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 - Got It? - Page 144: 3

Answer

The solution to this system of equations is $(4, 0)$.

Work Step by Step

We can solve this system of equations by the elimination method. We see that by adding the two equations, we can eliminate one variable and just work with the other one. $-2x + 8y = -8$ $ 5x - 8y = 20$ If we add the two equations, we can eliminate the $y$ terms from both equations: \begin{align*} (-2x+8y)+(5x-8y)&=-8+20\\ 3x&=12\\ x&=\frac{12}{3}\\ x&=4 \end{align*} We use this value for $x$ to substitute into one of the equations to find the value for $y$. Let us use the first equation: $-2(4) + 8y = -8$ $-8 + 8y = -8$ Add $8$ to both sides to isolate constants to the right side of the equation: $8y = 0$ Divide both sides by $8$ to solve for $y$: $y = 0$ The solution to this system of equations is $(4, 0)$.
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