8 Consider the system of equations below.
\[
\begin{array}{c}
x+2 y-z=1 \\
-x-3 y+2 z=0 \\
2 x-4 y+z=10
\end{array}
\]
What is the solution to the given system of equations
(1) $ (1,1,2) $
(2) $ (3,-1,0) $
(3) $ (5,-1,2) $
(4) $ (3,5,8) $

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Answer

\text{Solution: } (3,-1,-8)

Steps

Step 1 ：\begin{cases} x+2y-z=1\\ -x-3y+2z=0\\ 2x-4y+z=10 \end{cases}

Step 2 ：\text{Add Equation (1) and Equation (2): } 2y+z=1

Step 3 ：\text{Multiply Equation (1) by $-2$ and add to Equation (3): } -z = 8

Step 4 ：\text{Substitute $z=-8$ into Equation (2): } x=3

Step 5 ：\text{Substitute $z=-8$ and $x=3$ into Equation (1): } y=-1

Step 6 ：\text{Solution: } (3,-1,-8)