Practice adding and subtracting rational expressions with 20 problems, four worked examples, and a printable answer key. Each answer gives the simplified expression and the full restriction list.
Published September 13, 2026 · Last updated September 13, 2026 · Aligned to CCSS HSA.APR.D.7 and HSA.SSE.A.2
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Part of a nine guide unit. See all the rational expressions and rational functions worksheets, listed in a recommended order.
20 problems in four parts, plus a three problem challenge set, with the complete answer key on the back pages. Each answer shows the least common denominator that was used, the combined numerator before it was simplified, and the full restriction list. Print single sided if you want students to work without the key in view.
Both files are free for classroom and home use. If you post them, please link back to this page rather than rehosting the PDF, so students always land on the current version.
Multiplying two rational expressions requires nothing of the denominators. You factor, you cancel, you multiply across. Adding two rational expressions requires the denominators to match before anything can be combined, and that requirement is where the extra work in this topic comes from.
Two new jobs appear. The first is building a least common denominator out of factors and then rewriting both fractions so they sit over it. The second is handling the minus sign in a subtraction, which applies to the entire second numerator and not only to its first term.
One thing does not change. Restrictions still come from the denominators of the original expression, written down before any rewriting happens. If x = 3 makes a starting denominator zero, then x ≠ 3 belongs in the answer even when the finished expression shows no sign of it.
AC + BC = A + BC AC − BC = A − BC
The denominator is written once. The work in this topic sits in the steps that get the denominators to match and in the simplifying afterward.
The least common denominator is built from factors, so the denominators have to be factored first. I have students write a short inventory: list every distinct factor that appears anywhere, record the highest power it reaches in any one denominator, then multiply those together.
Here are the denominators x2 − 4 and x2 − 4x + 4, which factor as (x − 2)(x + 2) and (x − 2)2.
| Distinct factor | Highest power in any one denominator | Goes into the LCD as |
|---|---|---|
| (x − 2) | Squared, in the second denominator | (x − 2)2 |
| (x + 2) | First power, in the first denominator | (x + 2) |
The LCD is (x − 2)2(x + 2). Multiplying the two original denominators together would have given (x − 2)3(x + 2), which is a common denominator and will produce a correct answer, but it carries an extra factor of (x − 2) through every line and has to be canceled at the end. The inventory avoids that work.
Once the LCD is written, each fraction is multiplied above and below by whatever factors it is missing. A fraction that already sits over the full LCD needs no additional factor.
Factor every denominator completely. Greatest common factor first, then difference of squares, then trinomials, then grouping.
Write the restrictions from the original denominators. Set each factor equal to zero and solve. Do this before any rewriting.
Build the LCD. Each distinct factor, raised to the highest power it reaches in any single denominator.
Multiply each fraction up to the LCD. Whatever you multiply the denominator by, multiply the numerator by as well.
Combine the numerators over one copy of the LCD. On a subtraction, put parentheses around the second numerator, then distribute the minus sign to every term.
Simplify the numerator, factor it, and cancel any factor it shares with the denominator. Report the result with the restriction list from step 2.
Simplify and state the restrictions: 3x2 − x − 6 + 2x2 − 9
Step 1, factor the denominators. The first is a trinomial with roots 3 and −2. The second is a difference of squares.
3(x − 3)(x + 2) + 2(x − 3)(x + 3)
Step 2, restrictions. The factors appearing in denominators are (x − 3), (x + 2) and (x + 3), so x ≠ 3, x ≠ −2 and x ≠ −3.
Step 3, inventory and LCD. Three distinct factors, each at the first power. The LCD is (x − 3)(x + 2)(x + 3). Note that (x − 3) is shared, so it appears once and not twice.
Step 4, build each fraction up. The first fraction is missing (x + 3). The second is missing (x + 2).
3(x + 3)(x − 3)(x + 2)(x + 3) + 2(x + 2)(x − 3)(x + 2)(x + 3)
Step 5, combine the numerators. 3(x + 3) + 2(x + 2) = 3x + 9 + 2x + 4 = 5x + 13.
Step 6, simplify and report. The numerator 5x + 13 shares no factor with the denominator, so the expression is already in lowest terms.
5x + 13(x − 3)(x + 2)(x + 3), where x ≠ 3, x ≠ −2, x ≠ −3
Expanding (x − 3)(x + 2)(x + 3) into a cubic is extra work that hides the restrictions and makes the next step harder if the problem continues. Factored form also makes it obvious at a glance that nothing cancels.
In a subtraction, the minus sign applies to every term of the numerator that follows it. Put parentheses around the second numerator, then distribute the minus sign to every term.
Simplify and state the restrictions: 2xx2 − 9 − 2x − 3
Steps 1 to 3. The denominators factor as (x − 3)(x + 3) and (x − 3), giving x ≠ 3 and x ≠ −3. The inventory holds (x − 3) and (x + 3), each at the first power, so the LCD is (x − 3)(x + 3). The first fraction already sits over it.
Step 4, build up the second fraction by (x + 3), above and below.
2x(x − 3)(x + 3) − 2(x + 3)(x − 3)(x + 3)
Step 5, combine with the second numerator in parentheses.
2x − (2x + 6)(x − 3)(x + 3) = 2x − 2x − 6(x − 3)(x + 3) = −6(x − 3)(x + 3)
Step 6, report. The numerator is a constant, so nothing cancels.
−6(x − 3)(x + 3), where x ≠ 3, x ≠ −3
A common mistake is writing 2x − 2x + 6 instead of 2x − 2x − 6. Distribute the minus sign to both terms in 2(x + 3).
Two denominators like (x − 4) and (4 − x) are not different denominators. They differ by a factor of −1, since (4 − x) = −(x − 4). Factoring the −1 out of one of them makes the denominators match, and the LCD is then a single factor rather than a product.
For a denominator of the form a − x, rewrite it as −(x − a). Check the denominators for that pattern before building the LCD.
Simplify and state the restrictions: xx − 4 + 44 − x
Step 1, rewrite the opposite denominator. Since 4 − x = −(x − 4), the second fraction becomes −4/(x − 4). Moving the −1 to the front of the fraction turns the addition into a subtraction.
Step 2, restrictions. Both original denominators are zero at x = 4, so x ≠ 4.
Steps 3 to 5. The LCD is (x − 4). Neither fraction needs building up.
xx − 4 − 4x − 4 = x − 4x − 4
Step 6, simplify and report. The numerator and denominator are the same factor, so the expression reduces to 1, with x ≠ 4.
The answer is a constant and the restriction is invisible in it. It still belongs there. At x = 4 the original expression is undefined, and a value the starting expression never had cannot be created by simplifying.
Step 6 gets skipped because the numerator often does not factor, so students stop looking for a common factor at all. When it does factor, an unreduced answer loses credit on any key that asks for lowest terms.
Simplify and state the restrictions: x + 3x2 − 1 − 2x − 1
Steps 1 to 3. The denominators are (x − 1)(x + 1) and (x − 1), so x ≠ 1 and x ≠ −1, and the LCD is (x − 1)(x + 1).
Steps 4 and 5. Build the second fraction up by (x + 1), then combine with the second numerator in parentheses.
(x + 3) − (2x + 2)(x − 1)(x + 1) = x + 3 − 2x − 2(x − 1)(x + 1) = −x + 1(x − 1)(x + 1)
Step 6, factor the numerator. Written as −x + 1 the common factor is hidden. Factoring out −1 gives −(x − 1), which cancels against the (x − 1) below.
−(x − 1)(x − 1)(x + 1) = −1x + 1, where x ≠ 1, x ≠ −1
The finished expression shows only (x + 1) underneath, so a student reading the answer alone would report x ≠ −1 and stop. The restriction x ≠ 1 came from the original denominators and stays on the list. This is why step 2 sits where it does.
Pick one of the three problems below and play it. Each step appears one at a time, in the order I teach them, with the factoring, the restriction list, the LCD, the build up, the combination and the final reduction kept separate. Use the step buttons to move at your own pace, or pause on a step and try the next one on paper before revealing it.
Six steps per problem, the same six listed above. The three problems are fixed rather than generated, so the same solution appears every time and can be shown on a projector.
These are six errors I see regularly when a student brings me this homework, in rough order of how much credit they cost.
Writing 1/x + 1/y as 2/(x + y) comes from treating the fraction bar like a plus sign. Compare this with 1/2 + 1/3 = 5/6, not 2/5. Algebraic fractions follow the same rule: use a common denominator, then add the numerators.
In A/C − B/C, the minus sign applies to every term of B. When B is 2x + 5, the numerator becomes A − 2x − 5, not A − 2x + 5. Put parentheses around the second numerator, then distribute the minus sign to every term.
Multiplying a denominator by (x + 3) without multiplying its numerator by (x + 3) changes the value of the fraction. Every build up step is multiplication by a form of 1, which means the same factor goes above and below.
The product of the denominators is always a common denominator, so the answer that follows is not wrong, but it usually needs a final cancellation that gets skipped. A key that expects lowest terms marks the unreduced version down. The factor inventory is faster than canceling at the end.
When (x − 5) and (5 − x) both appear, the LCD is a single factor, not a product of two. Students who miss this build a denominator twice as long as it needs to be, and the extra algebra usually produces a sign error somewhere in the middle.
An answer like (5x + 13)/((x − 3)(x + 3)) tempts students to cancel a 3 or an x. Only whole factors cancel, and 5x + 13 is a single unfactorable piece. The check I use for everything on this page is numerical: pick a value of x that is not on the restriction list, evaluate the original expression and your answer separately, and compare. x = 0 is usually the easiest legal choice, and x = 1 or x = 10 work when 0 is restricted. If the two numbers disagree, the error is somewhere between the two lines, and you have found it before the grader did.
This is the same content as the printable cheat sheet, laid out here so it is readable on a phone and so you can check a rule without opening a PDF. The download button repeats at the bottom of this section.
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AC + BC = A + BC
AC − BC = A − BC
The denominators must match before anything is combined, and the shared denominator is written once.
Denominators x2 − 4 = (x − 2)(x + 2) and x2 − 4x + 4 = (x − 2)2.
| Distinct factor | Highest power seen | In the LCD |
|---|---|---|
| (x − 2) | 2 | (x − 2)2 |
| (x + 2) | 1 | (x + 2) |
LCD = (x − 2)2(x + 2). A shared factor is counted once, at its highest power.
Put parentheses around the second numerator, then distribute the minus sign to every term.
A − (2x + 5) = A − 2x − 5
Every term inside the parentheses changes sign, not only the first.
(a − b) = −(b − a)
For a denominator of the form a − x, rewrite it as −(x − a). Then (x − 5) and (5 − x) share one LCD factor, and the sign in front of that fraction flips.
Pick a value of x that is not on the restriction list, usually x = 0. Evaluate the original expression and your answer separately. The two numbers must match. If they do not, the error is between those two lines.
Download this cheat sheet (PDF, one page)
These are the same 20 problems that appear in the printable worksheet, in the same order, so a student can work on paper and check here, or work here and print the version with room to show work. The PDF adds a three problem challenge set after Part D. Simplify each expression completely and state every restriction on the variable. Answers for each part are behind the button under that part.
The worksheet targets two high school standards. HSA.APR.D.7 asks students to add, subtract, multiply and divide rational expressions and to recognize that they behave like the rational numbers. HSA.SSE.A.2 covers the structural reading that lets a student see x2 − 4x + 4 as (x − 2)2 while building the LCD. Parts C and D also depend on factoring skills from the polynomials unit.
If you want a second explanation in a different voice, these are the outside sources I point students toward:
This page sits inside a longer sequence. Simplifying a single rational expression comes first, multiplying and dividing comes next since it needs no common denominator, and this page is usually the last of the three before rational equations.
We can work through a few problems together to see which steps need practice. In my sessions, students who stall here are usually stuck on the factoring rather than on the fractions, and the two need different practice.
I tutor Algebra I, Algebra II, Precalculus and AP science in person in Fremont, Newark and Union City, and online elsewhere. Sessions happen at your table, so there is no drive at 4 pm.
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