Algebra 2 Common Core

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

Chapter 4 - Quadratic Functions and Equations - 4-9 Quadratic Systems - Practice and Problem-Solving Exercises - Page 262: 20

Answer

$(0, 1)$

Work Step by Step

We will use substitution to solve this system of equations. We substitute one of the expressions for $y$, which would mean that we are going to set the two equations equal to one another to solve for $x$ first: $x^2 + 5x + 1 = x^2 + 2x + 1$ We want to move all terms to the left side of the equation. $x^2 - x^2 + 5x - 2x + 1 - 1 = 0$ Combine like terms: $3x = 0$ Divide each side by $3$: $x = 0$ Now that we have the possible value for $x$, we can plug them into one of the original equations to find the corresponding $y$ value. Let's use the second equation: $y = x^2 + 2x + 1$ Substitute the solution $0$ for $x$: $y = 0^2 + 2(0) + 1$ Evaluate exponent first: $y = 0 + 2(0) + 1$ Multiply to simplify: $y = 0 + 0 + 1$ Add from left to right: $y = 1$ The solution is $(0, 1)$.
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.