Here's the question you clicked on:
Cubi-Cal
Add and simplify... x/(x-1) - 2x/(x^2-1)
Where are you stuck?
\[\frac{ x }{ (x-1) }-\frac{2x}{x^2-1}\]
I'm pretty sure there is some factoring int there, but I'm not sure.
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.
Right! So what do I multiply to get 1 and adds up to be 1?
Hint: the denominator of the second term is the difference of 2 perfect squares.
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.
So to start, can you factor x^2 - 1 ?
Another hint:\[a ^{2} - b ^{2} = (a + b)(a - b)\]
If you substitute a=x and b=1 into the equation I wrote out, you should be able to factor:\[x ^{2} - 1\]
Okay, so my final answer would be x^2-x/x^2-1?
No, but that's a good intermediate answer. You need to factor out "x-1" from the top and bottom.
\[\frac{ x }{ x - 1 } - \frac{ 2x }{ x ^{2} - 1 } = \frac{ x }{ x - 1 } - \frac{ 2x }{ (x - 1)(x + 1) } \] \[\frac{ x(x + 1) }{ (x - 1)(x + 1) } - \frac{ 2x }{ (x - 1)(x + 1) } = \frac{ x ^{2} + x - 2x }{ (x - 1)(x + 1) }\] \[\frac{ x ^{2} - x }{ (x - 1)(x + 1) } = \frac{ x\cancel{(x - 1)} }{ \cancel{(x - 1)}{(x + 1)} }\]
Would you mind helping me with some other problems? My teacher isn't available.
ok, see you in another question!