Unit 7 · Lesson 3 of 5
Adding and subtracting rational expressions
You'll be able to: add and subtract rational expressions, with the same or different denominators, and simplify the result.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first, then check.
Divide and simplify: \(\dfrac{x^2 - 1}{x} \div \dfrac{x + 1}{x^2}\)
Flip the second fraction and factor: \(\dfrac{(x-1)(x+1)}{x} \cdot \dfrac{x^2}{x+1}\).
Cancel \((x+1)\), and \(\dfrac{x^2}{x} = x\). Answer: \(x(x-1)\), the same as \(x^2 - x\).
\(\dfrac{5}{6} - \dfrac{3}{8}\)
The LCD of 6 and 8 is 24: \(\dfrac{5}{6} = \dfrac{20}{24}\) and \(\dfrac{3}{8} = \dfrac{9}{24}\).
\(\dfrac{20}{24} - \dfrac{9}{24} = \dfrac{11}{24}\).
Step 2 of 5The idea
Adding fractions only works when the bottoms match. That was true for \(\dfrac{5}{6} - \dfrac{3}{8}\) in the warm-up, and it is true for rational expressions.
If the denominators already match, combine the tops and keep the bottom. If they don't, build the LCD from the factors of the denominators.
Same denominator. \(\dfrac{A}{D} + \dfrac{B}{D} = \dfrac{A + B}{D}\) and \(\dfrac{A}{D} - \dfrac{B}{D} = \dfrac{A - B}{D}\).
Different denominators.
- Factor each denominator.
- LCD = every different factor, each used the most times it appears in any one denominator.
- Multiply each fraction's top and bottom by the factors its denominator is missing.
- Combine the tops over the LCD.
- Factor the top and cancel if you can.
Example of step 2: for \(x^2 - 4 = (x+2)(x-2)\) and \(x + 2\), the LCD is \((x+2)(x-2)\). The \((x+2)\) is shared, so it is used once.
Why this shows up on the placement test: combining rational expressions is a standard "simplify" question, and it is also the first step in many rational equations.
Step 3 of 5Worked examples
Example 1: \(\dfrac{3x+1}{x-2} - \dfrac{x+5}{x-2}\)
- Same denominator, so subtract the tops. Use parentheses: \(\dfrac{3x + 1 - (x + 5)}{x-2}\).
- Distribute the minus: \(3x + 1 - x - 5 = 2x - 4\).
- Factor the top: \(\dfrac{2(x-2)}{x-2}\). (Always check whether the top now shares a factor with the bottom.)
- Cancel \((x-2)\). Answer: \(2\).
Example 2: \(\dfrac{3}{x+2} + \dfrac{x}{x^2-4}\)
- Factor the denominators: \(x + 2\) and \((x+2)(x-2)\). So the LCD is \((x+2)(x-2)\).
- The first fraction is missing \((x-2)\): \(\dfrac{3}{x+2} \cdot \dfrac{x-2}{x-2} = \dfrac{3x-6}{(x+2)(x-2)}\).
- The second fraction already has the LCD. Add the tops: \(\dfrac{3x - 6 + x}{(x+2)(x-2)} = \dfrac{4x-6}{(x+2)(x-2)}\).
- Check for cancelling: \(4x - 6 = 2(2x-3)\) shares no factor with the bottom. Answer: \(\dfrac{4x-6}{(x+2)(x-2)}\).
Step 4 of 5Practice
Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. In problems 3 to 6 you type only the numerator (top) of the combined fraction, written over the LCD.
Simplify: \(\dfrac{x^2}{x-3} - \dfrac{9}{x-3}\)
Same denominator: \(\dfrac{x^2 - 9}{x-3}\).
Factor the top: \(\dfrac{(x-3)(x+3)}{x-3}\). Cancel \((x-3)\).
Answer: \(x + 3\).
Simplify: \(\dfrac{5x+1}{x+4} - \dfrac{2x-3}{x+4}\)
Subtract the tops, with parentheses: \(\dfrac{5x + 1 - (2x - 3)}{x+4}\).
Distribute the minus: \(5x + 1 - 2x + 3 = 3x + 4\).
Answer: \(\dfrac{3x+4}{x+4}\). Nothing cancels.
Write \(\dfrac{3}{x} + \dfrac{2}{x^2}\) as one fraction over the LCD. What is its numerator (top)?
The LCD is \(x^2\). The first fraction is missing one \(x\): \(\dfrac{3}{x} \cdot \dfrac{x}{x} = \dfrac{3x}{x^2}\).
Add the tops: \(\dfrac{3x + 2}{x^2}\). The numerator is \(3x + 2\).
Add \(\dfrac{2}{x+1} + \dfrac{3}{x-2}\). Over the LCD, what is the numerator (top) of the answer?
The LCD is \((x+1)(x-2)\).
\(\dfrac{2(x-2)}{(x+1)(x-2)} + \dfrac{3(x+1)}{(x+1)(x-2)}\).
Top: \(2x - 4 + 3x + 3 = 5x - 1\). The answer is \(\dfrac{5x-1}{(x+1)(x-2)}\), so the numerator is \(5x - 1\).
Subtract \(\dfrac{x}{x-4} - \dfrac{2}{x+3}\). Over the LCD, what is the numerator (top) of the answer?
The LCD is \((x-4)(x+3)\).
\(\dfrac{x(x+3)}{(x-4)(x+3)} - \dfrac{2(x-4)}{(x-4)(x+3)}\).
Top: \(x^2 + 3x - (2x - 8) = x^2 + 3x - 2x + 8 = x^2 + x + 8\).
The answer is \(\dfrac{x^2+x+8}{(x-4)(x+3)}\), so the numerator is \(x^2 + x + 8\). It does not factor, so nothing cancels.
Add \(\dfrac{x}{x^2-9} + \dfrac{2}{x+3}\). Over the LCD, what is the numerator (top) of the answer? Then check: does anything cancel?
Factor: \(x^2 - 9 = (x-3)(x+3)\). That is the LCD.
The second fraction is missing \((x-3)\): \(\dfrac{2(x-3)}{(x-3)(x+3)} = \dfrac{2x-6}{(x-3)(x+3)}\).
Top: \(x + 2x - 6 = 3x - 6 = 3(x-2)\). No factor matches the bottom.
The answer is \(\dfrac{3x-6}{(x-3)(x+3)}\), so the numerator is \(3x - 6\). Nothing cancels.
Simplify: \(\dfrac{2x}{x^2-1} - \dfrac{1}{x-1}\)
The LCD is \((x-1)(x+1)\). The second fraction is missing \((x+1)\).
\(\dfrac{2x - (x+1)}{(x-1)(x+1)} = \dfrac{x - 1}{(x-1)(x+1)}\).
Cancel \((x-1)\). Answer: \(\dfrac{1}{x+1}\).
Which expression is equivalent to \(\dfrac{x+7}{x+1} - \dfrac{2x+3}{x+1}\)?
- \(\dfrac{4-x}{x+1}\)
- \(\dfrac{10-x}{x+1}\)
- \(\dfrac{3x+10}{x+1}\)
- \(\dfrac{4-x}{2x+2}\)
Same denominator: \(\dfrac{x + 7 - (2x + 3)}{x+1} = \dfrac{x + 7 - 2x - 3}{x+1} = \dfrac{4 - x}{x+1}\). Choice A.
(B) forgot to give the minus to the 3. (C) added instead of subtracting. (D) added the bottoms, which you never do.
Step 5 of 5Wrap-up
You can now add and subtract rational expressions: factor the denominators, build the LCD, rewrite each fraction, combine the tops (parentheses when subtracting), and cancel at the end.
Read or watch next: OpenStax Intermediate Algebra 2e, 7.2 Add and Subtract Rational Expressions (free textbook, with more practice at the end of the section).
Also good: Khan Academy: Adding and subtracting rational expressions (short videos). Quick rules: Rational expressions cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "How do I know which factors go into the LCD?" It knows this course and can explain things another way.