Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.2 - Systems of Linear Equations in Three Variables - Exercise Set - Page 829: 12


The solution of the system is $\left( 1,1,2 \right)$.

Work Step by Step

Let us consider the given equation: $x+z=3$ ...... (I) $x+2y-z=1$ ...... (II) $2x-y+z=3$ ...... (III) Now, multiply equation (III) by 2 and add equation (II): $\begin{align} & x+2y-z=1 \\ & 4x-2y+2z=6 \end{align}$ $5x+z=7$ ...... (IV) And subtract equation (I) from equation (IV): $\begin{align} & 5x+z=7 \\ & x+z=3 \\ \end{align}$ Solve and we get: $\begin{align} & 4x=4 \\ & x=1 \end{align}$ Putting the value of $x=1$ in equation (I): $\begin{align} & 1+z=3 \\ & z=2 \end{align}$ Now, putting the value of $x=1$ and $z=2$ in equation (II): $\begin{align} & 1+2y-2=1 \\ & 2y=2 \\ & y=1 \end{align}$ Thus, the ordered triple $\left( 1,1,2 \right)$ satisfies the three systems.
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