$a$ and $b$ are solutions of $ \frac{1}{x^{2} - 10x-29} + \frac{1}{x^{2} - 10x-45} - \frac{2}{x^{2} - 10x-69} = 0 $, $a+b=?$ $a$ and $b$ are solutions of
$$ \frac{1}{x^{2} - 10x-29} + \frac{1}{x^{2} - 10x-45} - \frac{2}{x^{2} - 10x-69} = 0 $$
What is $a+b=?$
$$ $$
Are there better approaches than the one below?

Solution:
By letting $x^{2} - 10x = y$, then we have 
$$ \frac{1}{y-29} + \frac{1}{y-45} - \frac{2}{y-69} = 0, \:\: y \notin \{ 29,45,69 \} $$
and
$$ (y-45)(y-69) + (y-29)(y-69) - 2(y-29)(y -45) = 0 $$
$$ (y- 69)(y-37) = (y-29)(y-45) $$
$$ y^{2} - 106 y + 69 \cdot 37 = y^{2}-74y + 29 \cdot 45 $$
$$ -32y = 3 (29 \cdot 15 - 23 \cdot 37) = -1248 $$
$$  y = x^{2} - 10x = 39$$
Here are the roots:
$$ x^{2} -10x - 39 = 0 \implies (x-13)(x+3) = 0$$
So the answer is $a + b = 13 - 3 = 10$

Wait so at the start, you say $y=x^2-10x$ then proceed to say $y=x^2-10x-39$
Apart from that, this is probably the best solution.
