Simplifying Linear Fractional Expressions

Learn how to combine and simplify linear algebraic fractions with numerical denominators using the lowest common multiple (LCM), handle minus signs safely, and reduce fractions to lowest terms.

Before you start

Adding and subtracting numerical fractions: To add $\frac{1}{3}$ and $\frac{1}{4}$ , find the LCM of the denominators ( $12$ ): $\frac{4}{12} + \frac{3}{12} = \frac{7}{12}$ .
The Distributive Law: $a(b + c) = ab + ac$ and $-a(b - c) = -ab + ac$ .

1. What Is a Linear Fractional Expression?

In this lesson, a linear fractional expression (a linear expression with fractions) is made of fractions whose numerator is a linear expression (each term has at most one letter, raised to the power $1$ , or is a number) and whose denominator is a plain non-zero whole number.

Here are examples of linear fractional expressions:

- $\frac{2x}{3}$ (a single term with fractional coefficient $\frac{2}{3}$ )

- $\frac{x - 5}{2}$ (a two-term linear numerator over the numerical denominator $2$ )

- $\frac{2x}{3} - \frac{3(x - 5)}{2}$ (two linear fractions connected by a minus sign)

In Secondary 1, letters never appear in the denominator. Algebraic fractions with letters in the denominator (such as $\frac{2}{x}$ or $\frac{3}{x + 1}$ ) come in Secondary 2.

The Fraction Line Acts as a Bracket

The horizontal fraction line groups the terms above it together just like a pair of brackets.

Writing $\frac{a + b}{c}$ means $(a + b) \div c$ .

When subtracting a fraction, always put the numerator inside brackets before removing the fraction lines:

$-\frac{a - b}{c} = \frac{-(a - b)}{c} = \frac{-a + b}{c}$

2. Combining Fractions with the Same Denominator

When two algebraic fractions have the exact same denominator, combine them directly over that common denominator by adding or subtracting their numerators.

Fractions with Common Denominators

$\frac{A}{m} + \frac{B}{m} = \frac{A + B}{m}$

$\frac{A}{m} - \frac{B}{m} = \frac{A - (B)}{m}$

Worked Example 1: Same Denominators and the Minus Sign Rule

Express each of the following as a single fraction in its simplest form:
(a) $\frac{5x + 3}{7} + \frac{2x - 1}{7}$
(b) $\frac{3x + 1}{4} - \frac{x - 3}{4}$

Solution:

(a) The denominators are identical ( $7$ ). Add the numerators directly:

$\begin{aligned} \frac{5x + 3}{7} + \frac{2x - 1}{7} &= \frac{(5x + 3) + (2x - 1)}{7} \\ &= \frac{5x + 2x + 3 - 1}{7} \\ &= \frac{7x + 2}{7} \end{aligned}$

(b) The denominators are identical ( $4$ ). Place the second numerator in brackets to protect the minus sign:

$\begin{aligned} \frac{3x + 1}{4} - \frac{x - 3}{4} &= \frac{(3x + 1) - (x - 3)}{4} \\ &= \frac{3x + 1 - x + 3}{4} \\ &= \frac{2x + 4}{4} \end{aligned}$

Notice that every term in the numerator ( $2x$ and $4$ ) and the denominator ( $4$ ) is divisible by $2$ . Divide each term by $2$ to simplify:

$= \frac{2x \div 2 + 4 \div 2}{4 \div 2} = \frac{x + 2}{2}$

(a) $\frac{7x + 2}{7}$ (b) $\frac{x + 2}{2}$

3. Combining Fractions with Different Denominators (The LCM Method)

When fractions have different numerical denominators, we must first find their lowest common multiple (LCM) to express them with a single common denominator, exactly as with numerical fractions.

The 4-Step Strategy for Linear Fractions

Step 1: Find the LCM of the numerical denominators.

Step 2: Convert each fraction to an equivalent fraction with that common denominator. Always enclose any multi-term numerator in brackets when multiplying.

Step 3: Combine over one denominator. Write the whole expression as a single fraction: $\frac{\text{Numerator 1} \pm \text{Numerator 2}}{\text{LCM}}$ .

Step 4: Expand and simplify. Expand the brackets in the numerator, collect like terms, and reduce the fraction if all terms share a common factor greater than $1$ with the denominator.

Worked Example 2: Different Denominators and Multi-Term Numerators

Express as a single fraction in its simplest form:
$\frac{2x}{3} - \frac{3(x - 5)}{2}$

Solution:

Step 1: Find the LCM of denominators $3$ and $2$ . The LCM is $6$ .

Step 2 & 3: Express each fraction with denominator $6$ :

$\frac{2x}{3} = \frac{2x \times 2}{6} = \frac{4x}{6}$

$\frac{3(x - 5)}{2} = \frac{3(x - 5) \times 3}{6} = \frac{9(x - 5)}{6}$

Combine the two fractions into a single fraction over $6$ :

$= \frac{4x - 9(x - 5)}{6}$

Step 4: Expand the bracket $-9(x - 5)$ in the numerator, noting $(-9) \times (-5) = +45$ :

$= \frac{4x - 9x + 45}{6}$

Combine like terms ( $4x - 9x = -5x$ ):

$= \frac{-5x + 45}{6}$

This can also be written with the positive term first as $\frac{45 - 5x}{6}$ . Both forms are fully simplified.

$\frac{-5x + 45}{6}$ (or $\frac{45 - 5x}{6}$ )

Worked Example 3: Fractions with Two Variables

Express as a single fraction in its simplest form:
$\frac{2a - b}{3} + \frac{a + 3b}{4}$

Solution:

Step 1: The LCM of $3$ and $4$ is $12$ .

Step 2 & 3: Multiply the first fraction by $\frac{4}{4}$ and the second by $\frac{3}{3}$ :

$= \frac{4(2a - b) + 3(a + 3b)}{12}$

Step 4: Expand brackets in the numerator:

$= \frac{8a - 4b + 3a + 9b}{12}$

Group and combine like terms ( $8a + 3a = 11a$ , and $-4b + 9b = 5b$ ):

$= \frac{11a + 5b}{12}$

Since $11, 5,$ and $12$ share no common factor other than $1$ , this fraction cannot be reduced further.

$\frac{11a + 5b}{12}$

4. Reducing Fractions to Lowest Terms Correctly

Once the numerator is simplified into a single linear expression, check whether every term in the numerator and the denominator shares a common whole number factor greater than $1$ .

An Expression Is Not an Equation

An expression has no equals sign and no other side, so it is never multiplied through to clear the denominators: multiplying it by a number would change its value.

The answer keeps a denominator unless the denominator divides every term of the numerator, as in $\frac{5x + 5}{5} = x + 1$ . A smaller common factor only makes the denominator smaller, as in $\frac{2x + 4}{4} = \frac{x + 2}{2}$ .

The Golden Rule of Cancelling

A number can only be cancelled from a fraction if it divides every single term in the numerator and the denominator.

$\frac{ka + kb}{kc} = \frac{a + b}{c}$

You can divide through by $k$ because $k$ divides both $ka$ and $kb$ .

Never cancel from one term only: in $\frac{2x + 5}{2}$ , crossing out the $2$ s to get $x + 5$ divides only $2x$ by $2$ , not $5$ . To cancel the $2$ , it must divide every term, and $2$ does not divide $5$ , so $\frac{2x + 5}{2}$ is already in its simplest form. (Compare $\frac{2x + 6}{2} = x + 3$ , not $x + 6$ .)

Worked Example 4: Expanding, Combining and Reducing

Simplify each expression completely:
(a) $\frac{15x - 20}{10}$
(b) $\frac{5(p + q)}{6} - \frac{p + 3q}{2}$

Solution:

(a) Both terms in the numerator ( $15x$ and $-20$ ) and the denominator ( $10$ ) are divisible by $5$ :

$\frac{15x \div 5 - 20 \div 5}{10 \div 5} = \frac{3x - 4}{2}$

(b) The LCM of $6$ and $2$ is $6$ . The first fraction already has denominator $6$ ; multiply the second numerator and denominator by $3$ , keeping the numerator in brackets:

$= \frac{5(p + q) - 3(p + 3q)}{6}$

Expand brackets in the numerator, noting $(-3) \times 3q = -9q$ :

$= \frac{5p + 5q - 3p - 9q}{6}$

Combine like terms ( $5p - 3p = 2p$ and $5q - 9q = -4q$ ):

$= \frac{2p - 4q}{6}$

Every term, $2p$ , $-4q$ and $6$ , is divisible by $2$ . Divide each by $2$ :

$= \frac{2p \div 2 - 4q \div 2}{6 \div 2} = \frac{p - 2q}{3}$

(a) $\frac{3x - 4}{2}$ (b) $\frac{p - 2q}{3}$

Common Mistakes to Avoid

Cancelling terms instead of common factors: In $\frac{2x + 5}{2}$ , crossing out the $2$ s to write $x + 5$ . This divides only $2x$ by $2$ , not $5$ . A number can be cancelled only if it divides every term, and $2$ does not divide $5$ , so $\frac{2x + 5}{2}$ cannot be reduced. (Compare $\frac{2x + 6}{2} = x + 3$ , not $x + 6$ .)
Dropping brackets after a minus sign: Writing $-\frac{x - 3}{4}$ as $\frac{\dots - x - 3}{4}$ . The minus sign applies to both terms: $-(x - 3) = -x + 3$ .
Discarding the denominator: Writing $\frac{x}{4} + \frac{x}{6} = 3x + 2x = 5x$ . Multiplying the expression by $12$ makes it $12$ times as large, so it is no longer the same expression. The right answer is $\frac{3x}{12} + \frac{2x}{12} = \frac{5x}{12}$ .
Leaving separate fractions when a single fraction is required: Writing $\frac{3x + 1}{2}$ as $\frac{3}{2}x + \frac{1}{2}$ is mathematically correct, but if a question instructs you to "express as a single fraction", combining into $\frac{3x + 1}{2}$ is required.

Practice Questions

Work out each problem on paper before checking your solution. Questions 9 to 12 apply linear fractional expressions to lengths, petrol, costs and travel.

Question 1: Same Denominators with Addition and Subtraction

Express each as a single fraction in its simplest form:
(a) $\frac{3x + 1}{5} + \frac{2x + 4}{5}$
(b) $\frac{5a + 7}{4} - \frac{a + 3}{4}$

ShowHide solution

(a) Add numerators directly over $5$ :
$\frac{(3x + 1) + (2x + 4)}{5} = \frac{5x + 5}{5}$
Divide every term by $5$ :
$= \frac{5x \div 5 + 5 \div 5}{5 \div 5} = \frac{x + 1}{1} = x + 1$

(b) Place the second numerator in brackets and subtract over $4$ :
$\frac{(5a + 7) - (a + 3)}{4} = \frac{5a + 7 - a - 3}{4} = \frac{4a + 4}{4}$
Divide every term by $4$ :
$= a + 1$

(a) $x + 1$ (b) $a + 1$

Question 2: Same Denominator with Minus Sign Distribution

Express each as a single fraction in simplest form:
(a) $\frac{4x + 5}{3} - \frac{x - 2}{3}$
(b) $\frac{7a - 3b}{5} - \frac{2a - 8b}{5}$

ShowHide solution

(a) Subtract over $3$ , with the second numerator in brackets so that both of its signs change:
$\frac{(4x + 5) - (x - 2)}{3} = \frac{4x + 5 - x + 2}{3} = \frac{3x + 7}{3}$
$3$ does not divide $7$ , so the fraction cannot be reduced.

(b) Subtract over $5$ :
$\frac{(7a - 3b) - (2a - 8b)}{5} = \frac{7a - 3b - 2a + 8b}{5} = \frac{5a + 5b}{5} = a + b$

(a) $\frac{3x + 7}{3}$ (b) $a + b$

Question 3: Single-Term Numerators (a Warm-Up, as in Lesson 8.1)

Express each as a single fraction in simplest form:
(a) $\frac{3x}{4} + \frac{x}{6}$
(b) $\frac{4a}{5} - \frac{a}{2}$
(c) $\frac{7m}{6} - \frac{m}{4} + \frac{1}{3}$

ShowHide solution

(a) LCM of $4$ and $6$ is $12$ :
$\frac{9x}{12} + \frac{2x}{12} = \frac{11x}{12}$

(b) LCM of $5$ and $2$ is $10$ :
$\frac{8a}{10} - \frac{5a}{10} = \frac{3a}{10}$

(c) LCM of $6$ , $4$ and $3$ is $12$ :
$\frac{14m}{12} - \frac{3m}{12} + \frac{4}{12} = \frac{14m - 3m + 4}{12} = \frac{11m + 4}{12}$
$11$ , $4$ and $12$ share no common factor greater than $1$ , so the fraction cannot be reduced.

(a) $\frac{11x}{12}$ (b) $\frac{3a}{10}$ (c) $\frac{11m + 4}{12}$

Question 4: Two-Term Numerators with Different Denominators

Express each as a single fraction in simplest form:
(a) $\frac{x + 1}{2} + \frac{x - 3}{3}$
(b) $\frac{2x - 1}{4} - \frac{x - 3}{6}$

ShowHide solution

(a) LCM of $2$ and $3$ is $6$ :
$\frac{3(x + 1) + 2(x - 3)}{6} = \frac{3x + 3 + 2x - 6}{6} = \frac{5x - 3}{6}$

(b) LCM of $4$ and $6$ is $12$ :
$\frac{3(2x - 1) - 2(x - 3)}{12} = \frac{6x - 3 - 2x + 6}{12} = \frac{4x + 3}{12}$

(a) $\frac{5x - 3}{6}$ (b) $\frac{4x + 3}{12}$

Question 5: Fractions with Multipliers in Front of Brackets

Express as a single fraction in simplest form:
(a) $\frac{2(x - 1)}{3} - \frac{x + 2}{4}$
(b) $\frac{3(x + 2)}{5} - \frac{2(x - 1)}{3}$

ShowHide solution

(a) LCM of $3$ and $4$ is $12$ :
$\frac{4 \times 2(x - 1) - 3(x + 2)}{12} = \frac{8(x - 1) - 3(x + 2)}{12}$
Expand brackets in numerator:
$= \frac{8x - 8 - 3x - 6}{12} = \frac{5x - 14}{12}$

(b) LCM of $5$ and $3$ is $15$ :
$\frac{3 \times 3(x + 2) - 5 \times 2(x - 1)}{15} = \frac{9(x + 2) - 10(x - 1)}{15}$
Expand brackets in numerator:
$= \frac{9x + 18 - 10x + 10}{15} = \frac{-x + 28}{15} \quad (\text{or } \frac{28 - x}{15})$

(a) $\frac{5x - 14}{12}$ (b) $\frac{-x + 28}{15}$

Question 6: Two-Term Numerators and a Minus Sign

Simplify completely as a single fraction:
$\frac{3x - 1}{4} - \frac{2x + 3}{6}$

ShowHide solution

The LCM of $4$ and $6$ is $12$ :
$\frac{3(3x - 1) - 2(2x + 3)}{12}$
Expand brackets in the numerator. The minus sign in front of the second bracket makes $+3$ into $(-2) \times 3 = -6$ :
$= \frac{9x - 3 - 4x - 6}{12}$
Combine like terms:
$= \frac{5x - 9}{12}$

$\frac{5x - 9}{12}$

Question 7: Reducing Single Fractions to Lowest Terms

Reduce each fraction to its lowest terms by dividing numerator and denominator by common factors:
(a) $\frac{6x + 9}{3}$
(b) $\frac{14a - 21b}{7}$
(c) $\frac{8m + 12n - 16}{4}$

ShowHide solution

(a) Divide each term by $3$ :
$\frac{6x \div 3 + 9 \div 3}{3 \div 3} = 2x + 3$

(b) Divide each term by $7$ :
$\frac{14a \div 7 - 21b \div 7}{7 \div 7} = 2a - 3b$

(c) Divide each term by $4$ :
$\frac{8m \div 4 + 12n \div 4 - 16 \div 4}{4 \div 4} = 2m + 3n - 4$

(a) $2x + 3$ (b) $2a - 3b$ (c) $2m + 3n - 4$

Question 8: Multi-Term Fractions with Two Variables

Express as a single fraction in simplest form:
$\frac{5(2m - n)}{6} - \frac{m - 4n}{9}$

ShowHide solution

The LCM of $6$ and $9$ is $18$ :
$\frac{3 \times 5(2m - n) - 2(m - 4n)}{18} = \frac{15(2m - n) - 2(m - 4n)}{18}$
Expand brackets in the numerator:
$= \frac{30m - 15n - 2m + 8n}{18}$
Combine like terms:
$= \frac{28m - 7n}{18}$
$28m$ and $-7n$ share the factor $7$ , but $7$ does not divide $18$ , so the fraction cannot be reduced.

$\frac{28m - 7n}{18}$

Question 9: Ribbon Length and Average in a Setting

A craft ribbon is cut into two pieces:

The length of the first piece is $\frac{3x + 1}{4}\text{ m}$ .
The length of the second piece is $\frac{x + 3}{6}\text{ m}$ .
(a) Express the total length of the ribbon as a single fraction in terms of $x$ .
(b) Express the average length of the two pieces as a single fraction in terms of $x$ .
(c) If $x = 5$ , calculate the total length and the average length in metres.

ShowHide solution

(a) Add the two lengths over common denominator $12$ :
$\text{Total length} = \frac{3(3x + 1) + 2(x + 3)}{12} = \frac{9x + 3 + 2x + 6}{12} = (\frac{11x + 9}{12})\text{ m}$

(b) Average length is total length divided by $2$ :
$\text{Average length} = \frac{11x + 9}{12} \div 2 = (\frac{11x + 9}{24})\text{ m}$

(c) Substitute $x = 5$ :
$\text{Total length} = \frac{11(5) + 9}{12} = \frac{55 + 9}{12} = \frac{64}{12} = \frac{16}{3}\text{ m} \quad (\text{or } 5\frac{1}{3}\text{ m})$
$\text{Average length} = \frac{16}{3} \div 2 = \frac{8}{3}\text{ m} \quad (\text{or } 2\frac{2}{3}\text{ m})$

(a) $(\frac{11x + 9}{12})\text{ m}$ (b) $(\frac{11x + 9}{24})\text{ m}$ (c) Total $= \frac{16}{3}\text{ m}$ , Average $= \frac{8}{3}\text{ m}$

Question 10: Petrol Usage and Tank Balance in a Setting

A delivery van starts a day with $50\text{ litres}$ of petrol in its fuel tank.

On trip A, it uses $\frac{2k + 1}{5}\text{ litres}$ of petrol.
On trip B, it uses $\frac{k - 3}{4}\text{ litres}$ of petrol.
(a) Write an expression in terms of $k$ for the total petrol consumed on both trips as a single fraction in simplest form.
(b) Write an expression in terms of $k$ for the volume of petrol remaining in the fuel tank as a single fraction.
(c) If $k = 7$ , calculate the volume of petrol remaining in the tank.

ShowHide solution

(a) Add petrol used over common denominator $20$ :
$\text{Petrol used} = \frac{4(2k + 1) + 5(k - 3)}{20} = \frac{8k + 4 + 5k - 15}{20} = (\frac{13k - 11}{20})\text{ litres}$

(b) Subtract used petrol from $50\text{ litres}$ :
$\text{Remaining} = 50 - \frac{13k - 11}{20} = \frac{50 \times 20}{20} - \frac{13k - 11}{20} = \frac{1000 - (13k - 11)}{20} = (\frac{1011 - 13k}{20})\text{ litres}$

(c) Substitute $k = 7$ :
$\text{Remaining} = \frac{1011 - 13(7)}{20} = \frac{1011 - 91}{20} = \frac{920}{20} = 46\text{ litres}$
Check trips: Trip A $= \frac{2(7) + 1}{5} = \frac{15}{5} = 3\text{ litres}$ ; Trip B $= \frac{7 - 3}{4} = \frac{4}{4} = 1\text{ litre}$ . Total used $= 4\text{ litres}$ ; remaining $= 50 - 4 = 46\text{ litres}$ .

(a) $(\frac{13k - 11}{20})\text{ litres}$ (b) $(\frac{1011 - 13k}{20})\text{ litres}$ (c) $46\text{ litres}$

Question 11: Food and Drink Costs Shared in a Setting

Two CCA clubs hold a joint end-of-year gathering. The food costs $(\frac{5m + 30}{2})$ dollars and the cost of the drinks is $(\frac{9m + 12}{4})$ dollars.
(a) Express the total cost of the food and drinks as a single fraction in terms of $m$ .
(b) The total cost is shared equally between the $2$ clubs. Express the amount paid by each club as a single fraction in terms of $m$ .
(c) If $m = 20$ , find the amount paid by each club.

ShowHide solution

(a) The LCM of $2$ and $4$ is $4$ . Multiply the first numerator, in brackets, by $2$ :
$\text{Total cost} = \frac{2(5m + 30)}{4} + \frac{9m + 12}{4} = \frac{10m + 60 + 9m + 12}{4} = (\frac{19m + 72}{4})\text{ dollars}$
$19$ is odd, so $2$ does not divide every term, and the fraction cannot be reduced.

(b) Divide the total cost by $2$ :
$\text{Amount per club} = \frac{19m + 72}{4} \div 2 = \frac{19m + 72}{4 \times 2} = (\frac{19m + 72}{8})\text{ dollars}$

(c) Substitute $m = 20$ :
$\text{Amount per club} = \frac{19(20) + 72}{8} = \frac{380 + 72}{8} = \frac{452}{8} = \frac{113}{2}$
$\frac{113}{2} = 56.5$ , so each club pays $56.50.
Check: food $= \frac{5(20) + 30}{2} = \frac{130}{2} = 65$ , so $65; drinks $= \frac{9(20) + 12}{4} = \frac{192}{4} = 48$ , so $48. The total is $113, and half of $113 is $56.50.

(a) $(\frac{19m + 72}{4})$ dollars (b) $(\frac{19m + 72}{8})$ dollars (c) $56.50

Question 12: Journey Times in a Travel Setting

A cyclist rides two legs of a journey:

Leg 1: distance $(2x + 5)\text{ km}$ at an average speed of $15\text{ km/h}$ .
Leg 2: distance $(x - 1)\text{ km}$ at an average speed of $10\text{ km/h}$ .
Recall that $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$ .
(a) Write down an expression for the time taken for Leg 1 and for Leg 2 in hours.
(b) Express the total travel time as a single fraction in simplest form.
(c) If $x = 5$ , calculate the total time taken in hours and minutes.

ShowHide solution

(a) $\text{Time 1} = (\frac{2x + 5}{15})\text{ hours}$ ; $\text{Time 2} = (\frac{x - 1}{10})\text{ hours}$ .

(b) Common denominator of $15$ and $10$ is $30$ :
$\text{Total time} = \frac{2(2x + 5) + 3(x - 1)}{30} = \frac{4x + 10 + 3x - 3}{30} = (\frac{7x + 7}{30})\text{ hours}$
$7$ does not divide $30$ , so the fraction cannot be reduced.

(c) Substitute $x = 5$ :
$\text{Total time} = \frac{7(5) + 7}{30} = \frac{35 + 7}{30} = \frac{42}{30} = \frac{7}{5}\text{ hours}$
Converting $\frac{7}{5}$ hours: $1\text{ hour}$ and $\frac{2}{5} \times 60 = 24\text{ minutes}$ .
Total time $= 1\text{ hour } 24\text{ minutes}$ .
Check: Leg 1 is $15\text{ km}$ at $15\text{ km/h}$ , $1\text{ hour}$ ; Leg 2 is $4\text{ km}$ at $10\text{ km/h}$ , $\frac{4}{10}\text{ hour} = 24\text{ minutes}$ .

(a) Leg 1: $(\frac{2x + 5}{15})\text{ hours}$ , Leg 2: $(\frac{x - 1}{10})\text{ hours}$ (b) $(\frac{7x + 7}{30})\text{ hours}$ (c) $1\text{ hour } 24\text{ minutes}$

Key Points to Remember

Find the LCM of Denominators: Convert all fractions to equivalent fractions using their lowest common multiple as the common denominator.
Use Brackets for Numerators: Whenever subtracting a fraction, enclose the entire numerator in brackets: $-\frac{B}{m} = \frac{-(B)}{m}$ .
Keep the Denominator: An expression is not an equation, so never multiply through to clear the denominators. The answer keeps a denominator unless the denominator divides every term of the numerator; a smaller common factor only makes the denominator smaller.
Expand and Collect: Expand all brackets in the combined numerator and collect like terms together.
Cancelling Requires All Terms: You may only divide through by a number if it divides every single term in the numerator as well as the denominator.
The Unit Rule: When an algebraic fraction has a multi-term numerator and carries a unit, enclose the fraction in brackets, such as $(\frac{11x + 9}{12})\text{ m}$ .