How do you solve `3/(5-x)+2/(4-x) = 8/(x+2)` ?

The quickest way to find at least one solution is to "guess":

Note that `3/(5-x)` will divide evenly if `5-x=3`, i.e. when `x=color(blue)(2)`

Then we find:

`3/(5-color(blue)(2)) + 2/(4-color(blue)(2)) = 3/3+2/2 = 1+1 = 2 = 8/4 = 8/(color(blue)(2)+2)`

`color(white)()`
Are there any other solutions?

Given:

`3/(5-x)+2/(4-x) = 8/(x+2)`

Subtract `3/(5-x)+2/(4-x)` from both sides to get:

`0 = 8/(x+2)-3/(5-x)-2/(4-x)`

`color(white)(0) = 8/(x+2)+3/(x-5)+2/(x-4)`

Multiply through by `(x+2)(x-5)(x-4)` to get:

`0 = 8(x-5)(x-4)+3(x+2)(x-4)+2(x+2)(x-5)`

`color(white)(0) = 8(x^2-9x+20)+3(x^2-2x-8)+2(x^2-3x-10)`

`color(white)(0) = (8x^2-72x+160)+(3x^2-6x-24)+(2x^2-6x-20)`

`color(white)(0) = 13x^2-84x+116`

We can solve this quadratic by any method we like, but we already know that `x=2` is a zero of this quadratic, so `(x-2)` is a factor:

`13x^2-84x+116 = (x-2)(13x-58)`

So the other solution is `x = 58/13`

`3/(5-x)+2/(4-x)=8/(x+2)` can be simplified as

`(3(4-x)+2(5-x))/((5-x)(4-x))=8/(x+2)` or

`(12-3x+10-2x)/(5(4-x)-x(4-x))=8/(x+2)` or

`(22-5x)/(20-5x-4x+x^2)=8/(x+2)` or

`(22-5x)/(20-9x+x^2)=8/(x+2)` or

`8(20-9x+x^2)=(x+2)(22-5x)`

`160-72x+8x^2=x(22-5x)+2(22-5x)` or

`160-72x+8x^2=22x-5x^2+44-10x` or

`8x^2+5x^2-72x-22x+10x+160-44=0` or

`13x^2-84x+116=0` or

`13x^2-26x-58x+116=0` or

`13x(x-2)-58(x-2)=0` or

`(13x-58)(x-2)=0`

Hence `x=2` or `x=58/13`