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: 18

Answer

$r = -6$, $s = -6$

Work Step by Step

Given $r + s = -12$ and $4r - 6s = 12$. Solve for $r$ in the first equation by subtracting $s$ to give: $r = -s - 12$. Substitute $-s-12$ into the second equation: $4(-s - 12) - 6s = 12$ Now the equation is in terms of a single variable, solve for the variable: $4(-s - 12) - 6s = 12$ : distribute $4$ $-4s - 48 - 6s = 12$ : collect like terms $-10s - 48 = 12$ : add $48$ to both sides $-10s = 60$ : divide by $-10$ $s = -6$ Substitute $s = -6$ into earlier equation: $r + s = - 12$. $r + (-6) = -12$ : solve $r - 6 = -12$ : add 6 $r = -6$ Giving, $r = -6$, $s = -6$
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