# Thread: Solving linear equation, example of my problem

1. ## Solving linear equation, example of my problem

I understand how to solve problems with denominators that are variable only for example:

$(1 + x)/ (2x)$
but how does one solve a problem that has a constant integer in the denominator? For example :
$-1/(x+1) = 1 / (3x + 3) - 2 /(x-4)$

I dont know how to find the LCM for an equation like this to multiply on both sides.

2. ## Re: Solving linear equation, example of my problem

Notice that we have:

$\frac{-1}{x+1}=\frac{1}{3(x+1)}-\frac{2}{x-4}$

$3(x+1)(x-4)$ looks like a perfectly reasonable common denominator to me.

3. ## Re: Solving linear equation, example of my problem

Wow. I didn't see that you could simplify 3x + 3 down to 3(x + 1) this makes it much simpler since then you can cancel out x+1 on the other side. Great. Thanks, a ton of help.

4. ## Re: Solving linear equation, example of my problem

If by cancelling you mean multiplying both sides of the equation by it, then I agree. You can't just take them out of both sides as a common factor though.