Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 3 - Section 3.1 - Systems of Linear Equations in Two Variables - 3.1 Exercises: 61

Answer

$x=6$ $y=-\frac{5}{6}$

Work Step by Step

If we multiply the second equation by $-1$, then we add the two equations, we can eliminate the $x$ from the system: $\frac{1}{2}x+\frac{1}{3}y=\frac{49}{18}$ $-\frac{1}{2}x-2y=\frac{4}{3}$ $\frac{1}{3}y-2y=-\frac{5}{3}y=\frac{49}{18}-\frac{24}{18}=\frac{25}{18}$ $y=-\frac{5}{6}$ $\frac{1}{2}x+\frac{1}{3}(-\frac{5}{6})=\frac{49}{18}$ $x=\frac{54}{9}=6$ We have to check, if the solution is right. Substitute $x$ with 6 and $y$ with $-\frac{5}{6}$: $\frac{1}{2}(6)+\frac{1}{3}(-\frac{5}{6})=\frac{49}{18}$ $\checkmark$ $\frac{1}{2}(6)+2(-\frac{5}{6})=\frac{4}{3}$ $\checkmark$
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