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 263: 42

Answer

The solution is $(0, -5)$.

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: $-x^2 + x - 5 = x - 5$ We want to move all terms to the left side of the equation. $-x^2 + x - x - 5 + 5 = 0$ Combine like terms: $-x^2 = 0$ Divide both sides by $-1$: $x^2 = 0$ Take the square root of both sides of the equation: $x = 0$ Now that we have the value for $x$, we can plug it into one of the original equations to find the corresponding $y$ value. Let's use the second equation: $y = x - 5$ Substitute the solution $0$ for $x$: $y = 0 - 5$ Add to solve: $y = -5$ The solution is $(0, -5)$.
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