Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 3 - Linear Systems - 3-2 Solving Systems Algebraically - Practice and Problem-Solving Exercises - Page 147: 61

Answer

$(-1, -3)$

Work Step by Step

Simplify the first equation in order to be able to substitute in order it would be easier to plug in $y$ into the second equation: \begin{align*}-3y&=-6x+3\\ \frac{-3y}{-3}&=\frac{-6x+3}{-3}\\ y&=2x-1\end{align*} With $y=2x-1$, substitute $2x-1$ to the $y$ in the second equation to obtain: \begin{align*} 5x-5(2x-1)&=10\\ 5x-10x+5&=10\\ -5x+5&=10\\ -5x&=5\\ x&=-1 \end{align*} Substitute $x=-1$ into one of the equations to find y: \begin{align*} 6(-1)-3y&=3\\ -6-3y&=3\\ -3y&=9\\ y&=-3\end{align*} Thus, the solution is $x=-1$ and $y=-3$ or $(-1, -3)$.
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