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 850: 23


$(-2,-1)$, $(-2,1)$, $(2-1)$, $(2,1)$

Work Step by Step

Here, we have $3(3x^2+4y^2-16)=(3)(0) \implies 9x^2+12y^2-48=0$ ....(1) and $4(2x^2-3y^2-5)=(4)(0) \implies 8x^2-12y^2-20=0$ ...(2) After adding the above two equations, we get $17x^2-68 =0$ This gives: $x^2=4 \implies x=-2,2$ when $x=2$ then we have $3(2)^2+4(2)^2-16=0 \implies y^2=1$ This gives: $y=1, -1$ when $x=-2$ then we have $3(-2)^2+4(-2)^2-16=0 \implies y^2=1$ This gives: $y=1, -1$ Hence, our answers are: $(-2,-1)$, $(-2,1)$, $(2-1)$, $(2,1)$
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