 ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

##### Thank you for watching the video.

To unlock all 5,300 videos, start your free trial.

# Simplifying Complex Fractions - Problem 1

Carl Horowitz ###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

Share

Simplifying a complex fraction where we actually have more than one fraction in either the numerator or the denominator. So there’s two different ways of solving this. Often I refer to as method one and method two because I don’t know which way your teacher’s teaching it, we’ll do both saving off the method number just to not confuse you.

So the first way of simplifying these is to combine the numerator and the denominator separately, okay. We can solve a simple complex fraction by flipping and multiplying until we have a single fraction in both the numerator and the denominator. So the easiest way to do that is to combine the numerator into one fraction, combine the denominator into one fraction. Looking at this I see that my common denominator here is ab, so I need to multiply this first term by b/b and the second term by a/a so my new numerator is b plus a all over ab.

Doing the same idea in the denominator, need to combine these two into a single fraction so we need a common denominator. In this case my least common denominator is a² b². Multiplying this term by b² over b² and the first term by a² over a² I end up with my denominator being b² minus a² all over a²b².

Now that I have combined my numerator and my denominator and I have a single fraction at the top, single fraction at the bottom, I now can flip and multiply. Okay, so our numerator stays the same, b plus a over ab and our denominator gets flipped and we turn this into multiplication. So have a²b² in the numerator over b² minus a² in the denominator and we just multiply.

First thing we can do is cancel out some terms so we can cancel an a and a b from both, b² minus a², this can be factored, b minus a, b plus a, our b plus a's cancel leaving us with ab in the numerator and b minus a in the denominator. That’s one way of solving this out, by combining our numerator and denominator separately, flipping, multiplying.

Let’s do the other method as well and in general this is my favorite method just because I think it’s a little bit easier but either one is going to work just fine. How this works is looking at your entire problem, looking at both the numerator and the denominator we want to find the least common denominator of everything all together. In this case we’re looking at a, b, a² and b², our least common denominator is a²b².

Multiply the top by that least common denominator, a²b² and multiply the bottom by that same thing, a²b². We’re actually multiplying the top and the bottom by the same thing, so we’re multiplying by 1, not changing our problem.

So what happens when we distribute this in? This goes into both terms so our a cancel, one of our as cancel here leaving us with ab² for our first term. One of our b’s cancels here leaving us with a²b in our second term. Same idea for the denominator; distribute this a²b² in over here. Our a²s cancel leaving us with b²s. Distribute the a²b² in here b²s cancel leaving us with minus a². What we’ve done is we’ve multiplied by the least denominator everything and actually gotten rid of all our denominators all together.

From here we need to factor. so from the top we can take out, a, ab and leaving us with b plus a, divided by, here we have the difference of squares so we can just factor that, b minus a, b plus a. Our b plus a's cancel leaving us with ab over b minus a. The same solution we had for our other method.

What we have here is actually two different ways of solving that exact same problem, one that we referred to as method one, one that we referred to as method two. One of which combining our numerator and denominator separately, flip and multiply the other multiply by your least common denominator and then simplify it as you go.