Algebra 2 Common Core

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

Chapter 5 - Polynomials and Polynomial Functions - 5-1 Polynomial Functions - Practice and Problem-Solving Exercises - Page 287: 63

Answer

The solution is $(-20, 360)$.

Work Step by Step

To solve this system of equations, we set the two equations equal to one another because they are both equal to the same number, $y$: $x^2 + x - 20 = x^2 + 2x$ Subtract $x^2$ from both sides: $x - 20 = 2x$ Add $20$ to both sides to move constants to the right side of the equation: $x = 2x + 20$ Subtract $2x$ from both sides to move the variables to the left side of the equation: $-x = 20$ Divide each side by $-1$ to solve for $x$: $x = -20$ Now that we have the value for $x$, we can plug this value into either of the original equations to find the value for $y$. Let us use the second equation because we can deal with fewer terms: $y = (-20)^2 + 2(-20)$ Let's simplify by multiplying out the terms: $y = 400 - 40$ Subtract to solve for $y$: $y = 360$ The solution is $(-20, 360)$.
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