Basic algebra (composite functions)

Question

The question is

$h(x)=\frac{(1+x)}{(1-x)}$ find $h(1-x)$

I understand how to solve the question, it's:

$1+\frac{(1+x)}{(1-x)}$

What I can't seem to understand is why the denominator

$1-(1-x)$ equals $x$ and not $-x$

I'm guessing that it's because the $-$ cancels out, but since it's not multiplication (e.g $-2(1-3)=-2+6$) why does it still cancel out? Why can't it be answered as $1-1-x=-x$?

Answer 1

$$h(x)=\frac{1+x}{1-x} \Rightarrow h()=\frac{1+()}{1-()}$$

Now put into the brackets what you need ! Clearly $$1-(1-x)=1-1-(-x)=1-1+x$$

$$\Rightarrow h(1-x)=\frac{1+(1-x)}{1-(1-x)}=\frac{1+1-x}{1-1+x}=\frac{2-x}{x}$$

Answer 2

You can think of it like this: \begin{align} 1-(1-x) &= 1 + (-1)\cdot(1-x) = 1 + ((-1) \cdot 1 - (-1) \cdot x) \\ &= 1 + (-1 - (-x)) = 1 + (-1 +x) \\ &= x\,. \end{align}