## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Mid-Chapter Check Point - Page 920: 1

#### Answer

$(1,-1,2)$

#### Work Step by Step

Based on the given system, write an augmented matrix and perform row operations: $\begin{bmatrix} 1 & 2 & -3 & | & -7 \\ 3 & -1 & 2 & | & 8 \\ 2 & -1 & 1 & | & 5 \end{bmatrix} \begin{array} ..\\3R1-R2\to R2\\2R1-R3\to R3 \end{array}$ $\begin{bmatrix} 1 & 2 & -3 & | & -7 \\ 0 & 7 & -11 & | & -29 \\ 0 & 5 & -7 & | & -19 \end{bmatrix} \begin{array} ..\\..\\7R3-5R2\to R3 \end{array}$ $\begin{bmatrix} 1 & 2 & -3 & | & -7 \\ 0 & 7 & -11 & | & -29 \\ 0 & 0 & 6 & | & 12 \end{bmatrix} \begin{array} ..\\..\\.. \end{array}$ The last row gives $6z=12$ and $z=2$. Back-substitute to get $7y-11(2)=-29$, $y=-1$; and $x+2(-1)-3(2)=-7$, $x=1$. Thus the solution set for the system is $(1,-1,2)$

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