Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 13 - Section 13.3 - Solving Nonlinear Systems of Equations - Exercise Set - Page 948: 1

Answer

The solution is (-3, 4) and (3, -4). See Graph I for the 2 intersection points.

Work Step by Step

Now, $x^2 + y^2 = 25 \longrightarrow (i)$ $4x + 3y = 0 \longrightarrow (ii)$ From (ii), we have $4x + 3y = 0$ $x = \frac{-3y}{4} \longrightarrow (iii)$ Sub. (iii) into (i), we have $(\frac{-3y}{4})^2 + y^2 = 25$ $\frac{9y^2}{16} + y^2 = 25$ $9y^2 + 16y^2 = 400$ $25y^2 = 400$ $y^2 = 16$ $y = 4$ or $y = -4$ Sub. $y = 4$ into (iii), $x = \frac{-3(4)}{4} = -3$ Sub. $y = -4$ into (iii), $x = \frac{-3(-4)}{4} = 3$ The solution is (-3, 4) and (3, -4). See Graph I for the 2 intersection points.
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