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 - Lesson Check: 2

Answer

$x = -6$ $y = -6$

Work Step by Step

Equations: $2x-3y=6$ $x+y=-12$ Since $x$ and $y$ already have a coefficient of one in the second equation, we can pick either variable to substitute with. $x$ in the first equation has a smaller coefficient, so it's easier to use $x$. Getting $x$ alone: $x+y=-12$ $x=-12-y$ Substituting $x$ into the first equation: $2x-3y=6$ $2(-12-y)-3y=6$ $-24 -2y-3y=6$ $-2y-3y=30$ $-5y=30$ $y=-6$ Substituting $y$ into the second equation to find $x$ $x+y=-12$ $x+(-6)=-12$ $x-6=-12$ $x=-6$
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