## Algebra 1: Common Core (15th Edition)

Published by Prentice Hall

# Chapter 9 - Quadratic Functions and Equations - 9-8 Systems of Linear and Quadratic Equations - Practice and Problem-Solving Exercises - Page 599: 23

#### Answer

The solutions of the system are $(-4,-41)$ and $(\frac{1}{3},\frac{7}{3})$.

#### Work Step by Step

$-10x+y=-1 \rightarrow y=10x-1$ $y=3x^2+21x-5$ Substitute $10x-1$ for y $10x-1=3x^2+21x-5$ $3x^2+11x-4=0$ $(3x-1)(x+4)=0$ $x+4=0$ or $3x-1=0$ $x=-4$ or $x=\frac{1}{3}$ Find corresponding y-values. Use either original equation $y=10x-1$ $y=10(-4)-1$ or $y=10(\frac{1}{3})-1$ $y=-41$ or $y=\frac{7}{3}$ The solutions of the system are $(-4,-41)$ and $(\frac{1}{3},\frac{7}{3})$.

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