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: 38

Answer

$x = -3$, $y = 2$ or $(-3, 2)$

Work Step by Step

Given: $5x - 2y = -19$ $2x + 3y = 0$ Multiply both sides of the first equation, $5x - 2y = -19$, by $3$: $15x - 6y = -57$ Multiply both sides of the second equation, $2x + 3y = 0$, by $2$: $4x + 6y = 0$ Add the first equation, $15x - 6y = -57$, to the second equation, $4x + 6y = 0$: $15x - 6y + (4x + 6y) = -57 + (0)$ $15x + 4x - 6y + 6y = -57 + 0$ $19x = -57$ Divide both sides by $19$: $x = -3$ Substitute $x = -3$ into the first equation, $5x - 2y = -19$: $5x - 2y = -19$ $5(-3) - 2y = -19$ $-15 - 2y = -19$ Add $15$ to both sides: $-2y = -4$ Divide both sides by $-2$: $y = 2$ $x = -3$, $y = 2$
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