College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 5 - Systems of Equations and Inequalities - Exercise Set 5.4 - Page 559: 45

Answer

-2 or 2 and -1 or 1

Work Step by Step

We can write the system: $\begin{cases} x^2-y^2=3\\ 2x^2+y^2=9 \end{cases}$ We will use the addition method. Add Equation 2 to Equation 1 to eliminate $y$ and determine $x$: $x^2-y^2+2x^2+y^2=3+9$ $3x^2=12$ $x^2=\dfrac{12}{3}$ $x^2=4$ $x=\sqrt 4$ $x=\pm 2$ $x_1=-2$ $x_2=2$ Substitute each of the values of $x$ in Equation 1 to determine $y$: $x^2-y^2=3$ $x_1=-2\Rightarrow (-2)^2-y^2=3\Rightarrow y^2=1\Rightarrow y_1=-1,y_2=1$ $x_2=2\Rightarrow 2^2-y^2=3\Rightarrow y^2=1\Rightarrow y_1=-1,y_2=1$ The system's solutions are: $(-2,-1), (-2,1), (2,-1),(2,1)$ The numbers are -2 or 2 and -1 or 1
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