Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 9 - 9.3 - Multivariable Linear Systems - 9.3 Exercises - Page 659: 24


$x=\dfrac{5}{3}; y=\dfrac{1}{3}; z=3$

Work Step by Step

Subtract the second equation from equation $1$. $3y-3z=-8$ Subtract the equation from equation $3$. This yields $z=3$ Plug the value of $z$ into the second equation, giving: $y=\dfrac{1}{3}$ Now, we will plug the value of $y$ and $z$ into the first equation to get the value of $x$. $x+3+\dfrac{1}{3}=5 \\ \implies x=\dfrac{5}{3}$ Thus, $x=\dfrac{5}{3}; y=\dfrac{1}{3}; z=3$
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