Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 4 - Section 4.2 - Solving Systems of Linear Equations by Substitution - Practice: 3

Answer

$x = \frac{1}{3}, y = 4$ or $(\frac{1}{3}, 4)$

Work Step by Step

$\begin{cases} 3x + y = 5 \\ 3x - 2y = -7 \end{cases}$ We solve for $y$ from the first equation by subtracting $3x$ from both sides: $3x + y - 3x = 5 - 3x$ $y = 5 - 3x$ Now that we've solved for y we substitute $5-3x$ for y in the second equation: $3x - 2(5-3x)=-7$ $3x - 10 + 6x = -7$ $9x = 3$ $x = \frac{1}{3}$ Now that we've found x we substitute $\frac{1}{3}$ for x in $y = 5-3x$: $y = 5 - 3\times\frac{1}{3}$ $y = 5 - 1 = 4$ So the solution is $x = \frac{1}{3}, y = 4$ or $(\frac{1}{3}, 4)$
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