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: 69


$x=\frac{1}{3}$ $y=\frac{1}{2}$

Work Step by Step

If we multiply the first equation by $4$, then we add the two equations, we can eliminate the $y$ from the system: $3\times4x-2\times4y=0\times4$ $9x+8y=7$ $12x+9x=21x=7$ $x=\frac{1}{3}$ $3(\frac{1}{3})-2y=0$ $y=\frac{1}{2}$ We have to check, if the solution is right. Substitute $x$ with $\frac{1}{3}$ and $y$ with $\frac{1}{2}$: $3(\frac{1}{3})-2(\frac{1}{2})=1-1=0$ $\checkmark$ $9(\frac{1}{3})+8(\frac{1}{2})=3+4=7$ $\checkmark$
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