Answer
Solution set: $\{(4,-3),(-4,3)\}$
Work Step by Step
$x^{2}+y^{2}=25$ Equation $(1)$
$3x+4y=0$ Equation $(2)$
From Equation $(2)$
$4y = -3x$
$y = -\frac{3x}{4}$ Equation $(3)$
Substituting $y = -\frac{3x}{4}$ in Equation $(1)$
$x^{2}+y^{2}=25$
$x^{2}+( -\frac{3x}{4})^{2}=25$
$x^{2}+(\frac{9x^{2}}{16})=25$
Taking LCD,
$(\frac{16x^{2}+9x^{2}}{16})=25$
$9x^{2}+16x^{2}=25 \times16$
$25x^{2}=25 \times 16$
$x^{2}= 16$
$x = ±4$
$x = 4$ or $x=-4$
Substitute $ x$ values in Equation $(3)$ to get $y$,
Let $x = 4 $
$y = -\frac{3x}{4}$
$y = -\frac{3}{4} \times 4$
$y = -3$
Let $x = -4 $
$y = -\frac{3x}{4}$
$y = -\frac{3}{4} \times -4$
$y = 3$
$(4,-3)$ and $(-4,3)$ satisfies both equations.
So, the solution set is $\{(4,-3),(-4,3)\}$
