Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 4 - Sections 4.1-4.3 - Integrated Review - Systems of Linear Equations - Page 234: 14

Answer

Solution: $(3,\frac{3}{4})$

Work Step by Step

$y = \frac{1}{4} x$ Equation $(1)$ $2x-4y = 3$ Equation $(2)$ Substituting $y$ from Equation $(1)$ in Equation $(2)$ $2x-4y = 3$ $2x - 4 (\frac{1}{4}x) = 3$ $2x -x = 3$ $x = 3$ Substituting $x$ value in Equation $(1)$ $y = \frac{1}{4} x$ $y = \frac{1}{4} (3)$ $y = \frac{3}{4} $ Solution: $(3,\frac{3}{4})$
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