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 148: 68

Answer

$5$

Work Step by Step

It looks like using the elimination method is an easier way to solve this equation. First, we need to convert the equations so that one variable is the same in both equations but with opposite signs. If we add these two equations together, we can eliminate one variable and just deal with one variable instead of two: Since we are being asked for the value of $x$, we get rid of the $y$ terms in both equations. Multiply the first equation by $2$ to obtain the equivalent system : $2x + 2y = 14$ $3x - 2y = 11$ Add the equations together to eliminate $y$: $(2x+2y)+(3x-2y)=14+11\\ 5x = 25$ Divide each side by $5$ to solve for $x$: $x = 5$ Thus, the $x$-value of the solution of the given system is $5$.
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