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: 12

Answer

{($\frac{14}{5}$,$\frac{1}{5}$)}

Work Step by Step

-4x+y=-11 y=-11+4x Plug this value for y into the other equation. 2x-3y=5 2x-3(-11+4x)=5 2x-3(-11)-3(4x)=5 2x+33-12x=5 2x-12x=5-33 -10x=5-33 -10x=-28 x=$\frac{-28}{-10}$ x=$\frac{14}{5}$ Plug this back into the first equation to find y. y=-11+4x y=-11+4($\frac{14}{5}$) y=-11+$\frac{56}{5}$ y=-$\frac{55}{5}$+$\frac{56}{5}$ y=$\frac{1}{5}$ Therefore the solution is {($\frac{14}{5}$,$\frac{1}{5}$)}
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