College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 5, Systems of Equations and Inequalities - Section 5.2 - Systems of Linear Equations in Several Variables - 5.3 Exercises - Page 456: 46

Answer

$x=6$ $y=2$ $z=3$

Work Step by Step

Let $x$ denote Refrigerator Let $y$ denote Dishwashers Let $z$ denote Stove $\begin{cases} 8x+10y+14z=110\\ 16x+12y+10z=150\\ 10x+18y+6z=114 \end{cases}$ Multiplying Equation 1 by -2 and adding it to Equation 2, results in new Equation 1. $\begin{cases} -16x-20y-28z=-220\\ 16x+12y+10z=150\\ -- -- -- -- -- -\\ -8y-18z=-70 \end{cases}$ Multiplying Equation 3 by -8/5 and adding it to Equation 2, results in new Equation 3. $\begin{cases} 16x+12y+10z=150\\ -16x-28.8y-9.6z=-182.4\\ -- -- -- -- -- --\\ -16.8y+0.4z=-32.4 \end{cases}$ Multiplying the new Equation 3 by 45 and adding it to the new Equation 1. $\begin{cases} -8y-18z=-70\\ -756y+18z=-1458\\ -- -- -- -- --\\ -764y=-1528 \end{cases}$ Therefore, $y=2$, substituting back in, $z=3$ and $x=6$
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