Question

How do you solve `\frac { 3} { x + 1} - \frac { x + 5} { 1- x } = \frac { - 6} { x ^ { 2} - 1}`?

Answer 1

Note that in the right term `x^2-1=(x+1)(x-1)`
Also note that in the second term `-(1-x)=x-1`

Explanation:

We rewrite:
`3/(x+1)+(x+5)/(x-1)=(-6)/((x+1)(x-1))`

We multiply both left terms with differently "disguised" 1's
`3/(x+1)xx(x-1)/(x-1)+(x+5)/(x-1)xx(x+1)/(x+1)=(-6)/((x+1)(x-1))`

Work this out to:
`(3(x-1))/((x+1)(x-1))+((x+5)(x+1))/((x+1)(x-1))=(-6)/((x+1)(x-1))`

Provided we assume `x!=-1andx!=1` (division by zero):
We may now multiply by `(x+1)(x-1)`

`3(x-1)+(x+5)(x+1)=-6->`
`3x-3+x^2+x+5x+5=-6->`

Rework this to:

`x^2+9x+8=0->(x+1)(x+8)=0->`
`x=-1orx=-8`

Since `x!=-1` (see before) the only solution is `x=-8`