Question 997675
10/(x^2-25) - 1/(x-5) = 1 
<pre>{{{10/(x^2 - 25) - 1/(x - 5) = 1 }}}, with {{{x <> 5}}} and {{{x <> - 5}}}
10 - 1(x + 5) = 1(x - 5)(x + 5) -------- Multiplying by LCD, {{{x^2 - 25}}}, or (x - 5)(x + 5)
{{{10 - x - 5 = x^2 - 25}}}
{{{5 - x = x^2 - 25}}}
{{{x^2 + x - 5 - 25 = 0 }}}
{{{x^2 + x - 30 = 0}}}
(x + 6)(x - 5) = 0 
{{{highlight_green(x = - 6)}}}       OR     x = 5 (ignore)