Question

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

Answer 1

`x=7.5 or 15/2`

Explanation:

`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`