How to prove that the following system of equations has only one solution? $
\begin{cases}
(x - 1)^2 + (y + 1)^2 = 25 \\
(x + 5)^2 + (y + 9)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
I have to solve this system of equations. After substituting $y = -\frac{3}{4}x - \frac{13}{2}$ into $(x - 1)^2 + (y + 1)^2 = 25$, I got that $(x + 2)^2 = 0$, so $x = -2$, so $y = -\frac{3}{4} \cdot (-2) - \frac{13}{2} = -5$.
How to prove that this is the only solution?

The first two are equations of circles. Two circles may have no point in common, or only one tangency point, or two points of intersection or all points if they're actually the same circle.
It's easy to check this is not the first nor the last case. Now, subtract the first equation from the second one:
$
\begin{cases}
(x^2 - 2x + 1) + (y^2 + 2y + 1) = 25 \\
(x^2 + 10x + 25) + (y^2 + 18y + 81) = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
$
\begin{cases}
x^2 - 2x + 1 + y^2 + 2y + 1 = 25 \\
12x + 24 + 16y + 80 = 0 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
Now the second and third equation make a system of linear equations, so they may have only zero, one or infinity of satisfying points – but certainly not two. Q.E.D.
