Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 9 - 9.1 - Linear and Nonlinear Systems of Equations - 9.1 Exercises - Page 635: 17



Work Step by Step

Here, we have $2x+2=y$ Now, plug $y=2x+2$ into the second equation. $4x+2x+2-5=0$ $ 6x-3=0$ $6x=3 \implies x=\dfrac{1}{2}$ Use the $x$ value and the equation above to solve for $y$: $ y=2(1/2)+2=3$ Hence, $(x,y)=(\dfrac{1}{2},3)$
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