How do you solve $\frac{10}{x^{2}-25}-\frac{1}{x-5}=\frac{3}{x+5}$ ?

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Answer 1 · 2016-09-28T16:28:00

$x=7.5 or 15/2$

$10/(x^2-25) - 1/(x-5) = 3/(x+5)$

Rewrite in an equivalent form by factoring

$10/((x+5)(x-5)) - 1/(x-5) = 3/(x+5)$

Multiply $1/(x+5) " by " (x-5)/(x-5)$

And the whole thing becomes

$10/((x+5)(x-5)) - (x-5)/((x+5)(x-5) )= 3/(x+5)$

Simplify and get $(10-(x-5))/((x+5)(x-5)) = 3/(x+5)$

Keep going and get $(10-x+5)/((x+5)(x-5)) = 3/(x+5)$

Multiply both sides by $(x-5)(x+5)$

$(cancel((x-5)(x+5)) xx(10-x+5))/(cancel((x+5)(x-5))) = (3xx(x-5)cancel(x+5))/cancel(x+5)$

and get

$10-x+5 = 3(x-5)$

Simplify and get $10-x+5 = 3x-15$

Keep going to get $30=4x$

so $x=7.5 or 15/2$

How do you solve \frac{10}{x^{2}-25}-\frac{1}{x-5}=\frac{3}{x+5}\frac{10}{x^{2}-25}-\frac{1}{x-5}=\frac{3}{x+5}?