## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.2 - Systems of Linear Equations in Three Variables - Exercise Set - Page 829: 2

#### Answer

Yes, the ordered triple is a solution of the system.

#### Work Step by Step

We have to check if the provided point is a solution to the system, putting in the value $5$ for x, $-3$ for y and $-2$ for z in all the three equations: For the first equation: \begin{align} & 5-3-2=0 \\ & 5-5=0 \\ & 0=0 \end{align} This implies the point satisfies the first equation. For the second equation: \begin{align} & 5+2\left( -3 \right)-3\left( -2 \right)=5 \\ & 5-6+6=5 \\ & 5=5 \end{align} This implies the point satisfies the second equation. For the third equation: \begin{align} & 3x+4y+2z=-1 \\ & 3\left( 5 \right)+4\left( -3 \right)+2\left( -2 \right)=-1 \\ & 15-12-4=-1 \\ & -1=-1 \end{align} Thus, the ordered triple is a solution of the system.

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