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: 20



Work Step by Step

2x-y=4 2x-2x-y=4-2x -y=4-2x -1(-y)=-1(4-2x) y=-4+2x Plug this value for y into the other equation. 3x-5(-4+2x)=2 3x-5(-4)-5(2x)=2 3x+20-10x=2 3x-10x+20=2 -7x+20=2 -7x+20-20=2-20 -7x=-18 x=$\frac{-18}{-7}$ x=$\frac{18}{7}$ Plug this value for x into the first equation to solve for y. y=-4+2($\frac{18}{7}$) y=-4+$\frac{36}{7}$ y=-$\frac{28}{7}$+$\frac{36}{7}$ y=$\frac{8}{7}$ Therefore the solution is {($\frac{18}{7}$,$\frac{8}{7}$)}
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