
$x_3=6$
Work Step by Step
Write down the augmented matrix of the system of linear equations.
\begin{bmatrix}
2 & -1 & 3 & 24 \\
0 & 2 & -1 & 14 \\
7& -5 & 0 &6 & \\
\end{bmatrix}
$\frac{-7}{2} R_1+R_3 \rightarrow R_3$
\begin{bmatrix}
2 & -1 & 3 & 24 \\
0 & 2 & -1 & 14 \\
0& -3/2 & -21/2 &-78 & \\
\end{bmatrix}
$\frac{3}{4} R_2+R_3 \rightarrow R_3$
\begin{bmatrix}
2 & -1 & 3 & 24 \\
0 & 2 & -1 & 14 \\
0& 0 &-45/4 &-135/2 & \\
\end{bmatrix}
$4R_3 \rightarrow R_3$, getting rid of the fractions
\begin{bmatrix}
2 & -1 & 3 & 24 \\
0 & 2 & -1 & 14 \\
0& 0 & -45 &-270 & \\
\end{bmatrix}
Backsubstitution will be used to obtain the solution:
$x_3=-270/-45=6$
$2x_2-6=14 \Rightarrow x_2=20/2=10$
$2x_1-10+3*6=24 \Rightarrow 2x_1=16 \Rightarrow x_1=8$