Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 4 - Section 4.2 - Solving Systems of Linear Equations by the Substitution Method - Exercise Set - Page 300: 25



Work Step by Step

4x-y=100 4x-y+y=100+y 4x=100+y 4x-100=y y=4x-100 Plug this value for y into the other equation. 0.05x-0.06y=-32 100(0.05x-0.06y)=100(-32) 5x-6y=-3200 5x-6(4x-100)=-3200 5x-24x+600=-3200 -19x+600=-3200 -19x+600-600=-3200-600 -19x=-3800 x=$\frac{-3800}{-19}$ x=200 Plug this value for x into the first equation to find y. y=4(200)-100 y=800-100 y=700 Therefore the solution is {(200,700)}
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