Chapter 10 - Section 10.8 - Systems of Nonlinear Equations - 10.8 Exercises - Page 754: 4

{$\sqrt 5, 2 \sqrt 5$} and {$-\sqrt 5, -2 \sqrt 5$}

Given: $x^2+y^2=25$ and $y=2x$ This implies that $x^2+(2x)^2=25$ or, $x^2+4x^2=25 \implies 5x^2 =25$ This gives: $x=\pm \sqrt 5$ a) At $x= \sqrt 5$, we have $y=2x=2\sqrt 5$ b) At $x=- \sqrt 5$, we have $y=2x=-2\sqrt 5$ Thus, the solutions set are: {$\sqrt 5, 2 \sqrt 5$} and {$-\sqrt 5, -2 \sqrt 5$}