Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 8 - Systems of Linear Equations and Problem Solving - 8.2 Solving by Substitution or Elimination - 8.2 Exercise Set - Page 517: 35



Work Step by Step

Multiply each equation (don't forget: both sides) as denoted: $\left\{\begin{array}{ll} 5r-3s=24 & .../\times 5\\ 3r+5s=28 & .../\times 3 \end{array}\right.$ $\left\{\begin{array}{ll} 25r-15s=120 & ...\\ 9r+15s=84 & ... \end{array}\right.\quad$ adding eliminates $s$ $34r=204$ $r=\displaystyle \frac{204}{34}=6$ Back-substitute into one of the initial equations: $3r+5s=28$ $3(6)+5s=28$ $5s=28-18$ $5s=10$ $s=2$ Form an ordered pair $(r,s)$ as the solution to the system $(6,2)$
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