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## Algebra (all content)

### Course: Algebra (all content) > Unit 13

Lesson 4: Adding & subtracting rational expressions- Adding & subtracting rational expressions: like denominators
- Intro to adding & subtracting rational expressions
- Add & subtract rational expressions: like denominators
- Intro to adding rational expressions with unlike denominators
- Adding rational expression: unlike denominators
- Subtracting rational expressions: unlike denominators
- Add & subtract rational expressions (basic)
- Least common multiple
- Least common multiple: repeating factors
- Least common multiple
- Subtracting rational expressions: factored denominators
- Least common multiple of polynomials
- Adding & subtracting rational expressions
- Add & subtract rational expressions: factored denominators
- Subtracting rational expressions
- Add & subtract rational expressions

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# Subtracting rational expressions

CCSS.Math: ,

Sal subtracts and simplifies (a-2)/(a+2) - (a-3)/(a²+4a+4). Created by Sal Khan.

## Want to join the conversation?

- at3:00why did Sal make it into a difference of square, couldn't he have just left it as

(a+2)(a-2)(a-3)/(a+2)(a+2), then crossed out the (a+2)'s and simplified? (just curious)(8 votes)- No he couldn't, listen carefully around at4:05up to4:21to see why.

The final answer is:

a²-a-1/(a+2)²

Because if we factor a²-a-1 it will not turn to (a+2)(a-2)(a-3). In fact if solve this using the quadratic equation you will see that it will factor into complex numbers.(10 votes)

- how come he didn't change + a-3/(a+2)(a+2) to -a-3/(a+2)(a+2) like that's what her did in those previous videos but when I changed the sign my answer came out different ! help someone please explain 😩😥😭😭😭😭(3 votes)
- The minus in front of the fraction applies to the entire numerator: - (a - 3)

Distribute the minus: - (a - 3) = -a + 3

Hope this helps.(10 votes)

- The question asks for me to Simplify and expand the following expression:

(5z−2)/(2z+2)−(z/3z−5)

When I put in the answer in this exact format

(13z^2 −33z+10)/(6z^2 −4z−10)

It tells me "Your answer is not fully expanded and simplified.", when I go look up the answer for the question is display the same answer yet it keeps telling me it's incorrect. Is there something i'm doing incorrect in the input?(3 votes)- I am having the same issue with the last 2 tests in that section. I wrote it out the same way you did, the numbers in the numerator and denominator in parentheses divided by one another. I clicked on report the problem on both tests, so that they know there is a problem. You may want to try that too. If we are writing it out incorrectly, someone please reply.(5 votes)

- Good eve my given in denominator is the same how to solution them(3 votes)
- I had a question that was (7z^2)/(12z^3-4z^2) plus another rational expression. I pulled out 4z^2 to make it (7z^2)/ 4z^2(3z-1). Why is it wrong to cancel the two z^2 before adding?(3 votes)
- Based on the information you give, it isn't wrong to cancel them out --- BUT --- without seeing the entire problem and the rest of your work on it, I can't be sure why someone would have said it was wrong to cancel them out. ( Is it possible that not canceling them out would have made finding a common denominator easier?)(2 votes)

- how would i solve: 2x+1/x2+8x+16x-3/x2-16(2 votes)
- Hi Jennifer

If you are asking how to solve 2x+1/(x^2+8x+16x-3/x^2-16), than an example of how you would solve it goes like this where 2x+1 = a, x^2+24x-3 = b, and x^2-16 = c:

=a / (b/c)

=a/1 * c/b

=ac/b

So now let's re-arrange a, b, and c, using your numbers:

= ac / b

=(2x+1)(x^2-16) / x^2+8x+16

= (2x+1)(x+4)(x-4) / (x+4)(x+4)

cross out the x+4 in the numerator and the denominator because anything over itself = 1

=(2x+1)(x-4) / (x+4)

Hope this helps!(2 votes)

- 3 2 2

------- + ------- - -------

x (x-2) x^2

how would I solve this?(2 votes) - I have just now completed the Algebra II course. However, there are a couple of sections I need to review before I take the Unit Test because I started taking the older version of Algebra II and then switched to the newer version of Algebra II and there are more topics covered in the new version of Algebra II.

But bottom line I have two questions concerning my continued math journey.

1. After I complete my review of Algebra II should I go directly to Pre-Calculus or Calculus since I completed Algebra II?

2. I took Geometry many years ago and don’t remember very much, if any, about the course. Should I take Geometry on Khan Academy before I start taking Pre-Calculus or Calculus?

Thanks in advance.(1 vote)- Congrats! I would suggest just skimming the unit test of geometry before going into precalc/calc. Geometry is a very important part of math in my opinion, and you need to know the stuff. Also, I would first take the unit test on precalc and see how much I get correct before taking calc. If you feel like there were some things in there that you did not know, I would recommed taking precalc before calc, but otherwise, you for calc!

Also, there is a unit called "Trigonometry" that you should take as well (before calc/precalc). This should mostly review from alg-2 but there are also some other concepets there that I found helpful(2 votes)

- In4:29, the answer is a^2-a-1/(a+2)^2. However, the denominator which is (a+2)^2, when it is multiplied out, wouldn't it equal a^2+4a+4. So the a^2 in the numerator and the a^2 in the denominator would cancel out?(1 vote)
- not exactly. Let's use real numbers. if you had (2+3)/(2+5) the 2s do not cancel outthe only time something in the numerator and denominator cancel out is if everything in the numerator/ denominator are being multiplied by the same thing. Back to using variables.

a/a cancels out to 1

a(b+1)/(b+1) the b+1 cancels out so it's just a

[(b-2)(c-a+d)] /[(c+d-a)(x+y)] has c-a+d cancel out with c+d-a since they are equivalent, and the fraction simplifies to (b-2)/(x+y)

Let me know if that didn't help, I can try to explain differently, but basically two things have to be multiplied together to then have something int he denominator cancel out. Usually it's in parenthesis like that.(2 votes)

- Why did he get (a+2)(a+2) out of a^2+4a+4? Shouldn't it be something like (a+2)(a+2)+4a?(1 vote)
- He's doing "reverse" FOIL. If you use the FOIL method on (a+2)(a+2), you will find it is equal to a^2+4a+4.(1 vote)

## Video transcript

Find the difference. Express the answer as a
simplified rational expression, and state
the domain. We have two rational
expressions, and we're subtracting one from
the other. Just like when we first learned
to subtract fractions, or add fractions, we have to
find a common denominator. The best way to find a common
denominator, if were just dealing with regular numbers, or
with algebraic expressions, is to factor them out, and
make sure that our common denominator has all of the
factors in it-- that'll ensure that it's divisible by the
two denominators here. This guy right here is
completely factored-- he's just a plus 2. This one over here, let's see if
we can factor it: a squared plus 4a plus 4. Well, you see the pattern that
4 is 2 squared, 4 is 2 times 2, so a squared plus 4a plus 4
is a plus 2 times a plus 2, or a plus 2 squared. We could say it's a plus 2 times
a plus 2-- that's what a squared plus 4a plus 4 is. This is obviously divisible
by itself-- everything is divisible by itself, except, I
guess, for 0, is divisible by itself, and it's also divisible
by a plus 2, so this is the least common multiple of
this expression, and that expression, and it could be
a good common denominator. Let's set that up. This will be the same thing as
being equal to this first term right here, a minus 2 over
a plus 2, but we want the denominator now to be a plus 2
times a plus 2-- we wanted it to be a plus 2 squared. So, let's multiply this
numerator and denominator by a plus 2, so its denominator is
the same thing as this. Let's multiply both the
numerator and the denominator by a plus 2. We're going to assume that a
is not equal to negative 2, that would have made this
undefined, and it would have also made this undefined. Throughout this whole thing,
we're going to assume that a cannot be equal to negative 2. The domain is all real numbers,
a can be any real number except for negative 2. So, the first term is that--
extend the line a little bit-- and then the second term doesn't
change, because its denominator is already the
common denominator. Minus a minus 3 over-- and we
could write it either as a plus 2 times a plus 2, or
as this thing over here. Let's write it in the factored
form, because it'll make it easier to simplify later on:
a plus 2 times a plus 2. And now, before we-- let's set
this up like this-- now, before we add the numerators,
it'll probably be a good idea to multiply this out right
there, but let me write the denominator, we know what
that is: it is a plus 2 times a plus 2. Now this numerator: if we have
a minus 2 times a plus 2, we've seen that pattern
before. We can multiply it out if you
like, but we've seen it enough hopefully to recognize that this
is going to be a squared minus 2 squared. This is going to be
a squared minus 4. You can multiply it out, and the
middle terms cancel out-- the negative 2 times a cancels
out the a times 2, and you're just left with a squared minus
4-- that's that over there. And then you have this: you have
minus a minus 3, so let's be very careful here-- you're
subtracting a minus 3, so you want to distribute the negative
sign, or multiply both of these terms
times negative 1. So you could put a minus a here,
and then negative 3 is plus 3, so what does
this simplify to? You have a squared minus a
plus-- let's see, negative 4 plus 3 is negative 1, all
of that over a plus 2 times a plus 2. We could write that as
a plus 2 squared. Now, we might want to factor
this numerator out more, to just make sure it doesn't
contain a common factor with the denominator. The denominator is just 2a
plus 2 is multiplied by themselves. And you can see from inspection
a plus 2 will not be a factor in this top
expression-- if it was, this number right here would be
divisible by 2, it's not divisible by 2. So, a plus 2 is not one of the
factors here, so there's not going to be any more
simplification, even if we were able to factor this thing,
and the numerator out. So we're done. We have simplified the rational
expression, and the domain is for all a's, except
for a cannot, or, all a's given that a does not equal
negative 2-- all a's except for negative 2. And we are done.