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: 11



Work Step by Step

-3x+y=-1 y=-1+3x Plug this value for y into the other equation. x-2y=4 x-2(-1+3x)=4 x-2(-1)-2(3x)=4 x+2-6x=4 x-6x=4-2 -5x=4-2 -5x=2 x=$\frac{2}{-5}$ x=-$\frac{2}{5}$ Plug this back into the first equation to find y. y=-1+3x y=-1+3(-$\frac{2}{5}$) y=-1-$\frac{6}{5}$ y=-$\frac{5}{5}$-$\frac{6}{5}$ y=-$\frac{11}{5}$ Therefore the solution is {(-$\frac{2}{5}$,-$\frac{11}{5}$)}
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