Use Back-Substitution to Solve the Triangular System$$\left\{ \begin{array} { r }
x + 2 y + z = 0 \
y - 3 z = 1 \
4 z = - 4
\end{array} \right.$$

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Use Back-Substitution to Solve the Triangular System$$\left\{ \begin{array} { r } 
x + 2 y + z = 0 \
y - 3 z = 1 \
4 z = - 4
\end{array} \right.$$

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The Answer of Use Back-Substitution to Solve the Triangular System$$\left\{ \begin{array} { r } 
x + 2 y + z = 0 \
y - 3 z = 1 \
4 z = - 4
\end{array} \right.$$

Use Back-Substitution to Solve the Triangular System $\left\{ \begin{array} { r } x + 2 y + z = 0 \\y - 3 z = 1 \\4 z = - 4\end{array} \right.$

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Use Back-Substitution to Solve the Triangular System {x+2y+z=0y−3z=14z=−4\left\{ \begin{array} { r } x + 2 y + z = 0 \\y - 3 z = 1 \\4 z = - 4\end{array} \right.⎩⎨⎧​x+2y+z=0y−3z=14z=−4​

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Use Back-Substitution to Solve the Triangular System $\left\{ \begin{array} { r } x + 2 y + z = 0 \\y - 3 z = 1 \\4 z = - 4\end{array} \right.$

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