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: 26

Answer

$x=\frac{10}{7}$ $y=-\frac{5}{7}$

Work Step by Step

First, we have to solve the second equation for $x$ and after that, we can substitute this expression in the first equation: $3x+6y=0$ $3x=-6y$ $x=-2y$ $-3x+y=-5$ $-3(-2y)+y=7y=-5$ $y=-\frac{5}{7}$ $3x=-6y=\frac{(-6)\times (-5)}{7}\frac{30}{7}$ $x=\frac{10}{7}$ We need to check, if it is the right solution, we substitute x with $\frac{10}{7}$ and y with $-\frac{5}{7}$: $-3(\frac{10}{7})+(-\frac{5}{7})=\frac{-35}{7}=-5$ $\checkmark$ $3\frac{10}{7}+6(-\frac{5}{7})=0$ $\checkmark$
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