## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Section 8.2 - Inconsistent and Dependent Systems and Their Applications - Concept and Vocabulary Check - Page 902: 4

#### Answer

The provided statement is True.

#### Work Step by Step

Take an arbitrary value $z=1$; now substitute this value of z in the given solution set to find out the value of x, and y. Since, \begin{align} & x=2z+3 \\ & y=5z-1 \\ \end{align} Substitute value $z=1$ in $y=5z-1$ to get, \begin{align} & y=5\times 1-1 \\ & =5-1 \\ & =4 \end{align} Substitute value $z=1$ in $x=2z+3$ to get, \begin{align} & x=2\times 1+3 \\ & =2+3 \\ & =5 \end{align} So, the values of $x,y,z$ is: \begin{align} & x=5 \\ & y=4 \\ & z=1 \\ \end{align} Hence, these values are $\left( 5,\ 4,\ 1 \right)$, which is equal to the solution mentioned in the statement.

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