Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 7 - Algebra: Graphs, Functions, and Linear Systems - 7.3 Systems of Linear Equations in Two Variables - Concept and Vocabulary Check - Page 444: 3

Answer

The solution set while solving the equations by substitution method is \[y=-2\].

Work Step by Step

Given equations are\[3x-2y=5\ \text{and }y=\text{3}x\text{-3}\]and, \[x=\frac{1}{3}\]. Solve \[y\]using either of the two of the two linear equations by substituting the value of \[x=\frac{1}{3}\]as: \[\begin{align} & 3\left( \frac{1}{3} \right)-2y=5 \\ & 1-2y=5 \\ & 2y=-4 \\ & y=-2 \end{align}\] Also, using second equation, value of\[y\]is: \[\begin{align} & y=\text{3}x\text{-3} \\ & y=3\left( \frac{1}{3} \right)-3 \\ & y=1-3 \\ & y=-2 \end{align}\] Hence, the solution set while solving the equations by substitution method is \[y=-2\].
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