## Trying to simplify complex fraction: 1/[1/((x-1) - 2)]

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.

### Trying to simplify complex fraction: 1/[1/((x-1) - 2)]

Argh!

I'm stuck! Well, I guess that goes without saying or I wouldn't be posting.

Although I'm studying pre-calculus right now (implicit domains of composite functions), I am having trouble with an ugly fraction (well, I think so).

I can figure out the composite function with no problem, but they have simplified it to something else and I don't know how they got from this:

$(\frac{1}{\frac{1}{x-1}-2})$

to this:

$(\frac{x-1}{3-2x})$

I am sticking with the bottom first, and trying to find a common denominator to subtract $2$ from $(\frac{1}{x-1})$.

I tried using $x-1$ as a common denominator, but I end up getting $(\frac{1-2x-2}{x-1})$ which doesn't work (unless I messed up there). I know that if you have a fraction as the denominator, you can just multiply by its reciprocal instead. So I am thinking that they figured out how to subtract $2$ from $(\frac{1}{x-1})$ then just flipped it, which became the answer because you'd be multiplying by the numerator (1) anyway.

So how did they get from $(\frac{1}{x-1}-2)$ to $(\frac{x-1}{3-2x})$?
landgazr

Posts: 5
Joined: Mon Nov 14, 2011 4:26 am

landgazr wrote:I don't know how they got from this: $(\frac{1}{\frac{1}{x-1}-2})$ to this: $(\frac{x-1}{3-2x})$

I am...trying to find a common denominator.... I tried using $x-1$ as a common denominator,
but I end up getting $(\frac{1-2x-2}{x-1})$ which doesn't work....

What do you mean by this "not working"? (I'm assuming that you typoed the grouping symbols, since there should be parentheses around the "x - 1" when you take the -2 through.)

The steps for the simplification (using the proper grouping symbols for clarity) would be something like:

. . . . .$\frac{1}{x\, -\, 1}\, -\, \frac{2}{1}\left(\frac{x\, -\, 1}{x\, -\, 1}\right)$

. . . . .$\frac{1}{x\, -\, 1}\, -\, \frac{2(x\, -\, 1)}{x\, -\, 1}$

. . . . .$\frac{1}{x\, -\, 1}\, -\, \frac{2x\, -\, 2}{x\, -\, 1}$

. . . . .$\frac{1\, -\, (2x\, -\, 2)}{x\, -\, 1}$

. . . . .$\frac{1\, -\, 2x\, +\, 2}{x\, -\, 1}$

. . . . .$\frac{3\, -\, 2x}{x\, -\, 1}$

. . . . .$\frac{1}{\left(\frac{3\, -\, 2x}{x\,-\, 1}\right)}$

Flip-n-multiply, and you should be good to go!

stapel_eliz

Posts: 1715
Joined: Mon Dec 08, 2008 4:22 pm

### Re: Trying to simplify complex fraction: 1/[1/((x-1) - 2)]

Ah! I see what I did wrong.

It was here:

$\frac{1-2x+2}{x-1}$

My problem was I didn't add 2; I subtracted it. I messed up a sign. I should have distributed $-1$ through $(2x-2)$.

What I meant by it "not working" was that I didn't get $3-2x$ from $1-2x-2$.

I'm not sure I see your point about the grouping symbols, but I do see how you parenthesized $x-1$, both numerator and denominator, to multiply by $-2$ to get $2x-2$. Then, since that is its own term, and being subtracted from $1$, I need to make sure that $2$ has the correct sign, since subtracting a negative is adding its positive. I know that's very elementary, but it looks like I was on the right track - I just didn't get a sign right.
landgazr

Posts: 5
Joined: Mon Nov 14, 2011 4:26 am