Answer
$(-1,3),(2,6)$.
Work Step by Step
Use the first relation, substitute $y$ in the second equation to get
$x^2-(x+4)=-2$ or $x^2-x-2=0$
or $(x-2)(x+1)=0$
thus $x=-1, 2$, use to the first equation,
we have the solutions $(-1,3),(2,6)$.
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Section 13.4 The Tangent Problem; The Derivative - 13.4 Assess Your Understanding - Page 918: 52
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