Answer
$(3,-4),(-3,4)$
Work Step by Step
$x^{2}+y^{2}=25$ Equation $(1)$
$4x+3y=0$ Equation $(2)$
From Equation $(2)$
$3y = -4x$
$y = -\frac{4x}{3}$ Equation $(3)$
Substituting $y = -\frac{4x}{3}$ in Equation $(1)$
$x^{2}+y^{2}=25$
$x^{2}+(-\frac{4x}{3})^{2}=25$
$x^{2}+(\frac{16x^{2}}{9})=25$
Taking LCD,
$(\frac{9x^{2}+16x^{2}}{9})=25$
$9x^{2}+16x^{2}=25 \times 9$
$25x^{2}=25 \times 9$
$x^{2}= 9$
$x = ±3$
$x = 3 $ or $x=-3$
Substitute $ x$ values in Equation $(3)$ to get $y$.
Let $x = 3 $
$y = -\frac{4x}{3}$
$y = -\frac{4}{3} \times 3$
$y = -4$
Let $x = -3 $
$y = -\frac{4x}{3}$
$y = -\frac{4}{3} \times -3$
$y = 4$
$(3,-4)$ and $(-3,4)$ satisfies both equations.
So, the solution set is $\{(3,-4),(-3,4)\}$
