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 227: 32


$\frac{8}{21}=y$ $\frac{2}{7}=x$

Work Step by Step

The second equation is already solved for $x$, we have to substitute this expression into the first equation: $\frac{3}{4}y=6y-2$ $3y=24y-8$ $8=21y$ $\frac{8}{21}=y$ $x=\frac{3}{4}y=\frac{3}{4}\times\frac{8}{21}=\frac{2}{7}$ We need to check, if it is the right solution, we substitute x with $\frac{2}{7}$ and y with $\frac{8}{21}$: $\frac{2}{7}=6\frac{8}{21}-2=\frac{6}{21}$ $\checkmark$ $\frac{2}{7}=\frac{3}{4}\times\frac{8}{21}$ $\checkmark$
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