Adding fractions with algebraic denominators

[conwy]

Hey all,

So this one question in a practice exercise really has me stumped...

[MATH]\frac{1}{x-1}+\frac{1}{x}=3[/MATH]
I understand, in regular fractions without symbols, how to find the lowest common denominator.

But what on earth is the lowest common denominator of two algebraic denominators?

How do you know which one is lowest, or even what the common divisor is, when they're unknown quantities??

Or is there some fancy trick to somehow treat both expressions (x-1 and x) the same way?

I feel pretty terrible for not knowing how to do this already. Is there some concept that I missed out on learning? Is there a specific technique for dealing with algebraic fractions that I am ignorant of?

Help much appreciated. Thanks!

 

[R.M.]

First of all, you do not need the lowest denominator to add/subtract fractions; you simply need a common denominator. No fancy trick needed, other than your denominator will be their product. Just make sure you multiply the numerator by the appropriate factor as well.

 

[pka]

So this one question in a practice exercise really has me stumped...
[MATH]\frac{1}{x-1}+\frac{1}{x}=3[/MATH]

The LCD is \(x(x-1)\). Now multiply all three terms by that.
Post your results.

 

[JeffM]

You NEVER have to use the lowest common denominator; using the lowest merely simplifies things when you are doing arithmetic without a calculator.

The answer is to clear fractions, meaning to multiply both sides of the equation by the product of all the denominators. That eliminates all fractions.

[MATH]\dfrac{1}{x - 1} + \dfrac{1}{x} = 3 \implies[/MATH]
[MATH]x(x - 1) * \left ( \dfrac{1}{x - 1} + \dfrac{1}{x} \right ) = x(x - 1) * 3 \implies[/MATH]
[MATH]x * 1 + (x - 1) * 1 = 3(x^2 - x) \implies 2x - 1 = 3x^2 - 3x \implies WHAT?[/MATH]

 

[Dr.Peterson]

[MATH]\frac{1}{x-1}+\frac{1}{x}=3[/MATH]
I understand, in regular fractions without symbols, how to find the lowest common denominator.

But what on earth is the lowest common denominator of two algebraic denominators?

How do you know which one is lowest, or even what the common divisor is, when they're unknown quantities??

Or is there some fancy trick to somehow treat both expressions (x-1 and x) the same way?

As others have said, you don't need to worry too much about getting the lowest common denominator; but if you don't, you may find yourself with a considerably harder problem (for example, something complicated to factor later).

An important point to learn is that when two denominators have no common factors (that is, their GCF is 1), then the LCD is just their product (which I call the OCD, for Obvious Common Denominator). So in this case, you can use x(x-1) without any trouble.

I would expect your textbook to tell you a method for finding the LCD (unless this is review and they expect you to have learned it elsewhere). One way to think about it is to start with one of the denominators, and just tack on extra factors as needed to get a multiple of each denominator. For example, if the denominators had been x(x+1) and (x+1)^2(x-3), then you might start with the latter, (x+1)^2(x-3), and notice that you need an x to make it a multiple of x(x+1), but you don't need another (x+1). So the LCD is just x(x+1)^2(x-3).

Another way is to use the greatest power of each distinct factor that is found among all the denominators. This really amounts to the same thing.

 

[Steven G]

Here is a nice video to watch.