Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 11 - Systems of Equations and Inequalities - Chapter Review - Chapter Test - Page 796: 3


$\left\{\left(x,\dfrac{1-7x}{7},\dfrac{18-7x}{7}\right)|x\text{ is any real number}\right\}$

Work Step by Step

We are given the system of equations: $\begin{cases} x-y+2z=5\\ 3x+4y-z=-2\\ 5x+2y+3z=8 \end{cases}$ Multiply the second equation by 2 and add it to the first equation to eliminate $z$; then multiply the second equation by 3 and add it to the third equation to eliminate $z$: $\begin{cases} x-y+2z+2(3x+4y-z)=5+2(-2)\\ 5x+2y+3z+3(3x+4y-z)=8+3(-2) \end{cases}$ $\begin{cases} x-y+2z+6x+8y-2z=5-4\\ 5x+2y+3z+9x+12y-3z=8-6 \end{cases}$ $\begin{cases} 7x+7y=1\\ 14x+14y=2 \end{cases}$ Multiply the first equation by -2 and add it to the second: $-2(7x+7y)+14x+14y=-2(1)+2$ $-14x-14y+14x+14y=0$ $0=0$ We got an identity; therefore the system has infinitely many solutions. Express $y$ in terms of $x$: $x+y=\dfrac{1}{7}$ $y=\dfrac{1}{7}-x$ $y=\dfrac{1-7x}{7}$ Express $z$ in terms of $x$: $3x+4y-z=-2$ $3x+4\cdot \dfrac{1-7x}{7}-z=-2$ $z=3x+2+\dfrac{4(1-7x)}{7}$ $z=\dfrac{21x+14+4-28x}{7}$ $z=\dfrac{18-7x}{7}$ The solutions set is: $\left\{\left(x,\dfrac{1-7x}{7},\dfrac{18-7x}{7}\right)|x\text{ is any real number}\right\}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.