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 - Page 226: 25


$x=\frac{22}{9}$ $y=\frac{22}{3}$

Work Step by Step

First, we have to solve the second equation for $y$ and after that, we can substitute this expression in the first equation: $-3x+y=0$ $y=3x$ $3x-4y=-22$ $3x-4(3x)=-9x=-22$ $x=\frac{22}{9}$ $3x=y=3\frac{22}{9}=\frac{22}{3}$ We need to check, if it is the right solution, we substitute x with $\frac{10}{7}$ and y with $-\frac{5}{7}$: $3(\frac{22}{9})-4(\frac{22}{3})=\frac{-66}{3}=-22$ $\checkmark$ $-3\frac{22}{9}+(\frac{22}{3})=0$ $\checkmark$
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