Math Review · Fractions

½ Fractions & Properties

Learn why the rules work — not just how. Canceling, groupings, multiplying, dividing, adding, and equations with fractions.

Fractions lessons and practice topics

Why Canceling Works — Two Properties

  • Inverse Property of Multiplication: Any value divided by itself = 1.  \( \dfrac{b}{b} = 1 \) (where \(b \neq 0\))
  • Identity Property of Multiplication: Multiplying by 1 changes nothing.  \( 1 \cdot b = b \)
\(\dfrac{5xy}{5xy} = 1 \qquad \dfrac{-3}{-3} = 1\)

When you "cancel," you replace \(\frac{a}{a}\) with 1, then drop it (Identity Property). The value does not disappear — it becomes 1.

Worked Example — Simplify \(\dfrac{6x^2}{9x}\)

1
Factor fully: \(\dfrac{2 \cdot 3 \cdot x \cdot x}{3 \cdot 3 \cdot x}\)
2
Group shared factors: \(\dfrac{3}{3} \cdot \dfrac{x}{x} \cdot \dfrac{2x}{3}\)
3
Each shared pair = 1: \(1 \cdot 1 \cdot \dfrac{2x}{3}\)
4
Drop the 1s: \(\dfrac{2x}{3}\)

Practice — Intro Level

Simplify: \(-18 \cdot \dfrac{1}{-18}\)

Inverse Property: \(b \cdot \frac{1}{b} = 1\). Answer: \(\mathbf{1}\)

Simplify: \(\dfrac{3}{7} \cdot \dfrac{7}{3}\)

Reciprocal pair: product = 1. Answer: \(\mathbf{1}\)

Implicit Groupings — The Invisible Parentheses

A fraction bar creates two invisible groups. The entire numerator is one group; the entire denominator is another.

\(\dfrac{a - b}{c + d} = (a-b) \div (c+d)\)

You can only cancel factors of the whole numerator or denominator — not individual terms inside.

  • Wrong: \(\dfrac{x+5}{5} \ne x\)  — 5 is a term, not a factor of the numerator
  • Right: \(\dfrac{5(x+3)}{5} = x+3\)  — 5 is a factor of the whole numerator

Explicit vs. Implicit Groupings

  • Explicit: written parentheses or brackets — \(3x - [5x + 3(x+3)]\)
  • Implicit: created by how things are written — fraction bar, exponents, radicals

Properties that let you remove parentheses: Associative Property (multiplication or addition only), Distributive Property (when an outside value multiplies inside terms).

Worked Example — Simplify \(\dfrac{5x}{15x+5}\)

1
The denominator has invisible parentheses: \((15x+5)\). Factor it: \(5(3x+1)\)
2
Rewrite: \(\dfrac{5x}{5(3x+1)}\)
3
Cancel the factor of 5: \(\dfrac{x}{3x+1}\)

Practice

Write as division and simplify: \(\dfrac{7-3}{3}\)

Numerator implicit group: \((7-3) \div 3 = 4 \div 3 = \dfrac{4}{3}\)

Simplify: \(\dfrac{x-4}{4-x}\)

Factor \(-1\) from denominator: \(4-x = -(x-4)\). So \(\dfrac{x-4}{-(x-4)} = \mathbf{-1}\)

Simplify: \(\dfrac{(2x-1)(4x+3)}{6x^2-3x}\)

Factor denominator: \(3x(2x-1)\). Cancel \((2x-1)\):  \(\dfrac{4x+3}{3x}\)

Multiplying Fractions

\(\dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{ac}{bd}\)

Multiply numerators together; multiply denominators together; then simplify. Tip: cancel common factors before multiplying to keep numbers smaller.

Opposites and Negatives

\(-\dfrac{a}{b} = \dfrac{-a}{b} = \dfrac{a}{-b}\)   (avoid negative in denominator for final answers)
\(-(-a) = a \qquad -(a+b) = -a-b\)

The opposite \(-a\) has the same precedence as multiplication in the order of operations.

Worked Example — \(\dfrac{2x}{5} \cdot \dfrac{-5}{7x}\)

1
Write as one fraction: \(\dfrac{2x \cdot (-5)}{5 \cdot 7x}\)
2
Cancel \(x\) and \(5\): \(\dfrac{2 \cdot (-1)}{1 \cdot 7}\)
3
Result: \(-\dfrac{2}{7}\)

Practice

Multiply: \(\dfrac{4}{5} \cdot \dfrac{2}{3}\)

\(\dfrac{4\cdot2}{5\cdot3} = \dfrac{8}{15}\)

Multiply: \(\dfrac{-7}{6} \cdot \dfrac{6}{-7}\)

Reciprocal pair with sign: \(\dfrac{(-7)(6)}{(6)(-7)} = \dfrac{-42}{-42} = 1\)

Multiply and simplify: \(\dfrac{8x^3y}{-15} \cdot \dfrac{5}{4x^2y^2}\)

Cancel \(4\) into \(8\) (→2), \(5\) into \(15\) (→3), \(x^2\) into \(x^3\) (→\(x\)), \(y\) into \(y^2\) (→\(y\)):  \(\dfrac{2x}{-3y} = -\dfrac{2x}{3y}\)

Advanced — Watch implicit groupings: \(\dfrac{2}{3} \cdot \dfrac{x+3}{5}\)

\((x+3)\) is one group — do NOT distribute into it when multiplying fractions. Answer: \(\dfrac{2(x+3)}{15}\)

Dividing Fractions — Keep, Change, Flip

\(\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \cdot \dfrac{d}{c} = \dfrac{ad}{bc}\)

Dividing by a fraction is the same as multiplying by its reciprocal. Why? Dividing by \(\frac{c}{d}\) is the same as multiplying by \(\frac{d}{c}\), which cancels the original denominator.

Worked Example — \(\dfrac{3}{7} \div \dfrac{2}{5}\)

1
Keep \(\dfrac{3}{7}\), change ÷ to ×, flip \(\dfrac{2}{5}\) to \(\dfrac{5}{2}\)
2
Multiply: \(\dfrac{3 \times 5}{7 \times 2} = \dfrac{15}{14}\)

Practice

Divide: \(\dfrac{1}{9} \div \dfrac{1}{5}\)

\(\dfrac{1}{9} \cdot \dfrac{5}{1} = \dfrac{5}{9}\)

Divide: \(\dfrac{6}{5x} \div \dfrac{3}{10x^2}\)

\(\dfrac{6}{5x} \cdot \dfrac{10x^2}{3} = \dfrac{60x^2}{15x} = 4x\)

Finding a Common Denominator

To add or subtract fractions you need matching denominators. Multiply each fraction by a form of 1 (Identity Property) that supplies the missing factors.

Example:  \(\dfrac{3}{18x^3} + \dfrac{5}{24x^2y}\)
  • Numbers: 18 = 2·3·3 and 24 = 2·2·2·3. 18 is missing 4; 24 is missing 3. LCD = 72.
  • Variables: \(x^3\) is missing \(y\); \(x^2y\) is missing one \(x\). LCD letters = \(x^3y\).
  • LCD = \(72x^3y\). Multiply each fraction by what it's missing.
\(\dfrac{3}{18x^3}\cdot\dfrac{4y}{4y} + \dfrac{5}{24x^2y}\cdot\dfrac{3x}{3x} = \dfrac{12y}{72x^3y} + \dfrac{15x}{72x^3y} = \dfrac{12y+15x}{72x^3y}\)

Practice — Build the LCD

Rewrite \(\dfrac{2}{5}\) with denominator 20.

5 is missing a factor of 4. Multiply by \(\dfrac{4}{4}\): \(\dfrac{2\cdot4}{5\cdot4} = \dfrac{8}{20}\)

Rewrite \(\dfrac{7x}{5y}\) with denominator \(30xy\).

\(5y\) is missing \(6x\). Multiply by \(\dfrac{6x}{6x}\): \(\dfrac{42x^2}{30xy}\)

Rewrite \(\dfrac{6}{6x+3}\) with denominator \((5x-5)(6x+3)\).

Missing \((5x-5)\). Multiply by \(\dfrac{5x-5}{5x-5}\): \(\dfrac{6(5x-5)}{(5x-5)(6x+3)}\)

Adding & Subtracting Fractions

You can only add or subtract fractions when the denominators are the same. With a common denominator, add or subtract the numerators and keep the denominator.

\(\dfrac{a}{c} + \dfrac{b}{c} = \dfrac{a+b}{c} \qquad \dfrac{a}{c} - \dfrac{b}{c} = \dfrac{a-b}{c}\)

No common denominator? Build one using the Identity Property (multiply by a form of 1).

Worked Example — \(\dfrac{3}{4} + \dfrac{5}{6}\)

1
Find LCD: 4 = 2·2,  6 = 2·3.  LCD = 12.
2
Multiply \(\dfrac{3}{4}\) by \(\dfrac{3}{3}\):  \(\dfrac{9}{12}\).  Multiply \(\dfrac{5}{6}\) by \(\dfrac{2}{2}\):  \(\dfrac{10}{12}\).
3
Add: \(\dfrac{9}{12} + \dfrac{10}{12} = \dfrac{19}{12}\)

Worked Example — Subtract \(\dfrac{2}{3}\) from \(\dfrac{1}{5}\)  (means \(\dfrac{1}{5} - \dfrac{2}{3}\))

1
LCD = 15.
2
\(\dfrac{1}{5}\cdot\dfrac{3}{3} = \dfrac{3}{15}\)  and  \(\dfrac{2}{3}\cdot\dfrac{5}{5} = \dfrac{10}{15}\)
3
\(\dfrac{3}{15} - \dfrac{10}{15} = -\dfrac{7}{15}\)

Practice

Add: \(\dfrac{3}{5} + \dfrac{4}{5}\)

Same denominator: \(\dfrac{3+4}{5} = \dfrac{7}{5}\)

Add: \(\dfrac{1}{7} + \dfrac{2\tfrac{1}{2}}{1}\)  (i.e., \(\dfrac{1}{7}\) of \(2\tfrac{1}{2}\) means multiply: \(\dfrac{1}{7} \cdot \dfrac{5}{2}\))

\(\dfrac{1}{7} \cdot \dfrac{5}{2} = \dfrac{5}{14}\)

Subtract: \(\dfrac{3}{8} - 1\tfrac{1}{2}\)

Convert \(1\tfrac{1}{2} = \tfrac{3}{2}\). LCD = 8. \(\dfrac{3}{8} - \dfrac{12}{8} = -\dfrac{9}{8}\)

Add with variables: \(\dfrac{3}{18x^3} + \dfrac{5}{24x^2y}\)

LCD = \(72x^3y\).  \(\dfrac{12y + 15x}{72x^3y}\)

Equations with Fractions — Eliminate Using LCD

Multiplication Property of Equality: If \(a = b\), then \(a \cdot c = b \cdot c\). Multiply every term on both sides by the LCD to clear all fractions at once.

Worked Example — Solve \(\dfrac{x}{3} = \dfrac{1}{3}\)

1
LCD = 3. Multiply both sides by 3.
2
\(3 \cdot \dfrac{x}{3} = 3 \cdot \dfrac{1}{3}\) → \(x = 1\)

Worked Example — Solve \(\dfrac{4}{7}y - \dfrac{1}{9} = 5\)

1
LCD of 7 and 9 = 63. Multiply every term by 63.
2
\(36y - 7 = 315\)
3
\(36y = 322\) → \(y = \dfrac{161}{18}\)

Worked Example — Solve \(3x = \dfrac{9}{1}\) (multiply by inverse)

1
Divide both sides by 3 (or multiply by \(\tfrac{1}{3}\)).
2
\(x = 3\)

Expressions vs. Equations — Key Distinction

  • Expression (no = sign): find a common denominator, combine. Do NOT multiply through by LCD.
  • Equation (has = sign): multiply every term by LCD to eliminate fractions, then solve.

Practice

Solve: \(\dfrac{x}{5} = 2\)

Multiply both sides by 5: \(x = 10\)

Solve: \(-7y = 49\)

Multiply by \(\dfrac{1}{-7}\): \(y = \dfrac{49}{-7} = -7\)

Solve: \(\dfrac{1}{5}x = 2\)

Multiply both sides by 5: \(x = 10\)

Solve: \(7y - \dfrac{4}{9} = 5\)

Multiply all terms by 9: \(63y - 4 = 45\) → \(63y = 49\) → \(y = \dfrac{49}{63} = \dfrac{7}{9}\)

Solve: \(\dfrac{x}{2} + \dfrac{4}{5} = \dfrac{9}{10}\)

LCD = 10. Multiply: \(5x + 8 = 9\) → \(5x = 1\) → \(x = \dfrac{1}{5}\)

Practice Problems — From the Fractions Workbook (MCC TAAC)

Click "Show Answer" to reveal the solution and reasoning.

Intro Level — Using Properties

1. Simplify: \(-18 \cdot \dfrac{1}{-18}\)

Inverse Property: \(b \cdot \frac{1}{b} = 1\). Answer: \(\mathbf{1}\)

2. Simplify: \(\dfrac{3}{7} \cdot \dfrac{7}{3}\)

Reciprocal pair: product = 1. Answer: \(\mathbf{1}\)

3. Write as division: \(\dfrac{7-3}{3}\)

Numerator implicit grouping: \((7-3) \div 3 = 4 \div 3 = \mathbf{\dfrac{4}{3}}\)

4. Write as division: \(\dfrac{5-7+3}{x}\)

\((5-7+3) \div x = 1 \div x = \dfrac{1}{x}\)

Mid Level — Factor First, Then Simplify

5. Simplify: \(\dfrac{x^2-x}{x}\)

Factor numerator: \(x(x-1)\). Cancel \(x\): \(\mathbf{x-1}\)

6. Simplify: \(\dfrac{4x-16}{4}\)

Factor numerator: \(4(x-4)\). Cancel 4: \(\mathbf{x-4}\)

7. Multiply: \(\dfrac{2x}{5} \cdot \dfrac{-5}{7x}\)

Cancel \(x\) and 5: \(\dfrac{2 \cdot(-1)}{1 \cdot 7} = \mathbf{-\dfrac{2}{7}}\)

8. Find common denominator for \(\dfrac{7x}{5y}\) to get denominator \(30xy\).

Missing \(6x\): multiply by \(\dfrac{6x}{6x}\) → \(\dfrac{42x^2}{30xy}\)

Advanced Level — Implicit Groupings

9. Simplify: \(\dfrac{x-4}{4-x}\)

Factor \(-1\) from denominator: \(4-x = -(x-4)\). Result: \(\mathbf{-1}\)

10. Simplify: \(\dfrac{(x+3)(x-4)}{(x+3)}\)

\((x+3)\) cancels: \(\mathbf{x-4}\)

11. Simplify: \(\dfrac{(2x-1)(4x+3)}{6x^2-3x}\)

Factor denominator: \(3x(2x-1)\). Cancel \((2x-1)\): \(\dfrac{4x+3}{3x}\)

12. Agree or Disagree: \(\dfrac{3(x+6)}{9} = \dfrac{3}{3} \cdot \dfrac{x+6}{3}\)

Agree. You can split the fraction into two factors since the denominator 9 = 3·3. The result is \(\dfrac{x+6}{3}\).
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Question 1

Intro
Simplify:  \(\dfrac{-3}{-3}\)
Any non-zero value divided by itself = 1. That's the Inverse Property of Multiplication.

Question 2

Intro
Simplify:  \(\dfrac{5xy}{5xy}\)
The whole expression over itself = 1. The Inverse Property applies to any non-zero value.

Question 3

Mid
Simplify:  \(\dfrac{x^2 - x}{x}\)
Factor the numerator first: \(x^2 - x = x(x-1)\). Then cancel the \(x\).

Question 4

Mid
Multiply and simplify:  \(\dfrac{4}{5} \cdot \dfrac{2}{3}\)
Multiply numerators: \(4 \times 2 = 8\). Multiply denominators: \(5 \times 3 = 15\).

Question 5

Mid
Which correctly simplifies  \(\dfrac{x+5}{5}\)?
Canceling requires 5 to be a factor of the entire numerator. Here \(x+5\) is not equal to \(5 \cdot (\text{something})\).

Question 6

Mid
Add: \(\dfrac{3}{4} + \dfrac{5}{6}\)
LCD = 12. Convert: \(\frac{9}{12} + \frac{10}{12} = \frac{19}{12}\).

Question 7

Advanced
Simplify:  \(\dfrac{(x+3)(x-4)}{(x+3)}\)
\((x+3)\) is a factor of the entire numerator. Cancel it with the denominator: result = \(x-4\).

Question 8

Advanced
Solve for \(x\):  \(\dfrac{x}{2} + \dfrac{4}{5} = \dfrac{9}{10}\)
Multiply all terms by LCD = 10: \(5x + 8 = 9\) → \(5x = 1\) → \(x = \frac{1}{5}\).