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

Mar 12, 2018

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:

$1 = \frac{1}{1 - a} + \frac{a}{a - 1}$

$1 \textcolor{b l u e}{\setminus \cdot \left(a - 1\right)} = \frac{1}{1 - a} \textcolor{b l u e}{\setminus \cdot \left(a - 1\right)} + \frac{a}{\textcolor{red}{\cancel{\textcolor{b l a c k}{a - 1}}}} \textcolor{red}{\cancel{\textcolor{b l u e}{\cdot \left(a - 1\right)}}}$

$a - 1 = \frac{a - 1}{1 - a} + a$

$a - 1 = \frac{a - 1}{- a + 1} + a$

$a - 1 = \frac{a - 1}{- 1 \left(a - 1\right)} + a$

$a - 1 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{a - 1}}}}{- 1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(a - 1\right)}}}} + a$

$a - 1 = \frac{1}{- 1} + a$

$a - 1 = - 1 + a$

$a - 1 = a - 1$

$a = a$

Since this is true for all values of $a$, the solution should be $x \in \mathbb{R}$ ($x$ is all real numbers), but it isn't because of the original problem. Take a look at the denominator of those fractions:

$1 = \frac{1}{1 - a} + \frac{a}{a - 1}$

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.

Mar 12, 2018

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
$1 = \frac{1 - a}{1 - a} = \frac{1}{1 - a} - \frac{a}{1 - a} = \frac{1}{1 - a} + \frac{a}{a - 1}$