HSA.APR.D.7: Adding, Subtracting, Multiplying and Dividing Rational Expressions
In plain English: HSA.APR.D.7 is an advanced (+) Common Core algebra standard that asks students to add, subtract, multiply and divide rational expressions, and to understand that, like the rational numbers, rational expressions are closed under these operations as long as they never divide by zero. Students factor, find common denominators, simplify and state excluded values. It is usually taught in Algebra II or Precalculus.
(+) Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.
Common Core State Standards for Mathematics · Domain: Arithmetic with Polynomials and Rational Expressions (APR) · Cluster: Rewrite rational expressions Also written as HSA-APR.D.7 or A-APR.7 · Official standard
Students learn to add, subtract, multiply and divide rational expressions, and they see why these operations work: rational expressions are to polynomials what fractions are to integers. A rational number is a quotient a/b of integers with b ≠ 0; a rational expression is a quotient p/q of polynomials with q not the zero polynomial. The same rules apply to both: multiply across, multiply by the reciprocal to divide, and rewrite with a common denominator to add or subtract.
The lesson makes the closure idea explicit. Because polynomials are closed under addition, subtraction and multiplication (HSA.APR.A.1), each rule produces a new quotient of polynomials, so the result is again a rational expression. Division works too, as long as the divisor is not the zero expression, exactly as with rational numbers. Throughout, students factor first, simplify by dividing out common factors (never terms), and state the values of x that must be excluded.
Learning Objectives
By the end of this lesson, students will be able to:
Explain how rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication and division by a nonzero rational expression
Multiply and divide rational expressions by factoring, multiplying by the reciprocal of the divisor and dividing out common factors
Add and subtract rational expressions by rewriting them with a least common denominator
State the values of the variable that must be excluded, including the zeros of a divisor
Check a result by substituting a number and comparing it with the matching fraction computation
Prior Knowledge Required
Students should already be comfortable with:
Adding, subtracting, multiplying and dividing fractions and rational numbers 7.NS.A.2
Closure of polynomials under addition, subtraction and multiplication HSA.APR.A.1
Factoring polynomials, including the difference of squares and trinomials HSA.SSE.A.2
Simplifying a single rational expression by dividing out common factors HSA.APR.D.6
Post three fraction questions and give students four minutes:
Warm-Up Prompt
"(1) Compute 5/6 + 1/4. (2) Compute 2/3 ÷ 4/9. (3) Is the sum, difference, product or quotient of two fractions always a fraction? Is there any exception?"
Answers: 5/6 + 1/4 = 10/12 + 3/12 = 13/12, and 2/3 ÷ 4/9 = 2/3 · 9/4 = 18/12 = 3/2. For (3), students should reach the idea that the four operations always give another fraction, with one exception: you cannot divide by 0. Write "closed under +, -, ×, ÷ by a nonzero number" on the board and tell students that today they will show the same is true for rational expressions.
Direct Instruction20 minutes
Use Diagram 1 to put a fraction computation beside a rational-expression computation, then give the four rules for rational expressions p/q and r/s (Diagram 2):
Factor every numerator and denominator and record the excluded values: every zero of every denominator.
Multiply: p/q · r/s = pr/(qs). Divide out common factors before multiplying out.
Divide: p/q ÷ r/s = p/q · s/r, allowed only when r/s is not the zero expression. The zeros of r are also excluded.
Add or subtract: rewrite both expressions over the least common denominator, then add or subtract the numerators: p/q ± r/s = (ps ± qr)/(qs). Put the second numerator in parentheses when subtracting.
Simplify and state the result: factor the new numerator, divide out common factors, and list the excluded values next to the answer.
After the rules, make the closure argument: in every rule the new numerator and denominator are built from p, q, r and s using only +, - and ×, so they are polynomials, and the new denominator is not the zero polynomial. So the result is again a rational expression. Work through the examples:
Fraction analogy, subtraction
Compute 5/6 - 3/8 and then 5/(2x) - 3/x². Both use a least common denominator: 24 for the numbers and 2x² for the expressions.
After the last example, point out that the result 1 is still a rational expression (1 = 1/1), just as 6/6 = 1 is still a rational number. Ask why x = 2 is excluded even though x - 2 no longer appears in the answer: the original quotient divides by (x - 2)/(x + 3), which is 0 at x = 2, and dividing by zero is never allowed. This is the "nonzero" part of the standard.
Guided Practice15 minutes
Pairs work four problems, one operation each, and state the excluded values: (a) (x² + 3x)/(x² - 16) · (x - 4)/x = (x + 3)/(x + 4), x ≠ 0, 4, -4; (b) (2x - 6)/(x + 1) ÷ (x² - 9)/(x² + x) = 2x/(x + 3), x ≠ -1, 0, 3, -3; (c) 2/x + 3/(x + 4) = (5x + 8)/(x(x + 4)); (d) (x + 1)/(x - 3) - 4/(x² - 9) = (x² + 4x - 1)/((x - 3)(x + 3)). After each problem, one pair substitutes x = 5 into the original and into the answer to check that they agree. Listen for these errors: canceling terms instead of factors, flipping the first expression instead of the divisor, adding denominators, and forgetting to subtract every term of the second numerator.
Independent Practice15 minutes
Students work alone on five problems: (1) 6x²/(x² - 25) · (x + 5)/(3x) = 2x/(x - 5), x ≠ 0, 5, -5; (2) (x² - x - 6)/(x² - 4) ÷ (x - 3)/(x + 2) = (x + 2)/(x - 2), x ≠ 2, -2, 3; (3) 1/(x - 5) + 2/(x + 5) = (3x - 5)/((x - 5)(x + 5)); (4) 3x/(x + 2) - (x - 4)/(x + 2) = (2x + 4)/(x + 2) = 2, x ≠ -2; (5) explain why (x + 1)/(x - 1) ÷ x/(x - 1) is a rational expression, and simplify it ((x + 1)/x, x ≠ 0, 1). For each answer, students write which property of polynomials guarantees that the result is a rational expression.
Closure5 minutes
Exit ticket: (1) Subtract 4/(x + 1) - 2/(x - 1). (Answer: (2x - 6)/((x + 1)(x - 1)).) (2) Divide (x² - 1)/(2x) ÷ (x + 1)/(4x²). (Answer: 2x(x - 1), x ≠ 0, -1.) (3) Complete the sentence: "Rational expressions are like rational numbers because ..., and the one operation that is not allowed is ...".
Differentiation Strategies
For Struggling Students
Pair every expression problem with the same problem in numbers, for example 1/2 + 1/3 next to 1/x + 1/(x + 1), and do the number version first
Give a factoring reference card (difference of squares, common factor, trinomials) and require a fully factored line before any canceling
Use a two-column format: left column the work, right column the running list of excluded values
For Advanced Students
Ask students to prove that the set of rational expressions is closed under division by a nonzero rational expression, using only the closure of polynomials under multiplication
Challenge: simplify the complex fraction (1/x - 1/3)/(x - 3) and explain each step as one of the four operations
Ask students to find two rational expressions whose sum is a polynomial and two whose quotient is a constant
Assessment Guidance
What to Look For
Look for a fully factored form before any canceling, and ask students to point to the common factor they divided out: a student who cancels x in (x + 4)/4 is canceling a term, not a factor. In division, check that the divisor, not the first expression, is inverted and that the zeros of the divisor are excluded. In subtraction, check that parentheses surround the second numerator. For the closure part of the standard, students should be able to say why every answer is a quotient of polynomials and why division by the zero expression is excluded.
02
Classroom Activities
3 Activities
1
Numbers Beside Expressions
15 minPairs
Each problem on the handout comes in two versions: a fraction computation and a rational-expression computation that turns into it when x = 2. Pairs solve both versions and use the substitution to check their algebra.
The 4 Paired Problems
Add: 1/2 + 1/3 and 1/x + 1/(x + 1) (answers 5/6 and (2x + 1)/(x(x + 1)))
Multiply: 4/9 · 3/8 and 2x/(x + 7) · (x + 1)/(4x) (answers 1/6 and (x + 1)/(2(x + 7)))
Divide: 2/5 ÷ 4/3 and x/(x + 3) ÷ 2x/(x + 1) (answers 3/10 and (x + 1)/(2(x + 3)))
Procedure
Partner A solves the fraction version; Partner B solves the expression version
Partner B substitutes x = 2 into the simplified expression. It must equal Partner A's fraction
Together, the pair writes one sentence for each operation describing what is the same in the two versions
Discussion Questions
In the multiplication problem, why can you divide out x but not the 7 in x + 7?
Is there a value of x that you may not substitute in the division problem? Why?
Modification for Distance Learning
Post the paired problems on a shared slide. Pairs work in a breakout room, one partner per version, and paste a photo of both solutions with the x = 2 check.
2
Error Hunt
20 minGroups of 3-4
Groups receive 6 cards, each showing a worked solution with exactly one error. They find the error, explain it and write the correct solution.
(x + 5)/(x - 1) - (x - 2)/(x - 1) = 3/(x - 1). (The subtraction was not applied to -2; correct: 7/(x - 1).)
(x + 4)/4 = x. (A term was canceled, not a factor; the expression does not simplify.)
(x² - 9)/(x + 3) ÷ (x - 3) = (x - 3)². (The student multiplied by x - 3 instead of its reciprocal; correct: 1, x ≠ 3, -3.)
2/x - 3/x² = -1/x². (2/x was not rewritten over x²; correct: (2x - 3)/x².)
x/(x - 2) ÷ x/(x + 1) = (x + 1)/(x - 2), x ≠ 2, -1. (The simplification is right, but x = 0 must also be excluded because the divisor is 0 there.)
Procedure
Each student takes a card, finds the error alone for two minutes, then explains it to the group
The group checks each correction by substituting a number such as x = 3 into the original and the corrected answer
Groups sort the six errors into "operation errors" and "excluded value errors" and share one of each with the class
Challenge Variation
Each group writes a new error card for a classmate group, with a subtle mistake in a division problem, and a hidden answer key on the back.
3
Closure Investigation
20 minPairs
Pairs fill in a closure table for four number systems and justify each entry, which is the "understand" part of the standard: rational expressions behave like rational numbers, and polynomials behave like integers.
Procedure
Draw a table with rows Integers, Rational numbers, Polynomials, Rational expressions and columns +, -, ×, ÷ by a nonzero element
For each cell, write "closed" or give a counterexample. Expected result: integers and polynomials are not closed under division (7 ÷ 2 and x ÷ (x + 1)); rational numbers and rational expressions are closed under all four operations
For the rational expression row, write a one-line proof for each operation, for example: p/q · r/s = pr/(qs), where pr and qs are polynomials because polynomials are closed under multiplication, and qs is not zero because neither q nor s is the zero polynomial
Discussion Questions
Which system "fills the gap" for division in each case: what do you get by dividing two integers, and what by dividing two polynomials?
Why does the standard say "division by a nonzero rational expression"? Give an example of a rational expression you cannot divide by.
Is x² + 1 a rational expression? Is every polynomial one?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Adding Fractions and Adding Rational Expressions
The same procedure adds 3/4 + 5/7 and 3/(x + 2) + 5/(x - 1): rewrite over a common denominator, add the numerators and simplify. The rational expressions behave like fractions whose numerators and denominators are polynomials instead of integers.
Diagram 2: The Four Operations and Closure
Each operation turns two rational expressions into a new quotient of polynomials, so the set of rational expressions is closed under all four operations. The red entry marks the one extra condition: you may divide only by a nonzero rational expression, just as you may divide only by a nonzero rational number.
04
Homework Assignment
~30 min
HSA.APR.D.7 Homework: Operations on Rational Expressions
Directions: Factor completely before simplifying, show the least common denominator for every sum or difference, and list all excluded values next to each answer. Check at least one answer in each part by substituting a number.
Simplify (x + 1)/(x - 2) ÷ (x² - 1)/x and list every excluded value, including those that come from the divisor. Then explain, using the closure of polynomials, why the quotient of two rational expressions is always a rational expression when the divisor is not the zero expression, and why the quotient of two polynomials does not have to be a polynomial.
A cyclist rides 30 km to a lake at an average speed of v km/h and rides the 30 km back uphill at v - 5 km/h. (a) Write the total time as a single simplified rational expression in v. (b) Find the total time when v = 20. (c) The average speed for the round trip is 60 divided by the total time. Write it as a simplified rational expression and evaluate it at v = 20. Is it 17.5 km/h, the average of 20 and 15? Explain.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Factoring and Simplifying
Fully factored; only common factors divided out
One factoring or canceling error
Terms canceled or not factored
Operations
Correct rule for each operation, LCD shown, reciprocal of the divisor used
One operation carried out incorrectly
Rules not used correctly
Excluded Values
All excluded values listed, including zeros of the divisor
Some excluded values missing
No excluded values
Closure and Context
Closure explained with the closure of polynomials; context answers interpreted
Explanation or interpretation incomplete
Missing or incorrect
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again. State excluded values in your short answers.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
Factor: (x + 5)/((x - 3)(x + 3)) · (x - 3)/(x(x + 5)). Divide out x + 5 and x - 3 to get 1/(x(x + 3)), with x ≠ 0, 3, -3, -5. Choice B forgets to divide out x + 5. Choice C loses the factor x in x² + 5x = x(x + 5).
Question 2 of 20 · Multiple Choice
Divide and simplify: x/(x + 6) ÷ x²/(x² - 36).
Answer: C
Multiply by the reciprocal of the divisor: x/(x + 6) · (x - 6)(x + 6)/x² = (x - 6)/x, with x ≠ 0, 6, -6. Choice A multiplies without taking the reciprocal. Choice B is the reciprocal of the correct answer, which comes from inverting the first expression instead of the divisor.
Question 3 of 20 · Multiple Choice
Add: 2/(x + 5) + 3/(x - 2).
Answer: B
With the LCD (x + 5)(x - 2): 2(x - 2) + 3(x + 5) = 2x - 4 + 3x + 15 = 5x + 11. Choice A adds numerators and denominators separately, which is not how fractions add (1/2 + 1/2 is not 2/4). Choice C uses +4 instead of -4 when distributing. Choice D adds the numerators without rewriting them over the LCD.
Question 4 of 20 · Multiple Choice
Subtract: (3x + 2)/(x - 1) - (x + 4)/(x - 1).
Answer: D
Same denominator, so subtract the whole second numerator: (3x + 2 - x - 4)/(x - 1) = (2x - 2)/(x - 1) = 2(x - 1)/(x - 1) = 2, for x ≠ 1. Choice A subtracts only x and adds 4, a common sign error. Choice B adds the numerators. Choice C drops the denominator.
Question 5 of 20 · Multiple Choice
Subtract: 5/x - 2/(x + 3).
Answer: A
The LCD is x(x + 3): 5(x + 3) - 2x = 5x + 15 - 2x = 3x + 15 = 3(x + 5), so the difference is 3(x + 5)/(x(x + 3)), x ≠ 0, -3. Choice B subtracts the numerators without rewriting them over the LCD. Choice C adds 2x instead of subtracting it. Choice D subtracts numerators and denominators separately.
Question 6 of 20 · Multiple Choice
Which statement explains why the sum of two rational expressions p/q and r/s is always a rational expression?
Answer: B
Polynomials are closed under multiplication and addition, so ps + qr and qs are polynomials, and qs is nonzero because q and s are. That makes the sum a quotient of polynomials. Choice A is false: 1/x is not a polynomial. Choice D describes a common error, not the rule for adding.
Question 7 of 20 · Multiple Choice
Which of these sets is NOT closed under division by a nonzero element?
Answer: C
Dividing two polynomials does not always give a polynomial: x ÷ (x + 1) = x/(x + 1) is a rational expression but not a polynomial. Polynomials behave like the integers (7 ÷ 2 is not an integer), while rational expressions, like the rational numbers in choice A, are closed under division by a nonzero element.
Question 8 of 20 · Multiple Choice
Which values of x must be excluded from (x + 2)/(x - 1) ÷ (x - 4)/(x + 3)?
Answer: D
Exclude the zeros of both denominators, x = 1 and x = -3, and the zero of the divisor, x = 4, because at x = 4 you would divide by 0. Choice A misses the divisor condition. Choice B excludes -2, but the first expression is simply 0 there, and dividing 0 by a nonzero expression is allowed.
Question 9 of 20 · Multiple Choice
Multiply and simplify: (2x² - 8)/(x² + x - 6) · (x + 3)/(4x + 8).
Answer: A
Factor: 2(x - 2)(x + 2)/((x + 3)(x - 2)) · (x + 3)/(4(x + 2)). Everything except 2/4 divides out, so the product is 1/2, for x ≠ 2, -3, -2. Choice B inverts 2/4. Choices C and D leave a factor that should have been divided out.
Question 10 of 20 · Multiple Choice
Add: 1/(x² - 4) + 1/(x² + 2x).
Answer: B
Factor: x² - 4 = (x - 2)(x + 2) and x² + 2x = x(x + 2), so the LCD is x(x - 2)(x + 2). Then x/LCD + (x - 2)/LCD = (2x - 2)/(x(x - 2)(x + 2)). Choice A adds the denominators. Choice C adds the original numerators 1 + 1 without rewriting them over the LCD.
Question 11 of 20 · Multiple Choice
The identity 1/x - 1/(x + 1) = 1/(x(x + 1)) holds for x ≠ 0, -1. What true statement about fractions do you get by substituting x = 3?
Answer: C
At x = 3 the left side is 1/3 - 1/4 and the right side is 1/(3 · 4) = 1/12. Check: 4/12 - 3/12 = 1/12. This shows the analogy in the standard: rational expressions follow the same rules as fractions. Choice B adds the denominators (3 + 4 = 7), and choice D reverses the order of subtraction.
Multiply by the reciprocal: x(x + 4)/(x - 2) · (x - 2)(x + 2)/((x - 4)(x + 4)) = x(x + 2)/(x - 4), with x ≠ 2, -2, 4, -4. Choice A is the reciprocal of the answer, from inverting the wrong expression. Choice B loses the factor x + 2.
Question 13 of 20 · Multiple Choice
Is (x - 1)/(x + 2) · (x + 2)/(x - 1) equal to 1?
Answer: B
All factors divide out, so the product is 1, but the original expression is undefined at x = -2 and x = 1, so those values stay excluded. Choice A ignores the excluded values. Choice C multiplies without noticing that the second expression is the reciprocal of the first.
Question 14 of 20 · Multiple Choice
A student simplifies (x + 6)/(x + 2) - (x - 2)/(x + 2) and gets 4/(x + 2). What is the error?
Answer: A
(x + 6) - (x - 2) = x + 6 - x + 2 = 8, so the difference is 8/(x + 2), x ≠ -2. The student computed x + 6 - x - 2 = 4. Choice B treats the denominators like numerators. Choice D uses a denominator that is not the least common one and would still need 8(x + 2) on top.
Question 15 of 20 · Short Answer
Add and simplify: 3/(2x) + 5/(6x²).
The LCD is 6x². 3/(2x) = 9x/(6x²), so the sum is (9x + 5)/(6x²), x ≠ 0.
Factor: (x - 1)(x + 1)/(x(x + 3)) · (x + 3)/(x(x + 1)). Divide out x + 1 and x + 3: (x - 1)/x², with x ≠ 0, -3, -1.
Question 18 of 20 · Short Answer
Divide and simplify: 4x²/(x² - 9) ÷ 2x/(x + 3).
Multiply by the reciprocal: 4x²/((x - 3)(x + 3)) · (x + 3)/(2x) = 2x/(x - 3), with x ≠ 3, -3, 0. The value 0 is excluded because the divisor 2x/(x + 3) is 0 there.
Question 19 of 20 · Short Answer
Give an example showing that polynomials are not closed under division, and explain why rational expressions are closed under division by a nonzero rational expression.
Example: x² and x + 1 are polynomials, but x² ÷ (x + 1) = x²/(x + 1) is not a polynomial (it has a remainder 1 when divided). For rational expressions, p/q ÷ r/s = ps/(qr); ps and qr are polynomials because polynomials are closed under multiplication, and qr is not zero because q ≠ 0 and the divisor r/s is nonzero. This mirrors the rational numbers: integers are not closed under division, but fractions are, as long as you do not divide by 0.
Question 20 of 20 · Short Answer
Pipe A fills a tank in t hours and pipe B fills it in t + 2 hours. Write the fraction of the tank that both pipes fill together in 1 hour as a single rational expression, and evaluate it for t = 4.
In 1 hour, pipe A fills 1/t and pipe B fills 1/(t + 2), so together 1/t + 1/(t + 2) = (2t + 2)/(t(t + 2)), t > 0. For t = 4: 10/24 = 5/12 of the tank, which matches 1/4 + 1/6 = 5/12.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSA.APR.D.7 mean?
HSA.APR.D.7 means students can add, subtract, multiply and divide rational expressions, which are fractions whose numerator and denominator are polynomials, such as (x + 1)/(x - 3). It also asks students to understand why this works: rational expressions follow the same rules as fractions and are closed under the four operations, so the result is always another rational expression, as long as you never divide by zero.
Is HSA.APR.D.7 Algebra 2 or Precalculus?
It is usually taught in Algebra II, and reviewed in Precalculus before work with rational functions. It is marked (+), which Common Core uses for additional mathematics that students need for advanced courses such as calculus. Many Algebra II courses teach it together with HSA.APR.D.6 (rewriting a single rational expression) and HSA.REI.A.2 (solving rational equations).
What does "closed under an operation" mean?
A set is closed under an operation when combining any two members always gives another member of the set. The integers are closed under addition but not under division (7 ÷ 2 is not an integer). The rational numbers are closed under all four operations, except division by 0. HSA.APR.D.7 says rational expressions behave the same way: the sum, difference, product or quotient (by a nonzero expression) of two rational expressions is again a rational expression.
How are rational expressions like rational numbers?
A rational number is a quotient of integers, and a rational expression is a quotient of polynomials. The rules are the same: multiply numerators and denominators to multiply, multiply by the reciprocal to divide, and use a common denominator to add or subtract. Polynomials play the role of the integers: both are closed under +, - and × but not under ÷, and taking quotients fills that gap.
Why do you have to state excluded values?
A rational expression is undefined wherever a denominator is 0, and simplifying can hide those values. For example, (x² - 4)/(x² + 5x + 6) ÷ (x - 2)/(x + 3) simplifies to 1, but the original is undefined at x = -3, -2 and 2. Stating the excluded values keeps the simplified form equivalent to the original. In division, the zeros of the divisor's numerator are also excluded.
What are common mistakes with rational expressions?
Common errors are canceling terms instead of factors, as in (x + 4)/4 = x; adding the denominators when adding fractions; forgetting to distribute the minus sign to every term of the second numerator; inverting the first expression instead of the divisor; and leaving out excluded values. Substituting a number such as x = 3 into the original and the answer catches most of these.
How do you find the least common denominator of rational expressions?
Factor every denominator completely, then take each different factor the greatest number of times it appears in any one denominator. For 1/(x² - 16) and 1/(x² - 4x), the factors are (x - 4)(x + 4) and x(x - 4), so the LCD is x(x - 4)(x + 4). Multiplying the denominators always works too, but it gives a larger expression to simplify at the end.
Do you have to multiply out the answer?
No. A factored answer such as (8x + 7)/((x + 2)(x - 1)) is usually preferred, because it shows the excluded values and makes it easy to see that nothing more can be divided out. Teachers may ask for the numerator to be simplified, but the denominator is usually left factored.
Where is this used later?
Operations on rational expressions are needed to solve rational equations (HSA.REI.A.2), to combine rational functions (HSF.BF.A.1), to simplify difference quotients in calculus, and in rate problems such as combined work, round-trip speed and lens formulas in physics.
How can students check their answers?
Pick a number that is not excluded, such as x = 5, and evaluate both the original expression and the simplified answer. If the two values differ, there is an error. For a stronger check, try a second number. This also reinforces the analogy with fractions, since each substitution turns the problem into ordinary fraction arithmetic.
07
Related Standards
5 standards
These standards connect to HSA.APR.D.7: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.NS.A.2Prerequisite
Extend multiplication and division of fractions to rational numbers
