Precalculus (6th Edition) Blitzer

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

Chapter 7 - Cumulative Review Exercises - Page 880: 21


The solution of the equation is $\left\{ \left( 8,-2,-2 \right) \right\}$.

Work Step by Step

The system of equations is given below: $\left\{ \begin{align} & x+2y+3z=-2 \\ & 3x+3y+10z=-2 \\ & 2y-5z=6 \end{align} \right.$ And the equation $2y-5z=6$ has only two variables x and y: Now, consider the equations $ x+2y+3z=-2$ and $3x+3y+10z=-2$; eliminate $ z $ and obtain the equation containing the y and z variables in equation $2y-5z=6$. Multiply the equation $ x+2y+3z=-2$ by 3 and equation $3x+3y+10z=-2$ by 1. Therefore, $\begin{align} & \left( x+2y+3z \right)\times 3=-2\times 3 \\ & 3x+6y+9z=-6 \end{align}$ And $3x+3y+10z=-2$ And subtract the equation $3x+6y+9z=-6$ from $3x+3y+10z=-2$ to obtain: $\begin{align} & \left( 3x+6y+9z \right)-\left( 3x+3y+10z \right)=\left( -6 \right)-\left( -2 \right) \\ & 3y-z=-4 \end{align}$ Now, solve the equations $2y-5z=6$ and $3y-z=-4$ ; the values of $ z $ and $ y $ are $\begin{align} & z=-2 \\ & y=-2 \end{align}$ Put values of z and y in equation $ x+2y+3z=-2$, to obtain the value of $ x $ as given below: $\begin{align} & x+2y+3z=-2 \\ & x+2\left( -2 \right)+3\left( -2 \right)=-2 \\ & x-4-6=-2 \\ & x=8 \end{align}$ Thus, the values of x, y, z are $\left\{ \left( 8,-2,-2 \right) \right\}$.
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