Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 8 - Systems of Linear Equations and Problem Solving - 8.2 Solving by Substitution or Elimination - 8.2 Exercise Set - Page 516: 21


$(\displaystyle \frac{1}{2},\frac{1}{2})$

Work Step by Step

Isolate $y$ in eq.$2$ $ 3x-y=1\qquad$ ... add $y-1$ $3x-1=y$ $ y=3x-1\qquad$ (*) Replace $y$ with $3x-1$ in the other equation: $ 2x+2(3x-1)=2\qquad$ ... and solve for $x$. Simplify $2x+6x-2=2$ $ 8x-2=2\qquad$ ... add 2 $8x=4$ $x=\displaystyle \frac{1}{2}$ Now, back-substitute into (*) $y= 3(\displaystyle \frac{1}{2})-1=\frac{1}{2}$ Form an ordered pair $(x,y)$ as the solution to the system $(\displaystyle \frac{1}{2},\frac{1}{2})$
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