College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 8 - Section 8.1 - Systems of Linear Equations: Substitution and Elimination - 8.1 Assess Your Understanding - Page 555: 39


$x=\frac{4}{3}$ $y=\frac{1}{5}$

Work Step by Step

$3x-5y=3$ $(1)$ $15x+5y=21$ $(2)$ Add the two equations $(3x+15x) + (-5y+5y) = 3 + 21$ $18x=24$ $x=\frac{24}{18}$ $x=\frac{4}{3}$ Substitute into Eq $(1)$ $5y=3x-3$ $5y=3 \times \frac{4}{3}-3$ $5y=1$ $y=\frac{1}{5}$
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