Question

How do you solve `1=1/(1-a)+a/(a-1)`?

Answer 1

The solution is `x inRR;x!=1`.

(This means `x` can be any number but `1`.)

Explanation:

Get a common denominator, then simplify the expression:

`1=1/(1-a)+a/(a-1)`

`1color(blue)(\*(a-1))=1/(1-a)color(blue)(\*(a-1))+a/color(red)cancelcolor(black)(a-1)color(red)cancelcolor(blue)(*(a-1))`

`a-1=(a-1)/(1-a)+a`

`a-1=(a-1)/(-a+1)+a`

`a-1=(a-1)/(-1(a-1))+a`

`a-1=color(red)cancelcolor(black)(a-1)/(-1color(red)cancelcolor(black)((a-1)))+a`

`a-1=1/(-1)+a`

`a-1=-1+a`

`a-1=a-1`

`a=a`

Since this is true for all values of `a`, the solution should be `x inRR` (`x` is all real numbers), but it isn't because of the original problem. Take a look at the denominator of those fractions:

`1=1/(1-a)+a/(a-1)`

If `a` is `1`, then there is a division by zero, and that can't happen. Therefore, `x` can be any number but `1`.

The final solution set is `x inRR;x!=1`.

Answer 2

You can't solve this, because this is an identity, valid for all values of `a` (other than `a=1`) . You can, of course, prove this - see below

Explanation:

`1 = (1-a)/(1-a) = 1/(1-a) -a/(1-a) =1/(1-a)+a/(a-1)`