Precalculus (6th Edition) Blitzer

Published by Pearson

Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 850: 5

Answer

$(-3,11)$ or $(4,-10)$

Work Step by Step

Here, we have $y=x^2-4x-10$ Now, $(x^2-4x-10)=-x^2-2x+14$ This gives: $2x^2-2x-24=0$ This yields : $(x+3)(x-4)=0$ when $x=-3$ then we have $y=(-3)^2-4(-3)-10=11$ and when $x=4$ then we have $y=(-3)^2-4(-3)-10=-10$ Hence, our answers are: $(-3,11)$ or $(4,-10)$

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