# Evaluate: {\text{begin}array l x_{1}+x_{2}-x_{3}=2 }-x_{1}+3x_{3}=8 2x_{1}-x_{3}=4\text{end}array .

## Expression: $\left\{\begin{array} { l } x_1+x_2-x_3=2 \\ -x_1+3x_3=8 \\ 2x_1-x_3=4\end{array} \right.$

Rewrite the system as two systems, each consisting of two equations

$\begin{array} { l }\left\{\begin{array} { l } x_1+x_2-x_3=2 \\ -x_1+3x_3=8\end{array} \right.,\\\left\{\begin{array} { l } x_1+x_2-x_3=2 \\ 2x_1-x_3=4\end{array} \right.\end{array}$

Solve the system of equations

$\begin{array} { l }x_2+2x_3=10,\\\left\{\begin{array} { l } x_1+x_2-x_3=2 \\ 2x_1-x_3=4\end{array} \right.\end{array}$

Solve the system of equations

$\begin{array} { l }x_2+2x_3=10,\\-2x_2+x_3=0\end{array}$

Write as a system of equations

$\left\{\begin{array} { l } x_2+2x_3=10 \\ -2x_2+x_3=0\end{array} \right.$

Solve the system of equations

$\begin{array} { l }x_2=2,\\x_3=4\end{array}$

Substitute the given value of $x_3$ into the equation $-x_1+3x_3=8$

$-x_1+3 \times 4=8$

Solve the equation for $x_1$

$x_1=4$

The possible solution of the system is the ordered triple $\left( x_1, x_2, x_3\right)$

$\left( x_1, x_2, x_3\right)=\left( 4, 2, 4\right)$

Check if the given ordered triple is a solution of the system of equations

$\left\{\begin{array} { l } 4+2-4=2 \\ -4+3 \times 4=8 \\ 2 \times 4-4=4\end{array} \right.$

Simplify the equalities

$\left\{\begin{array} { l } 2=2 \\ 8=8 \\ 4=4\end{array} \right.$

Since all of the equalities are true, the ordered triple is the solution of the system

$\left( x_1, x_2, x_3\right)=\left( 4, 2, 4\right)$

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