Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 4 - Quadratic Functions and Equations - Get Ready! - Page 191: 9

Answer

The solution is $(10, -1)$.

Work Step by Step

We are asked to use the substitution method to solve this system of equations. The first thing we want to do is to isolate one of the variables in one of the equations. Let's try to isolate the $x$ in the first equation. First, we subtract $6y$ from both sides of the equation: $2x = -6y + 14$ Divide both sides by $2$: $x = -3y + 7$ Substitute this expression where we see $x$ in the second equation: $4(-3y + 7) - 8y = 48$ Use the distributive property: $-12y + 28 - 8y = 48$ Combine like terms on the left side of the equation: $-20y + 28 = 48$ Subtract $28$ from each side of the equation: $-20y = 20$ Divide each side of the equation by $-20$: $y = -1$ Now that we have the value of $y$, we plug in this number into one of the equations to find the value of $x$. Let's use the first equation: $2x + 6(-1) = 14$ Multiply to simplify: $2x - 6 = 14$ Add $6$ to each side of the equation: $2x = 20$ Divide each side of the equation by $2$: $x = 10$ The solution is $(10, -1)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.