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: 42

Answer

$x=6$ $y=-10$

Work Step by Step

If we multiply the first equation by $\frac{3}{2}$, the second one by $\frac{2}{5}$, then we add the two equations, we can eliminate the $y$ from the system: $-\frac{1}{3}\times \frac{3}{2}x+\frac{2}{5}\times \frac{3}{2}y=-6\times \frac{3}{2}$ $-\frac{1}{2}\times \frac{2}{5}x-\frac{3}{2}\frac{2}{5}y=12\times \frac{2}{5}$ $-\frac{1}{2}x-\frac{1}{5}x=-\frac{21}{5}$ $x=6$ $-\frac{1}{3}(6)+\frac{2}{5}y=-6$ $y=-10$ We have to check, if the solution is right. Substitute $x$ with 6 and $y$ with -10: $-\frac{1}{3}(6)+\frac{2}{5}(-10)=-2-4=-6$ $\checkmark$ $-\frac{1}{2}(6)-\frac{3}{2}(-10)=-3+15=12$ $\checkmark$
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