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: 23

Answer

{($\frac{43}{5}$,-$\frac{1}{5}$)}

Work Step by Step

x=2y+9 Plug this value for x into the other equation. x=7y+10 2y+9=7y+10 2y-7y+9=7y-7y+10 -5y+9=10 -5y+9-9=10-9 -5y=1 y=-$\frac{1}{5}$ Plug this value for y into the first equation to find x. x=2(-$\frac{1}{5}$)+9 x=-$\frac{2}{5}$+9 x=-$\frac{2}{5}$+$\frac{45}{5}$ x=$\frac{43}{5}$ Therefore the solution is {($\frac{43}{5}$,-$\frac{1}{5}$)}
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