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: 6

Answer

$(\frac{4}{3},\frac{16}{3})$

Work Step by Step

$x-y = -4$ Equation$(1)$ $y=4x$ Equation$(2)$ Substituting $y=4x$ in Equation$(1)$, we get $x-y = -4$ $x-4x= -4$ $-3x = -4$ $x= \frac{-4}{-3}$ $x= \frac{4}{3}$ Substituting $x$ value in Equation$(2)$, we get $y=4x$ $y=4 \times \frac{4}{3}$ $y=\frac{16}{3}$ Solution : $(\frac{4}{3},\frac{16}{3})$
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