Combining, simplifying and factoring fractions.

**Lemmy** · Apr 27th 2008, 03:18 PM

I have no idea what I'm doing as I haven't touched this topic in months. If anyone can help me, that would be great.

Edit: For some reason, all of the fractions are being messed up so I'm going to post a picture of how they should look.

**icemanfan** · Apr 27th 2008, 03:45 PM

Originally Posted by Lemmy

I have no idea what I'm doing as I haven't touched this topic in months. If anyone can help me, that would be great.

Edit: For some reason, all of the fractions are being messed up so I'm going to post a picture of how they should look.

For 1)
$\frac{3x + 1}{x^2 - 1} - \frac{1}{x + 1}$ notice that $x^2 - 1 = (x - 1)(x + 1)$ , so in order to subtract the two fractions you need the same denominator which means multiplying the second fraction by $\frac{x - 1}{x - 1}$ to get the difference $\frac{3x + 1}{x^2 - 1} - \frac{x - 1}{x^2 - 1}$ and then computing the difference as $\frac{4x}{x^2 - 1}$ .

**Lemmy** · Apr 27th 2008, 03:54 PM

Originally Posted by icemanfan

For 1)
$\frac{3x + 1}{x^2 - 1} - \frac{1}{x + 1}$ notice that $x^2 - 1 = (x - 1)(x + 1)$ , so in order to subtract the two fractions you need the same denominator which means multiplying the second fraction by $\frac{x - 1}{x - 1}$ to get the difference $\frac{3x + 1}{x^2 - 1} - \frac{x - 1}{x^2 - 1}$ and then computing the difference as $\frac{4x}{x^2 - 1}$ .

Thanks. That's pretty much the only one of the four that I was on the right track for.

**icemanfan** · Apr 27th 2008, 04:02 PM

Originally Posted by Lemmy

Thanks. That's pretty much the only one of the four that I was on the right track for.

The second problem on the list is actually the easiest. What happens if you multiply the fraction on the right by $\frac{-1}{-1}$ ?

**Lemmy** · Apr 27th 2008, 04:11 PM

Originally Posted by icemanfan

The second problem on the list is actually the easiest. What happens if you multiply the fraction on the right by $\frac{-1}{-1}$ ?

Yes, I was trying to figure out which way to go about doing that. Guess I haven't done these in a while.

**earboth** · Apr 27th 2008, 11:11 PM

Originally Posted by Lemmy

I have no idea what I'm doing as I haven't touched this topic in months. If anyone can help me, that would be great.

Edit: For some reason, all of the fractions are being messed up so I'm going to post a picture of how they should look.

to #4:

Keep in mind that

$\frac14-y^2 = -\left(y^2-\frac14\right)=-\left(y-\frac12\right) \left(y+\frac12\right)$

Now your fraction becomes:

$\frac{y-\frac12}{-\left(y-\frac12\right) \left(y+\frac12\right)} = \frac1{-\left(y+\frac12\right)} = -\frac1{y+\frac12}$

**earboth** · Apr 27th 2008, 11:18 PM

Originally Posted by Lemmy

I have no idea what I'm doing as I haven't touched this topic in months. If anyone can help me, that would be great.

Edit: For some reason, all of the fractions are being messed up so I'm going to post a picture of how they should look.

to #3:

$\frac4{x-1} = \frac5{2x-2} + \frac{3x}4~,~x \neq 1$

The common denominator of all fractions is $4(x-1)$
Multiply both sides of the equation by the denominator. You should get:

$16 = 10 + 3x(x-1)~\implies~3x^2-3x-6=0$

For confirmation only: x = -1 or x = 2

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