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: 44

Answer

$x=1$ $y=2$ $z=2$.

Work Step by Step

Let $x$ denote Toast Let $y$ denote Cottage cheese Let $z$ denote Fruit Thus, $\begin{cases} 100x+120y+60z=460\\ x+5y=11\\ 2x+2z=6 \end{cases}$ Multiplying Equation 2 by -2 and adding it to Equation 3, results in new Equation 3. $\begin{cases} -2x-10y=-22\\ 2x+2z=6\\ -- -- --\\ -10y+2z=-16 \end{cases}$ Dividing both sides by 2, $-5y+z=-8$. Multiplying Equation 2 by -100 and adding it to Equation 1, results in new Equation 1. $\begin{cases} -100x-500y=-1100\\ 100x+120y+60z=460\\ -- -- -- -- --\\ -380y+60z=-640 \end{cases}$ Dividing both sides by 20, $-19y+3z=-32$. Multiplying new Equation 3 by -3 and adding it to new equation 1. $\begin{cases} 15y-3z=24\\ -19y+3z=-32\\ -- -- -- --\\ -4y=-8 \end{cases}$ Thus, $y=2$, Substituting back in, $z=2$ and $x=1$
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