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.