I'm trying to figure out how to simplify a Rational Expression with subtraction, 3 equations, and they all have different denominators.

You would have to multiply each of the fractions by the other two denominators in order to make one same denominator For example, x/a + y/b + z/c would become (x*b*c + y*a*c + z*a*b) / a*b*c in order to be added or subtracted Does that make sense?

How do you do it when it's like this: m^3 - 3m (5m+3)(2m-1) - (3m-1)(2m-1) Because this is confusing me big time.

m^3/(5m+3)(2m-1) - 3m/(3m-1)(2m-1) 1. **find LCD**; LCD is (5m+3)(3m-1)(2m-1) 2. multiply both num and denom of both fractions by the missing factors to get LCD (first fraction by (3m-1) second fraction by (5m+3)) 3. expand numerator if needed 4. **distribute negative sign** to the numerator of the fraction you're subtracting from the other fraction -(15m^2+9m) = +(15m^2-9m) 5. simplify / add like terms we get: 3m^4-m^3-15m^2-9m/(5m+3)(2m-1)(3m-1) remember to expand the **numerator**

I dont get how you do the exponents! I am so confussed

(If you are still alive on Khan Academy) For the exponents, break it down into x times something else, like 9x^2 - 45 could be simplified into 9x(x-5). Then you just add/subtract the rational expressions. Hope this helps!

These 4 exercises have problems because they do not want to put the exponents

Use a caret: ^ (Shift + 6) For Example, if you wanted 5x squared, type 5x^2 into the answer box, it should automatically change it.

In the Check Your Understandings,there needs to be a bottom row of buttons when answering, which includes one to show an exponent.

Actually, there usually is, only on some problems you don't need it to answer the question, so they don't put it.

I have a problem that looks like this: 5/(x-1) + 8/(x-1)^2 - 3/(x-1)^3. How should I start this? Should I multiple the first term by (x-1)^2 and (x-1)^3 since those are its two missing denominators, or is there another step I should take?

You need a common denominator. Since all your denominators have the same factor of (x-1), you common denominator = the one with the highest exponent: LCD = (x-1)^3 Thus, you multiply the first fraction's numerator & denominator by (x-1)^2 to get it to a denominator of (x-1)^3. And, you multiply the 2nd fraction's numerator & denominator by (x-1) to get it to a denominator of (x-1)^3. Hope this helps. If you have more questions, comment back.

If I have a number like this X^2/x^3 +x^2 + x^2-1/x^4-1 - 1/x-1 how do i proceed? I tried to multiply each of the fractions by the other denominators, is it the right way to start?

Typically you should factor the denominators first, as it will make getting the least common multiple easier. For the first denominator, you can factor out x^2, the second denominator is a perfect square, (x^2+1) (x^2-1) and the last denominator is already simplified. It appears the denominators don't have a common factor though, so yes, you would have to multiply each of the denominators out.

I believe you did something wrong. In the answer above you said the final answer was 5x^2+14x+12/(x+2)(x+3), however, this could further be simplified. We could do 5x^2 + 10x + 4x + 12, and that would factor too (5x+4)(x+2)/(x+2)(x+3). We then could cancel out the (x+2) and we would get (5x+4)/(x+3), where x can't be -2 or -3.

Sorry, but your factors don't work. (5x+4)(x+2) = 5x^2 +10x +4x + 8 = 5x^2+14x+8 You need factors that create 5x^2+14x+12. To factor 5x^2+14x+12, you would need 2 factors of 5(12) = 60, that add to 14. There are none. So, it is not factorable. Hope this helps.

Main content

Adding & subtracting rational expressions

Google Classroom

Have you learned the basics of rational expression addition/subtraction? Great! Now dig deeper with some advanced examples.

What we need to know before this lesson

A rational expression is a ratio of two polynomials.

To add or subtract two rational expressions with the same denominator, we simply add or subtract the numerators and write the result over the common denominator.

When the denominators are not the same, we must manipulate them so that they become the same. In other words, we must find a common denominator.

If this is new to you, you may want to check out the following articles first:

What you will learn in this lesson

In this lesson, you will practice adding and subtracting rational expressions with different denominators. You will use the least common denominator as your common denominator in these examples and explore why it is beneficial to do so.

Warm-up: start fraction, 3, divided by, x, minus, 2, end fraction, minus, start fraction, 2, divided by, x, plus, 1, end fraction

To subtract two rational expressions, each fraction must have the same denominator.

In this example, we can create a common denominator by multiplying the first fraction by left parenthesis, start fraction, x, plus, 1, divided by, x, plus, 1, end fraction, right parenthesis and the second fraction by left parenthesis, start fraction, x, minus, 2, divided by, x, minus, 2, end fraction, right parenthesis.

[Why can we do this?]

Then, we can subtract the numerators and write the result over the common denominator.

\begin{aligned} &\phantom{=}{\dfrac{3}{\blueE{x-2}}-\dfrac{2}{\greenE{x+1}}} \\\\ &=\dfrac{3}{\blueE{x-2}}{\left(\greenE{\dfrac{x+1}{x+1}}\right)}-\dfrac{2}{\greenE{x+1}}{\left(\blueE{\dfrac{x-2}{x-2}}\right)} \\\\ &=\dfrac{3(x+1)}{(x-2)(x+1)}-\dfrac{2(x-2)}{(x+1)(x-2)} \\\\ &=\dfrac{3(x+1)-2(x-2)}{(x-2)(x+1)} \\\\ &=\dfrac{3x+3-2x+4}{(x-2)(x+1)} \\\\ &=\dfrac{x+7}{(x-2)(x+1)} \end{aligned}

Check your understanding

Problem 1

Add.
The numerator should be expanded and simplified. The denominator should be either expanded or factored.

start fraction, 5, x, divided by, x, plus, 3, end fraction, plus, start fraction, 4, divided by, x, plus, 2, end fraction, equals

Least common denominators

Numerical fractions

Sometimes, the denominators of two fractions are different but have some shared factors.

For example, consider start fraction, 3, divided by, 4, end fraction, plus, start fraction, 1, divided by, 6, end fraction:

\begin{aligned} &\phantom{=}\dfrac{3}{4}+\dfrac{1}{6} \\\\ &=\dfrac{3}{\blueE2\cdot \greenE2}+\dfrac{1}{\blueE2\cdot \goldE3} \\\\ &=\dfrac{3}{\blueE2\cdot \greenE2} {\left(\dfrac{\goldE3}{\goldE3}\right)}+\dfrac{1}{\blueE2\cdot\goldE3}{\left(\dfrac{\greenE2}{\greenE2}\right)} \\\\ &=\dfrac{9}{12}+\dfrac{2}{12} \\\\ &=\dfrac{11}{12} \end{aligned}

Notice that the common denominator used in this example was not the product of the two individual denominators (24). Instead it was the least common multiple of 4 and 6 (12).

[What does least common multiple mean?]

The least common multiple of the denominators in two or more fractions is called the least common denominator.

[How did you know that 12 was the least common denominator here?]

Variable expressions

Now let's apply this reasoning to perform the following addition:

start fraction, 2, divided by, left parenthesis, x, minus, 2, right parenthesis, left parenthesis, x, plus, 1, right parenthesis, end fraction, plus, start fraction, 3, divided by, left parenthesis, x, plus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, end fraction

First, let's find the least common denominator:

\underbrace{\dfrac{2}{\blueE{(x-2)}\greenE{(x+1)}}}_{\begin{aligned} &\scriptsize\text{Needs a} \\ &\scriptsize\text{factor of } \\ &\scriptsize \purpleD{(x+3)} \end{aligned}}+\underbrace{\dfrac{3}{\greenE{(x+1)}\purpleD{(x+3)}}}_{\begin{aligned} &\scriptsize\text{Needs a} \\ &\scriptsize\text{factor of } \\ &\scriptsize \blueE{(x-2)} \end{aligned}}

So the least common denominator is start color #0c7f99, left parenthesis, x, minus, 2, right parenthesis, end color #0c7f99, start color #0d923f, left parenthesis, x, plus, 1, right parenthesis, end color #0d923f, start color #7854ab, left parenthesis, x, plus, 3, right parenthesis, end color #7854ab.

We can add the rational expressions as follows:

\begin{aligned} &\phantom{=}\dfrac{2}{\blueE{(x-2)}\greenE{(x+1)}}+\dfrac{3}{\greenE{(x+1)}\purpleD{(x+3)}} \\\\ &=\dfrac{2}{\blueE{(x-2)}\greenE{(x+1)}}{\left(\purpleD{\dfrac{x+3}{x+3}}\right)}+\dfrac{3}{\greenE{(x+1)}\purpleD{(x+3)}}{\left(\blueE{\dfrac{x-2}{x-2}}\right)} \\\\ &=\dfrac{2(x+3)}{(x-2)(x+1)(x+3)}+\dfrac{3(x-2)}{(x+1)(x+3)(x-2)} \\\\ &=\dfrac{2(x+3)+3(x-2)}{(x-2)(x+1)(x+3)} \\\\ &=\dfrac{2x+6+3x-6}{(x-2)(x+1)(x+3)} \\\\ &=\dfrac{5x}{(x-2)(x+1)(x+3)} \end{aligned}

Check your understanding

Problem 2

Add.
The numerator should be expanded and simplified. The denominator should be either expanded or factored.

start fraction, 1, divided by, x, left parenthesis, x, minus, 6, right parenthesis, end fraction, plus, start fraction, 3, divided by, left parenthesis, x, plus, 1, right parenthesis, left parenthesis, x, minus, 6, right parenthesis, end fraction, equals

Problem 3

Subtract.
The numerator should be expanded and simplified. The denominator should be either expanded or factored.

start fraction, 3, x, divided by, 2, left parenthesis, x, minus, 1, right parenthesis, end fraction, minus, start fraction, 4, divided by, left parenthesis, x, minus, 1, right parenthesis, left parenthesis, x, plus, 2, right parenthesis, end fraction, equals

Challenge problem

Add.
The numerator should be expanded and simplified. The denominator should be either expanded or factored.

start fraction, 2, divided by, x, squared, minus, 1, end fraction, plus, start fraction, 1, divided by, x, squared, minus, 3, x, minus, 4, end fraction, equals

Why use the least common denominator?

You may be wondering why it is so important to use the least common denominator to add or subtract rational expressions.

After all, this is not a requirement, and it is easy enough to use other denominators with numerical fractions.

For example, the table below calculates start fraction, 3, divided by, 4, end fraction, plus, start fraction, 1, divided by, 6, end fraction using two different common denominators: one using the least common denominator (12) and the other using the product of the two denominators (24).

Least common denominator (12)	Common denominator (24)
$\begin{aligned}~\dfrac{3}{4}+\dfrac{1}{6}&=\dfrac{3}{\blueE4} {\left(\dfrac{\goldE3}{\goldE3}\right)}+\dfrac{1}{\greenE6}{\left(\dfrac{\purpleD2}{\purpleD2}\right)}\\\\&=\dfrac{9}{12}+\dfrac{2}{12}\\\\&=\dfrac{11}{12}\\\\&\phantom{\dfrac{1}{2}}\end{aligned}$	$\begin{aligned}\dfrac{3}{\blueE4}+\dfrac{1}{\greenE6}&=\dfrac{3}{\blueE4}\left(\greenE{\dfrac{6}{6}}\right)+\dfrac{1}{\greenE6}\left(\blueE{\dfrac{4}{4}}\right)\\\\&=\dfrac{18}{24}+\dfrac{4}{24}\\\\&=\dfrac{22}{24}\\\\&=\dfrac{11}{12}\end{aligned}$

Notice that when using 24 as the common denominator, more work was required. The numbers were larger and the resulting fraction needed to be simplified.

This will also happen if you do not use the least common denominator when adding or subtracting rational expressions.

However, with rational expressions, this process is much more difficult because the numerators and denominators will be polynomials instead of integers! You will have to perform arithmetic with higher degree polynomials and factor polynomials to simplify the fraction.

[I'd like to see an example of this with rational expressions]

All of this extra work can be avoided by using the least common denominator when adding or subtracting rational expressions.

Want to join the conversation?

Sort by:

Nadia Parker
Posted 7 years ago. Direct link to Nadia Parker's post “I'm trying to figure out ...”
I'm trying to figure out how to simplify a Rational Expression with subtraction, 3 equations, and they all have different denominators.
Button navigates to signup pageButton navigates to signup page
(10 votes)
Answer
- marce.jandrita
  Posted 7 years ago. Direct link to marce.jandrita's post “You would have to multipl...”
  You would have to multiply each of the fractions by the other two denominators in order to make one same denominator
  For example,
  x/a + y/b + z/c would become (x*b*c + y*a*c + z*a*b) / a*b*c in order to be added or subtracted
  Does that make sense?
  Comment on marce.jandrita's post “You would have to multipl...”
  (22 votes)
JaridClark
Posted 6 years ago. Direct link to JaridClark's post “How do you do it when it'...”
How do you do it when it's like this:
m^3 - 3m
(5m+3)(2m-1) - (3m-1)(2m-1)

Because this is confusing me big time.
Button navigates to signup pageComment on JaridClark's post “How do you do it when it'...”
(5 votes)
Answer
- may lin
  Posted 2 years ago. Direct link to may lin's post “m^3/(5m+3)(2m-1) - 3m/(3m...”
  m^3/(5m+3)(2m-1) - 3m/(3m-1)(2m-1)
  
  1. *find LCD*; LCD is (5m+3)(3m-1)(2m-1)
  2. multiply both num and denom of both fractions by the missing factors to get LCD (first fraction by (3m-1) second fraction by (5m+3))
  3. expand numerator if needed
  4. *distribute negative sign* to the numerator of the fraction you're subtracting from the other fraction -(15m^2+9m) = +(15m^2-9m)
  5. simplify / add like terms
  
  we get: 3m^4-m^3-15m^2-9m/(5m+3)(2m-1)(3m-1)
  
  remember to expand the *numerator*
  Button navigates to signup page
  (5 votes)
lillianfollowell
Posted 4 years ago. Direct link to lillianfollowell's post “I dont get how you do the...”
I dont get how you do the exponents! I am so confussed
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Rohan
  Posted 9 months ago. Direct link to Rohan's post “(If you are still alive o...”
  (If you are still alive on Khan Academy) For the exponents, break it down into x times something else, like 9x^2 - 45 could be simplified into 9x(x-5). Then you just add/subtract the rational expressions. Hope this helps!
  Button navigates to signup page
  (5 votes)
Aliyah Austin
Posted 5 years ago. Direct link to Aliyah Austin's post “I still don't understand ...”
I still don't understand the technique here.
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(6 votes)
Answer
Art Coe
Posted 3 months ago. Direct link to Art Coe's post “Not a question but a comm...”
Not a question but a comment: the square function is not showing up when I try and enter an answer.
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(6 votes)
Answer
- junwooshin2005
  Posted 2 months ago. Direct link to junwooshin2005's post “unless you put it for den...”
  unless you put it for denominator, you can use the up arrow key instead
  Comment on junwooshin2005's post “unless you put it for den...”
  (1 vote)
raidely.682021
Posted 5 years ago. Direct link to raidely.682021's post “These 4 exercises have pr...”
These 4 exercises have problems because they do not want to put the exponents
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Eddie Campbell
  Posted 3 years ago. Direct link to Eddie Campbell's post “Use a caret: ^ (Shift + 6...”
  Use a caret: ^ (Shift + 6) For Example, if you wanted 5x squared, type 5x^2 into the answer box, it should automatically change it.
  Comment on Eddie Campbell's post “Use a caret: ^ (Shift + 6...”
  (4 votes)
Neil Chase
Posted 3 years ago. Direct link to Neil Chase's post “In the Check Your Underst...”
In the Check Your Understandings,there needs to be a bottom row of buttons when answering, which includes one to show an exponent.
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Esther Hernandez
  Posted 3 years ago. Direct link to Esther Hernandez's post “Actually, there usually i...”
  Actually, there usually is, only on some problems you don't need it to answer the question, so they don't put it.
  Comment on Esther Hernandez's post “Actually, there usually i...”
  (4 votes)
Eric Williams
Posted 5 years ago. Direct link to Eric Williams's post “I have a problem that loo...”
I have a problem that looks like this: 5/(x-1) + 8/(x-1)^2 - 3/(x-1)^3. How should I start this? Should I multiple the first term by (x-1)^2 and (x-1)^3 since those are its two missing denominators, or is there another step I should take?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Kim Seidel
  Posted 5 years ago. Direct link to Kim Seidel's post “You need a common denomin...”
  You need a common denominator. Since all your denominators have the same factor of (x-1), you common denominator = the one with the highest exponent: LCD = (x-1)^3
  Thus, you multiply the first fraction's numerator & denominator by (x-1)^2 to get it to a denominator of (x-1)^3. And, you multiply the 2nd fraction's numerator & denominator by (x-1) to get it to a denominator of (x-1)^3.
  
  Hope this helps. If you have more questions, comment back.
  Button navigates to signup page
  (2 votes)
dhruvrk14
Posted 4 months ago. Direct link to dhruvrk14's post “I believe you did somethi...”
I believe you did something wrong. In the answer above you said the final answer was 5x^2+14x+12/(x+2)(x+3), however, this could further be simplified. We could do 5x^2 + 10x + 4x + 12, and that would factor too (5x+4)(x+2)/(x+2)(x+3). We then could cancel out the (x+2) and we would get (5x+4)/(x+3), where x can't be -2 or -3.
Button navigates to signup pageComment on dhruvrk14's post “I believe you did somethi...”
(1 vote)
Answer
- Kim Seidel
  Posted 4 months ago. Direct link to Kim Seidel's post “Sorry, but your factors d...”
  Sorry, but your factors don't work.
  (5x+4)(x+2) = 5x^2 +10x +4x + 8 = 5x^2+14x+8
  You need factors that create 5x^2+14x+12.
  
  To factor 5x^2+14x+12, you would need 2 factors of 5(12) = 60, that add to 14. There are none. So, it is not factorable.
  
  Hope this helps.
  Button navigates to signup page
  (4 votes)
Simon
Posted 7 years ago. Direct link to Simon's post “If I have a number like t...”
If I have a number like this X^2/x^3 +x^2 + x^2-1/x^4-1 - 1/x-1 how do i proceed? I tried to multiply each of the fractions by the other denominators, is it the right way to start?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Sarah Shores
  Posted 6 years ago. Direct link to Sarah Shores's post “Typically you should fact...”
  Typically you should factor the denominators first, as it will make getting the least common multiple easier. For the first denominator, you can factor out x^2, the second denominator is a perfect square, (x^2+1) (x^2-1) and the last denominator is already simplified. It appears the denominators don't have a common factor though, so yes, you would have to multiply each of the denominators out.
  Button navigates to signup page
  (2 votes)

Precalculus

Course: Precalculus > Unit 4

Adding & subtracting rational expressions

What we need to know before this lesson

What you will learn in this lesson

Warm-up: start fraction, 3, divided by, x, minus, 2, end fraction, minus, start fraction, 2, divided by, x, plus, 1, end fraction

Check your understanding

Least common denominators

Numerical fractions

Variable expressions

Check your understanding

Why use the least common denominator?

Want to join the conversation?

Precalculus

Course: Precalculus > Unit 4

What we need to know before this lesson

What you will learn in this lesson

Warm-up: 3x−2−2x+1\dfrac{3}{x-2}-\dfrac{2}{x+1}x−23​−x+12​start fraction, 3, divided by, x, minus, 2, end fraction, minus, start fraction, 2, divided by, x, plus, 1, end fraction

Check your understanding

Least common denominators

Numerical fractions

Variable expressions

Check your understanding

Why use the least common denominator?

Want to join the conversation?

Warm-up: start fraction, 3, divided by, x, minus, 2, end fraction, minus, start fraction, 2, divided by, x, plus, 1, end fraction