## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.2 - Systems of Linear Equations in Three Variables - Exercise Set - Page 829: 6

#### Answer

The solution of the system of equations is $\left( 1,-1,1 \right)$

#### Work Step by Step

It is given that: $2x+y-2z=-1$ ...... (I) $3x-3y-z=5$ ...... (II) $x-2y+3z=6$ ...... (III) Multiply equation (III) by 2 and subtract it from equation (I): \begin{align} & 2x+y-2z=-1 \\ & 2x-4y+6z=1 \\ \end{align} Solve and get: $5y-8z=-13$ ...... (IV) Multiply equation (III) by $3$ and subtract from equation (II): \begin{align} & 3x-3y-z=5 \\ & 3x-6y+9z=18 \\ \end{align} Solve and get: $3y-10z=-13$ ...... (V) Multiply equation (V) by $5$ and equation (IV) by $3$ and subtract each other: \begin{align} & 15y-24z=-39 \\ & 15y-50z=-65 \\ \end{align} Solve and we get: \begin{align} & 26z=26 \\ & z=1 \end{align} Putting the value of $z=1$ in equation (IV): \begin{align} & 5y-8\left( 1 \right)=-13 \\ & 5y=-5 \\ & y=-1 \\ \end{align} Putting the value of $z=1$ and $y=-1$ in equation (I): \begin{align} & 2x+\left( -1 \right)-2\left( 1 \right)=-1 \\ & 2x-1-2=-1 \\ & 2x=2 \\ & x=1 \end{align} Thus, the order triple $\left( 1,-1,1 \right)$ satisfies the three systems.

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