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: 48

Answer

$x = 6, y = 4$ or $(6, 4)$

Work Step by Step

First, rewrite both equations such that the variables are on one side of the equation and the constant is on the other: Equation 1: $3x - 6y = -6$ Equation 2: $4x - 5y = 4$ Rewrite the equations such that one of the variables in the two equations is the same but differing in sign. To do this, multiply the first equation by $4$ and the second equation by $-3$: Equation 1: $12x - 24y = -24$ Equation 2: $-12x + 15y = -12$ Add the equations together to eliminate the $x$ variable: $$-9y = -36$$ Divide both sides of the equation by $-9$ to solve for $y$: $$y = 4$$ Substitute this value of $y$ into one of the equations to solve for $x$. Use Equation 2: $$4x - 5(4) = 4$$ Multiply to simplify: $$4x - 20 = 4$$ Add $20$ to each side of the equation to isolate the $x$ term: $$4x = 24$$ Divide both sides of the equation by $4$ to solve for $x$: $$x = 6$$
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