Answer
The solution is (-4, 3) and (4, -3).
See Graph I for the 2 intersection points.
Work Step by Step
Now,
$x^2 + y^2 = 25 \longrightarrow (i)$
$3x + 4y = 0 \longrightarrow (ii)$
From (ii), we have
$3x + 4y = 0$
$x = \frac{-4y}{3} \longrightarrow (iii)$
Sub. (iii) into (i), we have
$(\frac{-4y}{3})^2 + y^2 = 25$
$\frac{16y^2}{9} + y^2 = 25$
$16y^2 + 9y^2 = 225$
$25y^2 = 225$
$y^2 = 9$
$y = 3$ or $y = −3$
Sub. $y = 3$ into (iii), $x = −4$
Sub. $y = −3$ into (iii), $x = 4$
The solution is (-4, 3) and (4, -3).
See Graph I for the 2 intersection points.