###### 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!

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# Adding and Subtracting Rational Expressions - Problem 3

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!

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Whenever we are adding and subtracting any kind of fraction what we need is a common denominator. And so whenever we are dealing with rational expressions what we have to do is first factor our denominators to see what are least common denominators can be.

So for this one what we want to do is factor them both out, this gives us b plus 5, b plus 1 and this gives us b plus 4, b plus 1. So what I see is that my least common denominator is b plus 1 which is in both terms b plus 5 and b plus 4.

So what we need then is to get a b plus 4 into this denominator and a b plus 5 in that one. But we have an expression we don’t have an equation so we really have to multiply everything by 1. So we have to multiply the first one by b plus 4 over b plus 4 and the second term by b plus 5 over b plus 5.

So I’m now going to strictly focus on my numerators, my denominators are the same which means I can add and subtract as I would with any other fraction. So what we end up with is 4b times b plus 4 so that gives me 4b² plus 16b and then minus 3b times b plus 5. Careful with subtraction because we do have to distribute this negative sign through.

So what we have is 4b² plus 16 b minus 3b² minus 15b. Combining like terms what this gives me is 4b² minus 3b² is just going to be b² and then 16b minus 15b is just going to be plus b. What I see is that these actually can factor, I can factor out a b leaving me with b and b plus 1.

Now remember this was just focusing in the numerator, so I still have to go back and plug in my denominator which is b plus 4, b plus 5 and b plus 1. Now hopefully you notice that we have some common factors so we go ahead and cancel those, leaving me with the final answer of b over b plus 4 times b plus 5.

So just like any fractions in order to subtract them what we need is a common denominator, going to the process simplifying up our numerator and then when we got our numerator factored what I ended up seeing is that something in our numerator and denominator cancelled leaving us with this is our answer.

And I like I said before, always make sure you bring back your denominator into play or so either keep it in there the entire time or if you just want to focus on your numerator that’s okay too, at least in my book your teacher might not be so friendly, but make sure you bring it back in at the end as well you don’t want to just have your numerator as your answer.