Question

Add and simplify... x/(x-1) - 2x/(x^2-1)

Answer 1

Where are you stuck?

Answer 2

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

Answer 3

I'm pretty sure there is some factoring int there, but I'm not sure.

Answer 4

You will want to factor the denominator of that second term. Then you will want to get the LCM (Least Common Multiple) for both terms.

Answer 5

Right! So what do I multiply to get 1 and adds up to be 1?

Answer 6

Hint: the denominator of the second term is the difference of 2 perfect squares.

Answer 7

Once you factor that denominator of that second term, you will see one of the factors as the denominator in the first term. Take the OTHER factor of the denominator of the second term and make a factor of "1" out of it (by putting the factor in the top and bottom) and multiply the first term by that. Then you will have a common denominator.

Answer 8

So to start, can you factor

x^2 - 1

?

Answer 9

Another hint:\[a ^{2} - b ^{2} = (a + b)(a - b)\]

Answer 10

Think of 1 as 1^2

Answer 11

If you substitute a=x and b=1 into the equation I wrote out, you should be able to factor:\[x ^{2} - 1\]

Answer 12

Okay, so my final answer would be x^2-x/x^2-1?

Answer 13

No, but that's a good intermediate answer. You need to factor out "x-1" from the top and bottom.

Answer 14

uh.. x/x-1?._.

Answer 15

Or would it be +1?

Answer 16

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

Answer 17

Ohhhh! Okay thanks!

Answer 18

uw!

Answer 19

Would you mind helping me with some other problems? My teacher isn't available.

Answer 20

ok, see you in another question!

Answer 21

Okay!