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: 6

Answer

$x=2$ $y=1$

Work Step by Step

Equations: $3x+4y=10$ $2x+3y=7$ Multiply $2$ to the first and $-3$ to the second equation equations to have the equivalent system ofequations: $2(3x+4y)=2(10)$ $-3(2x+3y)=-3(7)$ $6x+8y=20$ $-6x -9y=-21$ Eliminate $x$ by adding the equations together, then solve for $y$: $6x+8y=20$ $+$ $-6x -9y=-21$ $-y=-1$ $y=1$ Substitute $1$ to to the $y$ of the second original equation to solve for $x$: $2x+3(1)=7$ $2x + 3 = 7$ $2x = 4$ $x = 2$
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