# Calculate: {\text{begin}array l-5x+2y+z=6 } x+4y+z=0 2x-2y+z=-3\text{end}array .

## Expression: $\left\{\begin{array} { l } -5x+2y+z=6 \\ x+4y+z=0 \\ 2x-2y+z=-3\end{array} \right.$

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

$\begin{array} { l }\left\{\begin{array} { l } -5x+2y+z=6 \\ 2x-2y+z=-3\end{array} \right.,\\\left\{\begin{array} { l } x+4y+z=0 \\ 2x-2y+z=-3\end{array} \right.\end{array}$

Solve the system of equations

$\begin{array} { l }-3x+2z=3,\\\left\{\begin{array} { l } x+4y+z=0 \\ 2x-2y+z=-3\end{array} \right.\end{array}$

Solve the system of equations

$\begin{array} { l }-3x+2z=3,\\5x+3z=-6\end{array}$

Write as a system of equations

$\left\{\begin{array} { l } -3x+2z=3 \\ 5x+3z=-6\end{array} \right.$

Solve the system of equations

$\begin{array} { l }x=-\frac{ 21 }{ 19 },\\z=-\frac{ 3 }{ 19 }\end{array}$

Substitute the given values of $\begin{array} { l }x,& z\end{array}$ into the equation $-5x+2y+z=6$

$-5 \times \left( -\frac{ 21 }{ 19 } \right)+2y+\left( -\frac{ 3 }{ 19 } \right)=6$

Solve the equation for $y$

$y=\frac{ 6 }{ 19 }$

The possible solution of the system is the ordered triple $\left( x, y, z\right)$

$\left( x, y, z\right)=\left( -\frac{ 21 }{ 19 }, \frac{ 6 }{ 19 }, -\frac{ 3 }{ 19 }\right)$

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

$\left\{\begin{array} { l } -5 \times \left( -\frac{ 21 }{ 19 } \right)+2 \times \frac{ 6 }{ 19 }+\left( -\frac{ 3 }{ 19 } \right)=6 \\ -\frac{ 21 }{ 19 }+4 \times \frac{ 6 }{ 19 }+\left( -\frac{ 3 }{ 19 } \right)=0 \\ 2 \times \left( -\frac{ 21 }{ 19 } \right)-2 \times \frac{ 6 }{ 19 }+\left( -\frac{ 3 }{ 19 } \right)=-3\end{array} \right.$

Simplify the equalities

$\left\{\begin{array} { l } 6=6 \\ 0=0 \\ -3=-3\end{array} \right.$

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

$\left( x, y, z\right)=\left( -\frac{ 21 }{ 19 }, \frac{ 6 }{ 19 }, -\frac{ 3 }{ 19 }\right)$

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