Why must you state excluded values even after cancellation?

Excluded values must be stated based on the ORIGINAL denominators, not the simplified result. When a factor cancels during simplification, it disappears from the final expression — but the original expression was still undefined at that value. For example, if (x−2) cancels, x=2 is still an excluded value because the original fractions had (x−2) in their denominators.

What is the most common mistake when subtracting rational expressions?

The most common mistake is failing to distribute the negative sign to ALL terms of the subtracted numerator. Students often apply the negative only to the first term of the polynomial in the numerator. For example: A/(x) − (2x−3)/(x) should give (A − 2x + 3)/x, NOT (A − 2x − 3)/x.

Adding and Subtracting Rational Expressions — 12 Worked Examples with Steps

Q: How do you add rational expressions with different denominators?

To add rational expressions with different denominators: (1) Factor all denominators completely. (2) Find the LCD (Least Common Denominator) — the product of the highest power of each distinct factor. (3) Rewrite each fraction with the LCD as denominator by multiplying numerator and denominator by the missing factor. (4) Add the numerators and keep the LCD as the denominator. (5) Simplify the result and state excluded values.

Q: How do you distribute the negative sign when subtracting rational expressions?

When subtracting rational expressions, the negative sign must be distributed to ALL terms of the subtracted numerator. For example: P/Q − (ax² + bx + c)/Q = [P − ax² − bx − c]/Q. The most common error is distributing the negative to only the first term, giving P − ax² + bx + c, which is incorrect.

Adding and subtracting rational expressions is one of the most tested skills in algebra. This page walks through 12 fully worked practice problems — from basic same-denominator cases to multi-fraction problems with quadratic denominators. Every solution shows the complete LCD method, step-by-step, with special attention to the sign distribution error that costs students the most exam points.

📋 What’s on This Page

Same Denominator (Problems 1–2)
Different Denominators — Simple (Problems 3–4)
One Is a Factor of the Other (Problems 5–6)
Quadratic Denominators (Problems 7–8)
Three Rational Expressions (Problem 9)
Sign Distribution Traps (Problems 10–11)
Challenge Problem (Problem 12)
LCD Quick Reference Table
Common Mistakes
How to Check Your Answer
Frequently Asked Questions

⚡ Adding and Subtracting Rational Expressions — Quick Rules

1
Factor ALL denominators completely — before doing anything else, factor every denominator into its prime polynomial factors.
2
Find the LCD (Least Common Denominator)
LCD = product of the highest power of each distinct factor
3
Rewrite each fraction with the LCD as denominator
Multiply numerator AND denominator by the missing factor
4
Add or subtract the numerators — keep the LCD as the denominator
⚠️ For subtraction: distribute the negative sign to ALL terms in the subtracted numerator
5
Simplify the resulting expression — factor the numerator and cancel any common factors with the denominator.
6
State excluded values — identify all x-values where ANY original denominator equals zero.

📐 Key Rules

Same denominator:	`P/Q + R/Q = (P+R)/Q`
Different denominators:	`P/Q + R/S = (PS + RQ)/(LCD)` — use LCD, not Q×S when possible
Subtraction sign:	`P/Q − (R+S)/Q = (P − R − S)/Q` — negative applies to ALL terms
Excluded values:	State ALL x-values from original denominators, even after cancellation

Section 1 Same Denominator — Add Numerators Directly

When adding and subtracting rational expressions that share an identical denominator, the process is straightforward: combine the numerators and keep the denominator. The critical step is applying the subtraction sign correctly — it must be distributed to every term in the subtracted numerator.

Addition — Same Denominator

Add: 3x + 2x² − 4 + 5 − xx² − 4

Denominators are identical — add numerators directly

(3x + 2 + 5 - x) / (x² - 4) = (2x + 7) / (x² - 4)

Factor denominator and check for common factors

x² - 4 = (x + 2)(x - 2) 2x + 7 cannot be factored to share (x+2) or (x-2) \to Already in simplest form

State excluded values
x² − 4 = 0 ⟹ x = ±2
x ≠ 2 and x ≠ −2

✓ Final Answer

(2x + 7) / [(x+2)(x−2)]

x ≠ ±2

Subtraction — Same Denominator · ⚠️ Sign Distribution Trap

Subtract: 4x² + 3xx² − 1 − 3x² + 3x − 2x² − 1

Same denominator — subtract numerators
⚠️ CAUTION: distribute the negative sign to ALL terms of the second numerator:

[(4x² + 3x) - (3x² + 3x - 2)] / (x² - 1)

Distributing the negative:

= [4x² + 3x - 3x² - 3x + 2] / (x² - 1)

The −2 becomes +2 because we multiply the entire bracket by −1.

= (x² + 2) / (x² - 1)

Check for simplification

x² - 1 = (x+1)(x-1) x² + 2 cannot be factored over the reals \to no cancellation

Excluded values: x ≠ ±1

⚠️ Most Common Wrong Answer

4x² + 3x − 3x² + 3x − 2 = x² + 6x − 2

❌ Error: forgot to distribute the negative to BOTH the 3x term AND the −2 term.
The bracket (3x² + 3x − 2) means ALL three terms flip sign.

✓ Correct: −(3x² + 3x − 2) = −3x² − 3x + 2 ⟹ numerator = x² + 2

✓ Final Answer

(x² + 2) / [(x+1)(x−1)]

x ≠ ±1

Section 2 Different Denominators — Simple Cases

When adding and subtracting rational expressions with different denominators, the LCD method is essential. For simple cases where the denominators share no common factors, the LCD is simply the product of the two denominators.

Addition — Different Denominators, No Common Factors

Add: 1x + 1x + 1

Factor denominators
Den 1: x (already factored)
Den 2: x+1 (already factored)
No common factors ⟹ LCD = x(x+1)

Rewrite each fraction with the LCD

1/x = (x+1) / [x(x+1)] \leftarrow multiply top and bottom by (x+1) 1/(x+1) = x / [x(x+1)] \leftarrow multiply top and bottom by x

Add numerators

[(x+1) + x] / [x(x+1)] = (2x+1) / [x(x+1)]

Check simplification — 2x+1 shares no factors with x(x+1) ✓

Excluded values: x ≠ 0 and x ≠ −1

✓ Final Answer

(2x+1) / [x(x+1)]

x ≠ 0, −1

Subtraction — Different Denominators, Expand Both Numerators

Subtract: x + 3x − 2 − x − 1x + 4

Denominators (x−2) and (x+4) share no common factors
LCD = (x−2)(x+4)

Rewrite with the LCD

(x+3)/(x-2) = (x+3)(x+4) / [(x-2)(x+4)] (x-1)/(x+4) = (x-1)(x-2) / [(x-2)(x+4)]

Set up the subtraction

[(x+3)(x+4) - (x-1)(x-2)] / [(x-2)(x+4)]

Expand each product

(x+3)(x+4) = x² + 7x + 12 (x-1)(x-2) = x² - 3x + 2

Subtract — distribute negative to ALL terms

[x² + 7x + 12 - (x² - 3x + 2)] / [(x-2)(x+4)] = [x² + 7x + 12 - x² + 3x - 2] / [(x-2)(x+4)] = (10x + 10) / [(x-2)(x+4)] = 10(x+1) / [(x-2)(x+4)]

Check simplification — 10(x+1) shares no factors with (x−2)(x+4) ✓

Excluded values: x ≠ 2 x ≠ −4

✓ Final Answer

10(x+1) / [(x−2)(x+4)]

x ≠ 2, −4

Section 3 Different Denominators — One Is a Factor of the Other

A key situation in adding and subtracting rational expressions arises when one denominator divides evenly into the other. Factoring first reveals this relationship, and the LCD is simply the larger denominator — not their product. This saves significant algebraic work.

Addition — One Denominator Divides the Other

Add: 3x + 2 + 2xx² − 4

Factor denominators

Den 1: x+2 (already factored) Den 2: x²-4 = (x+2)(x-2)

Since (x+2) is already a factor of (x+2)(x−2), the LCD is LCD = (x+2)(x−2)
NOT (x+2)²(x−2) — because (x+2) appears only once in any single denominator.

Rewrite with the LCD

3/(x+2) = 3(x-2) / [(x+2)(x-2)] \leftarrow multiply by (x-2)/(x-2) 2x/(x²-4) = 2x / [(x+2)(x-2)] \leftarrow already has LCD

Add numerators

[3(x-2) + 2x] / [(x+2)(x-2)] = [3x - 6 + 2x] / [(x+2)(x-2)] = (5x - 6) / [(x+2)(x-2)]

Check simplification — 5x−6 = 0 at x = 6/5, which is NOT (x+2) or (x−2) → no cancellation ✓

Excluded values: x ≠ ±2

Key insight: Because (x+2) is already a factor of x²−4, the LCD is just x²−4, NOT (x+2)(x²−4). This is why factoring denominators first — before setting up the LCD — is always the right first move in adding and subtracting rational expressions. It prevents multiplying out unnecessarily large expressions.

✓ Final Answer

(5x − 6) / [(x+2)(x−2)]

x ≠ ±2

Subtraction — LCD Uses Highest Power

Subtract: 4x − 3x²

Factor denominators

Den 1: x Den 2: x² = x \cdot x

Highest power of x is x² ⟹ LCD = x²

Rewrite with the LCD

4/x = 4x/x² \leftarrow multiply by x/x 3/x² = 3/x² \leftarrow already has LCD

Subtract numerators

(4x - 3) / x²

Check simplification — 4x−3 shares no factors with x² ✓

Excluded values: x ≠ 0

✓ Final Answer

(4x − 3) / x²

x ≠ 0

Need to simplify the result of your calculation? Simplify Rational Expressions →

Section 4 Factoring to Find the LCD — Quadratic Denominators

The most important step in adding and subtracting rational expressions with quadratic denominators is complete factoring. Without factoring, it is impossible to identify the correct LCD. These problems demonstrate why the LCD is the product of the highest power of each distinct factor — never a simple product of all denominators.

Addition — Quadratic Denominators, Repeated Factor

Add: x + 1x² + 3x + 2 + xx² + 4x + 4

Factor denominators completely

x²+3x+2 = (x+1)(x+2) x²+4x+4 = (x+2)²

Find the LCD
Distinct factors: (x+1) and (x+2)
Highest powers: (x+1)¹ and (x+2)²
LCD = (x+1)(x+2)²

Rewrite each fraction with the LCD

(x+1)/[(x+1)(x+2)] = (x+1)(x+2) / [(x+1)(x+2)²] \leftarrow multiply by (x+2)/(x+2) x/(x+2)² = x(x+1) / [(x+1)(x+2)²] \leftarrow multiply by (x+1)/(x+1)

Add numerators

[(x+1)(x+2) + x(x+1)] / [(x+1)(x+2)²] = [(x+1)(x+2+x)] / [(x+1)(x+2)²] \leftarrow factor (x+1) from numerator = [(x+1)(2x+2)] / [(x+1)(x+2)²] = [2(x+1)²] / [(x+1)(x+2)²] = 2(x+1) / (x+2)² \leftarrow cancel one (x+1)

Excluded values: x ≠ −1 x ≠ −2

✓ Final Answer

2(x+1) / (x+2)²

x ≠ −1, −2

Subtraction — Quadratic Denominators, Shared Repeated Factor

Subtract: 3x² − 4 − 2x² − 4x + 4

Factor denominators completely

x²-4 = (x+2)(x-2) x²-4x+4 = (x-2)²

Find the LCD
Distinct factors: (x+2) and (x−2)
Highest powers: (x+2)¹ and (x−2)²
LCD = (x+2)(x−2)²

Rewrite with the LCD

3/[(x+2)(x-2)] = 3(x-2) / [(x+2)(x-2)²] \leftarrow multiply by (x-2)/(x-2) 2/(x-2)² = 2(x+2) / [(x+2)(x-2)²] \leftarrow multiply by (x+2)/(x+2)

Subtract — distribute negative to ALL terms of 2(x+2)

[3(x-2) - 2(x+2)] / [(x+2)(x-2)²] = [3x - 6 - 2x - 4] / [(x+2)(x-2)²] \leftarrow -2(x+2) = -2x - 4, not -2x + 4 = (x - 10) / [(x+2)(x-2)²]

Excluded values: x ≠ 2 x ≠ −2

✓ Final Answer

(x − 10) / [(x+2)(x−2)²]

x ≠ ±2

Section 5 Three Rational Expressions

Adding three rational expressions follows the same LCD method. The only difference is that three numerators must each be rewritten over the common denominator before combining. The LCD is still determined by the distinct factors across all three denominators.

Addition — Three Fractions, Three Distinct Denominators

Add: 1x + 1x + 1 + 1x − 1

Denominators: x, (x+1), (x−1) — all distinct, no common factors
LCD = x(x+1)(x−1) = x(x²−1)

Rewrite each fraction with the LCD

1/x = (x+1)(x-1) / [x(x+1)(x-1)] = (x²-1) / [x(x+1)(x-1)] 1/(x+1) = x(x-1) / [x(x+1)(x-1)] 1/(x-1) = x(x+1) / [x(x+1)(x-1)]

Add all three numerators

Numerator = (x²-1) + x(x-1) + x(x+1) Expand each term: x²-1 = x² - 1 x(x-1) = x² - x x(x+1) = x² + x Sum = x²-1 + x²-x + x²+x = 3x² - 1 \leftarrow the -x and +x cancel to zero

Excluded values: x ≠ 0 x ≠ 1 x ≠ −1

✓ Final Answer

(3x² − 1) / [x(x²−1)]

x ≠ 0, ±1

Section 6 Subtraction With Sign Distribution Traps

The sign distribution trap is the single most common error in adding and subtracting rational expressions. When subtracting a fraction whose numerator contains more than one term, the negative sign must be distributed to every term — not just the first. The problems below show the wrong answer first, so you can recognise exactly what goes wrong.

Subtraction — Sign Trap With Different Denominators · ⚠️

Subtract: 2x + 3x − 1 − x² + 2x + 5x² − 1

Factor denominators

x-1 = (x-1) x²-1 = (x+1)(x-1)

(x−1) is already a factor of x²−1 ⟹ LCD = (x+1)(x−1) = x²−1

Rewrite first fraction with the LCD

(2x+3)/(x-1) = (2x+3)(x+1) / [(x+1)(x-1)] (x²+2x+5)/(x²-1) already has LCD

Expand first numerator

(2x+3)(x+1) = 2x² + 2x + 3x + 3 = 2x² + 5x + 3

Subtract — CAREFUL: distribute negative to ALL three terms of second numerator

[2x²+5x+3 - (x²+2x+5)] / (x²-1) = [2x²+5x+3 - x²-2x-5] / (x²-1) \leftarrow each term flips sign = (x²+3x-2) / (x²-1)

Check simplification — x²+3x−2 does not factor to share (x+1) or (x−1) ✓

Excluded values: x ≠ 1 x ≠ −1

⚠️ Most Common Wrong Answer

2x²+5x+3 − x²+2x+5 = x²+7x+8

❌ Error: the negative was applied only to x² (first term) but NOT to +2x and +5.
The bracket (x²+2x+5) means ALL three terms must flip sign when subtracted.

✓ Correct: −(x²+2x+5) = −x²−2x−5 ⟹ numerator = x²+3x−2

✓ Final Answer

(x²+3x−2) / (x²−1)

x ≠ ±1

Mixed Operations — Same Denominator · ⚠️ Classic Sign Trap

Simplify: xx + 3 − 2x − 1x + 3 + 4x + 3

All three fractions share the same denominator — combine numerators directly

[x - (2x-1) + 4] / (x+3)

Distribute the negative to (2x−1):

= [x - 2x + 1 + 4] / (x+3) \leftarrow the -1 becomes +1 = (-x + 5) / (x+3) = (5-x) / (x+3)

Excluded values: x ≠ −3

⚠️ Most Common Wrong Answer — The Classic Sign Trap

x − 2x − 1 + 4 = (3−x)/(x+3)

❌ Error: the negative sign was applied to 2x but NOT to the −1 inside the bracket.
The bracket (2x−1) has two terms. Subtracting the whole bracket means:
−(2x−1) = −2x+1, NOT −2x−1.

✓ Correct: x − 2x + 1 + 4 = (5−x)/(x+3)

✓ Final Answer

(5 − x) / (x + 3)

x ≠ −3

Section 7 Challenge Problem

This multi-step problem combines everything: factoring quadratic denominators, building a three-factor LCD, careful sign distribution, and the critical rule about excluded values after cancellation. It appears on MCAT, AP Calculus, and university entrance exams.

Mixed Operations — Three Quadratic Denominators · MCAT / AP Level

Simplify: 1x² − 5x + 6 − 1x² − 4x + 3 + 1x² − 3x + 2

Factor all three denominators completely

x²-5x+6 = (x-2)(x-3) x²-4x+3 = (x-1)(x-3) x²-3x+2 = (x-1)(x-2)

Find the LCD
Distinct factors across all three denominators: (x−1), (x−2), (x−3)
Each appears to the first power only
LCD = (x−1)(x−2)(x−3)

Rewrite each fraction with the LCD

1/[(x-2)(x-3)] = (x-1) / [(x-1)(x-2)(x-3)] 1/[(x-1)(x-3)] = (x-2) / [(x-1)(x-2)(x-3)] 1/[(x-1)(x-2)] = (x-3) / [(x-1)(x-2)(x-3)]

Combine — subtract middle numerator, add last numerator

[(x-1) - (x-2) + (x-3)] / [(x-1)(x-2)(x-3)] Distribute the negative to (x-2): = [x-1 - x+2 + x-3] / [(x-1)(x-2)(x-3)] = (x - 2) / [(x-1)(x-2)(x-3)]

Cancel the common factor (x−2) from numerator and denominator

(x-2) / [(x-1) (x-2) (x-3)] = 1 / [(x-1)(x-3)]

⚠️ CRITICAL — Excluded values from the ORIGINAL denominators
Even though (x−2) cancelled from the final answer, x=2 made the ORIGINAL expression undefined.
The first fraction had (x−2)(x−3) in its denominator — so x=2 must still be excluded.
x ≠ 1 x ≠ 2 x ≠ 3
x = 2 remains excluded even though (x−2) no longer appears in the simplified expression.

Why x=2 stays excluded after cancellation: The simplified form 1/[(x−1)(x−3)] is defined at x=2 — but the original expression was NOT. Cancellation doesn’t retroactively make the original expression defined. Excluded values are always determined by the original denominators, not the simplified result.

✓ Final Answer

1 / [(x−1)(x−3)]

x ≠ 1, 2, 3 (x=2 remains excluded despite cancellation)

Quick Reference — LCD Rules for Rational Expressions

Use this table as a fast lookup when adding and subtracting rational expressions. The key principle: the LCD uses the highest power of each distinct factor, never a simple product of all denominators.

Denominators	LCD	Reasoning
x and x+1	x(x+1)	No common factors — multiply both
x and x²	x²	x divides x² — use highest power
(x+a) and (x+b)	(x+a)(x+b)	Distinct factors — multiply both
(x+a) and (x+a)²	(x+a)²	Use highest power of repeated factor
(x+a)(x+b) and (x+b)	(x+a)(x+b)	(x+b) is already in first denominator
(x²−a²) and (x+a)	(x+a)(x−a)	Factor x²−a² = (x+a)(x−a) first
(x+a)² and (x²−a²)	(x+a)²(x−a)	Highest power (x+a)², plus unique (x−a)

Common Mistakes — What NOT to Do

These five mistakes account for the majority of lost marks when adding and subtracting rational expressions. Each example shows the incorrect approach first, then explains exactly what went wrong.

❌ Mistake 1 — Adding Numerators AND Denominators

❌ Wrong

(x+2)/(x+3) + 1/(x+4) = (x+3)/(2x+7)

Adding both numerators (x+2+1=x+3) AND both denominators (x+3+x+4=2x+7) — this is how you add arithmetic fractions in elementary school, which is INCORRECT for algebra.

✓ Correct

LCD = (x+3)(x+4)
= (x+2)(x+4)/[(x+3)(x+4)] + (x+3)/[(x+3)(x+4)]
= (x²+6x+8 + x+3) / [(x+3)(x+4)]
= (x²+7x+11) / [(x+3)(x+4)]

❌ Mistake 2 — Distributing Negative to First Term Only

❌ Wrong

(x+1)/x − (2x−3)/x = (x+1 − 2x−3)/x = (−x−2)/x

The negative sign was applied to 2x but NOT to −3. The −3 should become +3, not remain −3.

✓ Correct

(x+1)/x − (2x−3)/x = (x+1 − 2x+3)/x = (−x+4)/x

❌ Mistake 3 — Multiplying Out Before Finding LCD

❌ Wrong Approach

3/(x+2) + 2x/(x²−4)
→ Multiply out: LCD = (x+2)(x²−4) = x³+2x²−4x−8 ← WRONG

Multiplying without factoring first produces a wrong (too large) LCD and unnecessary algebraic complexity.

✓ Correct Approach

Factor first: x²−4 = (x+2)(x−2)
LCD = (x+2)(x−2) = x²−4 ← just the larger denominator, NOT (x+2)(x²−4)

❌ Mistake 4 — Removing a Cancelled Factor from Excluded Values

❌ Wrong

After cancelling (x−2), student writes: x ≠ 1, 3 (forgets x=2)

Even though (x−2) cancelled, x=2 made the original expression undefined. Excluded values come from original denominators, always.

✓ Correct

Always list ALL values from original denominators: x ≠ 1, 2, 3
State clearly: “x=2 remains excluded despite cancellation”

❌ Mistake 5 — Using Product of Denominators Instead of LCD

❌ Works But Creates Unnecessary Work

For denominators (x+2) and (x²−4):
Using CD = (x+2)(x²−4) instead of LCD = x²−4
→ Creates larger numerators that must be simplified later

While not mathematically wrong, using the product instead of the LCD makes simplification much harder and increases the chance of arithmetic errors.

✓ Better Practice

Always factor denominators first → identify LCD → use LCD
This keeps numerators as small as possible throughout the calculation.

How to Check Your Answer

After adding and subtracting rational expressions, use these three verification methods to confirm your result is correct before moving on.

1 Substitute a Specific x Value

Choose a simple value of x (such as x = 1 or x = 3) that is not an excluded value. Substitute into the original expression and your simplified result — both should give the same numerical answer. If they match, your answer is almost certainly correct.

Example: to check (2x+1)/[x(x+1)] for Problem 3, substitute x=2: Original: 1/2 + 1/3 = 5/6 Simplified: (2\cdot2+1)/[2\cdot3] = 5/6 ✓

2 Re-add to Verify Subtraction

If you computed A − B = C, verify by confirming that C + B = A. Add your answer back to what you subtracted — the result should equal the first fraction. This catches sign distribution errors immediately.

If A/Q - B/Q = C/Q, then verify: C/Q + B/Q = A/Q Add the numerators: C + B should equal A

3 Verify Excluded Values

For every excluded value you stated, substitute it into each original denominator and confirm the result is zero. If any denominator does not equal zero at that x, you have incorrectly identified an excluded value. Also check that you haven’t missed any by going back through all original denominators.

For x \neq 2 and x \neq -2: Check: x²-4 at x=2 \to 4-4 = 0 ✓ Check: x²-4 at x=-2 \to 4-4 = 0 ✓

Frequently Asked Questions

How do you add rational expressions with different denominators? +

Adding rational expressions with different denominators requires the LCD method. First, factor every denominator completely. Then find the least common denominator — the product of the highest power of every distinct factor. Rewrite each fraction over the LCD by multiplying numerator and denominator by the missing factors. Finally, add the numerators, keep the LCD, and simplify. Adding and subtracting rational expressions always follows this same sequence: factor → LCD → rewrite → combine → simplify → state excluded values.

What is the LCD of rational expressions and how do you find it? +

The LCD (Least Common Denominator) of rational expressions is the product of the highest power of each distinct factor that appears in any denominator. To find it: (1) factor every denominator completely; (2) list every distinct factor; (3) use the highest power of each. For example, if denominators are (x+2) and (x+2)², the LCD is (x+2)² — not (x+2)³. If denominators are (x+1)(x+2) and (x+2)², the LCD is (x+1)(x+2)². The LCD is always the smallest expression that every denominator divides into evenly.

How do you distribute the negative sign when subtracting rational expressions? +

When subtracting a rational expression, the negative sign must be distributed to every term in the subtracted numerator — not just the first term. Mentally treat the subtracted numerator as a bracketed group: P/Q − (ax² + bx + c)/Q = [P − ax² − bx − c]/Q. This is the single most common error in adding and subtracting rational expressions. A useful check: count the number of terms in the subtracted numerator and confirm the same number of sign changes in your working.

Why must you state excluded values even after a factor cancels out? +

Excluded values are determined by the original expression, not the simplified result. When a factor such as (x−2) cancels during simplification, it disappears from the final answer — but the original expression was undefined at x=2. Cancellation does not retroactively make the original expression defined there. In formal mathematics, the simplified form is considered equivalent to the original only on the domain where the original was defined. Problem 12 on this page demonstrates this exactly: after (x−2) cancels, x=2 must still be listed as an excluded value.

What is the most common mistake when adding and subtracting rational expressions? +

The most common mistake — appearing in Problems 2, 10, and 11 on this page — is failing to distribute the subtraction sign to all terms of the subtracted numerator. Students correctly subtract the first term but forget to flip the signs of remaining terms. The second most common mistake is using the product of denominators as the LCD instead of finding the true least common denominator by factoring first. This produces correct results but forces unnecessary simplification work that increases the risk of arithmetic errors.

When can you cancel factors in a rational expression? +

You can only cancel factors that appear in both the numerator AND the denominator of the same fraction — and only entire factors, never individual terms. For example, (x+2)/(x+2)(x−1) simplifies to 1/(x−1) because (x+2) is a factor of both. But (x+2)/(x+3) cannot be simplified because 2 and 3 are not common factors — they are terms inside addition, not multiplicative factors. Never cancel across a plus or minus sign.