Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 851: 41

Answer

$(\dfrac{12}{5},-\dfrac{29}{5})$, $(-2,3)$

Work Step by Step

After using the substitution method, we get $x^2+(-2x-1)^2+3(-2x-1)=22$ or, $x^2+4x^2+4x+1-6x-3=22$ or, $5x^2-2x-24=0$ Thus, we have $(5x-12)(x+2)=0$ when $x=12/5$ then we have $y=-2(12/5)-1$ This gives: $y=-\dfrac{29}{5}$ when $x=-2$ then we have $y=-2(-2)-1$ This gives: $y=3$ Hence, our answers are: $(\dfrac{12}{5},-\dfrac{29}{5})$, $(-2,3)$
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