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: 20



Work Step by Step

Here, we have $0.5x+3.2y=9.0$ ...(1) and $0.2x-1.6y=-3.6$ ...(2) Re-arrange the second equation as: $x=[ 0.2x-1.6y=-3.6] \times 5 \implies x=8y-18$ Now, plug $x=\dfrac{1}{3}+\dfrac{2}{3} y$ into the first equation. $0.5( 8y-18)+3.2y=9.0$ $4y-9+3.2y=9.0$ This gives $ 7.2y=18 \implies y=2.5$ Use the $y$ value and the equation above to solve for $x$: $ x=8(2.5)-18=2$ Hence, $(x,y)=(2,2.5)$
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