Question

What is the simplified form of (1/x)-(2/x^2-x) ?

Answer 1

Checking this one as well. Did I get it right?

Answer 2

fourth choice

Answer 3

I know that you're supposed to multiply the right term by (-1).
\[\frac{ 1 }{ x }+\frac{ -2 }{ x ^{2}-x }\]

Then I combined them.
\[\frac{ -1 }{ x + x(x - 1)}\]

And now I'm stuck.

Answer 4

I appreciate the answer but I'd like to know how to do it so I don't have to keep coming back here for help :)

Answer 5

You cannot combine them as you tried to do. To add them they have to have the same denominator. So the first step is to express them with a common denominator.

Answer 6

I'm not sure how to arrive at a common denominator...

Answer 7

Here is the general pattern:

\[\large {a \over b} + {c \over d} = {ad \over bd} + {cb \over bd} = {{ad+cb} \over bd}\]

Of course a the plus sign could be a minus too.

Answer 8

\[\frac{ 1 }{ x }+\frac{ -2 }{ x(x - 1) } = \frac{ 1(x(x - 1)) }{ x(x(x - 1)) } + \frac{ -2(x) }{ x(x(x - 1))} = \frac{ x(x - 1) - 2x }{ x ^{2}(x - 1) }\]

So like this?

Answer 9

That's right. Now you simplify.

Answer 10

\[\frac{ x - 2x }{ x ^{2} }\]
You cancel out the (x - 1) on both top and bottom right?

Answer 11

No. To cancel x-1, x-1 would have to appear in all the terms of the numerator. But it doesn't in the -2x term. There is something else that is in all of the terms of the nominator and is in the denominator (slightly disguised. What is another of writing \(x^2\)?

Answer 12

(x)(x) ?

Answer 13

Yes. So what can you cancel in ALL of the terms of the numerator and in the denominator?

Answer 14

\[\frac{ (x - 1) - 2x }{ x(x - 1)}\]
The x's?

Answer 15

Wait hold on

Answer 16

\[\frac{ (x - 1) - 2 }{ (x - 1) }\]

Answer 17

No. You can only cancel one x in each term of the numerator and in the denominator.

Answer 18

You have the numerator right. But you only burned one x in the denominator.

Answer 19

\[\frac{ (x - 1) - 2x }{ x(x - 1) }\]

So it's still this?

Answer 20

No you had to take one x out of BOTH terms in the numerator and out of the denominator.

Answer 21

\[\frac{ -1 - 2 }{ (x - 1) }\]
I am so confused

Answer 22

Everything was fine here.

\[\frac{ x(x - 1) - 2x }{ x ^{2}(x - 1) }\]

Answer 23

now cancel one x out of x(x-1) and 2x with one x in the denominator.

Answer 24

\[\frac{ (x - 1) - 2 }{ x(x - 1) }\]

Answer 25

Yes. Now it should be looking suspiciously like one of your choices. Why not bring the -2 into the parentheses, or lose the parentheses in the numerator altogether.

Answer 26

\[\frac{ x - 3 }{ x(x -1) }\]

I think this is the answer