Simple Fractional Equations

Solve linear equations with fractions by clearing denominators with the lowest common multiple, apply cross-multiplication, and solve equations with the unknown in the denominator.

Before you start

In Lesson 8.3, you simplified algebraic fractions by finding the lowest common multiple (LCM) of numerical denominators.
When solving linear equations, you multiplied both sides of $\frac{m}{4} = 3$ by $4$ to remove the fraction. This lesson does the same when there is more than one fraction.
With an expression such as $\frac{x}{3} + \frac{x}{4}$ you could not remove the denominators: multiplying it by $12$ would change its value. In an equation we may multiply both sides by $12$ , because doing the same to both sides keeps the equation balanced, and the denominators disappear.
Remember that a fraction bar acts as a bracket: $\frac{x - 3}{4}$ means $\frac{(x - 3)}{4}$ .

Equations with Numerical Denominators

A fractional equation has the unknown in a fraction (in a numerator or a denominator). When the denominators are numbers, the most efficient method is to clear the fractions by multiplying every term on both sides by the LCM of all the denominators.

Every single term — whether a fraction or an integer — must be multiplied by the LCM to keep the equation balanced.

The LCM Method for Clearing Fractions

1. Find the lowest common multiple (LCM) of all denominators.

2. Multiply every term on both sides of the equation by this LCM.

3. Cancel each denominator with the LCM to obtain an equation without fractions.

4. Solve the resulting linear equation and verify your answer by substitution.

Example 1: Numerical Denominators

Solve each equation and check the solution:
(a) $\frac{x}{3} + \frac{x}{4} = 7$
(b) $\frac{2y}{5} - 3 = 1$

Solution

(a) Solve $\frac{x}{3} + \frac{x}{4} = 7$ :

The denominators are $3$ and $4$ . The LCM of $3$ and $4$ is $12$ .

Multiply every term by $12$ :

$12\left(\frac{x}{3}\right) + 12\left(\frac{x}{4}\right) = 12(7)$

Cancel each fraction:

$4x + 3x = 84$

Combine like terms:

$7x = 84$

Divide by $7$ :

$x = \frac{84}{7} = 12$

Check: $\text{LHS} = \frac{12}{3} + \frac{12}{4} = 4 + 3 = 7 = \text{RHS} \quad \checkmark$

(b) Solve $\frac{2y}{5} - 3 = 1$ :

Add $3$ to both sides first:

$\frac{2y}{5} - 3 + 3 = 1 + 3$

$\frac{2y}{5} = 4$

Multiply both sides by $5$ :

$2y = 4 \times 5 = 20$

Divide by $2$ :

$y = 10$

(Multiplying every term by $5$ also works: $2y - 15 = 5$ , so $2y = 20$ and $y = 10$ . Here adding $3$ first is quicker.)

Check: $\text{LHS} = \frac{2(10)}{5} - 3 = \frac{20}{5} - 3 = 4 - 3 = 1 = \text{RHS} \quad \checkmark$

(a) $x = 12$ ; (b) $y = 10$ .

Numerators with More than One Term

When a numerator has more than one term (such as $x + 2$ or $2x - 5$ ), place brackets around the numerator before multiplying by the LCM.

This prevents sign errors, particularly when a fraction is preceded by a minus sign.

Brackets for Numerators with Two Terms

Write $-\frac{x - 3}{4}$ as $-\frac{(x - 3)}{4}$ .

When multiplying by the LCM, the minus sign applies to every term inside the brackets: $-(x - 3) = -x + 3$ .

Example 2: Numerators with Two Terms and a Minus Sign

Solve $\frac{x + 1}{3} - \frac{x - 2}{4} = 2$ .

Solution

Insert brackets around each numerator:

$\frac{(x + 1)}{3} - \frac{(x - 2)}{4} = 2$

The LCM of $3$ and $4$ is $12$ . Multiply every term by $12$ :

$12\left(\frac{x + 1}{3}\right) - 12\left(\frac{x - 2}{4}\right) = 12(2)$

Cancel denominators:

$4(x + 1) - 3(x - 2) = 24$

Expand brackets, taking care with $-3(x - 2) = -3x + 6$ :

$4x + 4 - 3x + 6 = 24$

Combine like terms on LHS ( $4x - 3x = x$ and $4 + 6 = 10$ ):

$x + 10 = 24$

Subtract $10$ from both sides:

$x + 10 - 10 = 24 - 10$

$x = 14$

Check: $\text{LHS} = \frac{14 + 1}{3} - \frac{14 - 2}{4} = \frac{15}{3} - \frac{12}{4} = 5 - 3 = 2 = \text{RHS} \quad \checkmark$

$x = 14$ .

Cross-Multiplication for Proportions

When an equation consists of a single fraction on the left equal to a single fraction on the right, we have a proportion: $\frac{A}{B} = \frac{C}{D}$ .

Multiplying both sides by the common denominator $BD$ gives the cross-multiplication rule: $AD = BC$ .

The Cross-Multiplication Rule

If $\frac{A}{B} = \frac{C}{D}$ (where $B \neq 0$ and $D \neq 0$ ), then:

$A \times D = B \times C$

Warning: Cross-multiplication applies only when each side of the equation consists of a single fraction. If there is an extra added or subtracted term, find the LCM first.

Example 3: Cross-Multiplication

Solve $\frac{2x + 1}{3} = \frac{x + 5}{2}$ .

Solution

Both sides are single fractions. Multiply both sides by $6$ (the LCM of $3$ and $2$ ):

$6\left(\frac{2x + 1}{3}\right) = 6\left(\frac{x + 5}{2}\right)$

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

Cross-multiplication is this step done at once: the numerator $2x + 1$ is multiplied by the other denominator $2$ , and the numerator $x + 5$ by the other denominator $3$ .

Expand brackets on both sides:

$4x + 2 = 3x + 15$

Subtract $3x$ from both sides:

$4x + 2 - 3x = 3x + 15 - 3x$

$x + 2 = 15$

Subtract $2$ from both sides:

$x + 2 - 2 = 15 - 2$

$x = 13$

Check: $\text{LHS} = \frac{2(13) + 1}{3} = \frac{27}{3} = 9$ ; $\text{RHS} = \frac{13 + 5}{2} = \frac{18}{2} = 9 = \text{LHS} \quad \checkmark$

$x = 13$ .

Equations with the Unknown in the Denominator

When the unknown appears in the denominator, such as $\frac{12}{x} = 4$ , we must state the restriction that the denominator cannot be zero ( $x \neq 0$ ), because dividing by zero has no meaning.

We then multiply both sides by the denominator (or cross-multiply) so that the unknown is no longer in a denominator. At the end we check that the solution is not the value that the restriction excludes.

Example 4: The Unknown in the Denominator

Solve each equation and state any restrictions:
(a) $\frac{15}{x} = 5$
(b) $\frac{3}{x - 2} = 6$

Solution

(a) Solve $\frac{15}{x} = 5$ :

Restriction: $x \neq 0$ .

Multiply both sides by $x$ :

$\frac{15}{x} \times x = 5 \times x$

$15 = 5x$

Divide both sides by $5$ :

$x = 3$

Since $3 \neq 0$ , the solution is valid.

Check: $\text{LHS} = \frac{15}{3} = 5 = \text{RHS} \quad \checkmark$

(b) Solve $\frac{3}{x - 2} = 6$ :

Restriction: $x - 2 \neq 0$ , so $x \neq 2$ .

Multiply both sides by $(x - 2)$ :

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

$3 = 6(x - 2)$

Expand the bracket:

$3 = 6x - 12$

Add $12$ to both sides:

$3 + 12 = 6x - 12 + 12$

$15 = 6x$

Divide both sides by $6$ :

$x = \frac{15}{6} = \frac{5}{2} = 2\frac{1}{2}$

Since $2\frac{1}{2} \neq 2$ , the solution is valid.

Check: $\text{LHS} = \frac{3}{2\frac{1}{2} - 2} = \frac{3}{\frac{1}{2}} = 3 \times 2 = 6 = \text{RHS} \quad \checkmark$

(a) $x = 3$ ; (b) $x = 2\frac{1}{2}$ .

Common Mistakes

Forgetting to multiply non-fraction terms by the LCM: In $\frac{x}{3} + 2 = 5$ , multiplying by $3$ gives $x + 6 = 15$ , not $x + 2 = 15$ . Every single term must be multiplied.
Sign errors when a minus sign stands before a numerator with two terms: In $\frac{x}{3} - \frac{x - 3}{4} = 1$ , multiplying every term by $12$ gives $4x - 3(x - 3) = 12$ , so $4x - 3x + 9 = 12$ . Forgetting the brackets gives $4x - 3x - 3 = 12$ , which is wrong.
Using cross-multiplication when there are extra terms: In $\frac{x}{2} + 1 = \frac{3}{4}$ , cross-multiplying directly is invalid because of the added $1$ . Clear fractions by LCM or combine terms first.
Ignoring division by zero: In $\frac{3}{x - 2} = 6$ , $x$ cannot equal $2$ , because then the denominator would be $0$ . Always state the restriction before solving, and check that the solution is not the excluded value.

Practice Questions

Work each question on paper using the LCM method or cross-multiplication, then check your solution.

Question 1

Solve $\frac{3x}{7} + \frac{x}{7} = 8$ .

ShowHide solution

The fractions have the same denominator, so combine them:

$\frac{4x}{7} = 8$

Multiply both sides by $7$ :

$\frac{4x}{7} \times 7 = 8 \times 7$

$4x = 56$

Divide both sides by $4$ :

$x = \frac{56}{4} = 14$

Check: $\text{LHS} = \frac{3(14)}{7} + \frac{14}{7} = 6 + 2 = 8 = \text{RHS} \quad \checkmark$

$x = 14$

Question 2

Solve $\frac{x}{2} - \frac{x}{5} = 3$ .

ShowHide solution

The LCM of $2$ and $5$ is $10$ . Multiply every term by $10$ :

$10\left(\frac{x}{2}\right) - 10\left(\frac{x}{5}\right) = 10(3)$

$5x - 2x = 30$

$3x = 30$

Divide both sides by $3$ :

$x = 10$

Check: $\text{LHS} = \frac{10}{2} - \frac{10}{5} = 5 - 2 = 3 = \text{RHS} \quad \checkmark$

$x = 10$

Question 3

Solve $\frac{3x - 1}{4} = 5$ .

ShowHide solution

Multiply both sides by $4$ :

$\frac{3x - 1}{4} \times 4 = 5 \times 4$

$3x - 1 = 20$

Add $1$ to both sides:

$3x - 1 + 1 = 20 + 1$

$3x = 21$

Divide both sides by $3$ :

$x = 7$

Check: $\text{LHS} = \frac{3(7) - 1}{4} = \frac{20}{4} = 5 = \text{RHS} \quad \checkmark$

$x = 7$

Question 4

Solve $\frac{x}{4} + 3 = \frac{x}{3}$ .

ShowHide solution

The LCM of $4$ and $3$ is $12$ . Multiply every term on both sides by $12$ , including the $3$ :

$12\left(\frac{x}{4}\right) + 12(3) = 12\left(\frac{x}{3}\right)$

$3x + 36 = 4x$

Subtract $3x$ from both sides:

$3x + 36 - 3x = 4x - 3x$

$36 = x$

Check: $\text{LHS} = \frac{36}{4} + 3 = 9 + 3 = 12$ ; $\text{RHS} = \frac{36}{3} = 12 = \text{LHS} \quad \checkmark$

$x = 36$

Question 5

Solve $\frac{2x - 1}{3} + 2 = \frac{x}{2}$ .

ShowHide solution

Insert brackets around the numerator with two terms: $\frac{(2x - 1)}{3} + 2 = \frac{x}{2}$ .

The LCM of $3$ and $2$ is $6$ . Multiply every term on both sides by $6$ :

$6\left(\frac{2x - 1}{3}\right) + 6(2) = 6\left(\frac{x}{2}\right)$

$2(2x - 1) + 12 = 3x$

Expand the bracket:

$4x - 2 + 12 = 3x$

$4x + 10 = 3x$

Subtract $3x$ from both sides:

$4x + 10 - 3x = 3x - 3x$

$x + 10 = 0$

Subtract $10$ from both sides:

$x = -10$

Check: $\text{LHS} = \frac{2(-10) - 1}{3} + 2 = \frac{-21}{3} + 2 = -7 + 2 = -5$ ; $\text{RHS} = \frac{-10}{2} = -5 = \text{LHS} \quad \checkmark$

$x = -10$

Question 6

Solve $\frac{x - 4}{3} = \frac{2x + 1}{5}$ .

ShowHide solution

Both sides are single fractions. Cross-multiply (multiply both sides by $15$ ):

$5(x - 4) = 3(2x + 1)$

Expand brackets on both sides:

$5x - 20 = 6x + 3$

The variable term with the smaller coefficient is $5x$ . Subtract $5x$ from both sides:

$5x - 20 - 5x = 6x + 3 - 5x$

$-20 = x + 3$

Subtract $3$ from both sides:

$-20 - 3 = x + 3 - 3$

$-23 = x$

Writing the unknown on the left:

$x = -23$

Check: $\text{LHS} = \frac{-23 - 4}{3} = \frac{-27}{3} = -9$ ; $\text{RHS} = \frac{2(-23) + 1}{5} = \frac{-45}{5} = -9 = \text{LHS} \quad \checkmark$

$x = -23$

Question 7

Solve $\frac{y - 2}{5} - \frac{y - 4}{2} = 3$ .

ShowHide solution

Insert brackets around numerators: $\frac{(y - 2)}{5} - \frac{(y - 4)}{2} = 3$ .

The LCM of $5$ and $2$ is $10$ . Multiply every term by $10$ :

$10\left(\frac{y - 2}{5}\right) - 10\left(\frac{y - 4}{2}\right) = 10(3)$

$2(y - 2) - 5(y - 4) = 30$

Expand brackets, taking care with $-5(y - 4) = -5y + 20$ :

$2y - 4 - 5y + 20 = 30$

Combine like terms on LHS ( $2y - 5y = -3y$ and $-4 + 20 = 16$ ):

$-3y + 16 = 30$

Subtract $16$ from both sides:

$-3y + 16 - 16 = 30 - 16$

$-3y = 14$

Divide both sides by $-3$ :

$y = -\frac{14}{3} = -4\frac{2}{3}$

Check: $y - 2 = -\frac{14}{3} - \frac{6}{3} = -\frac{20}{3}$ , and $-\frac{20}{3} \div 5 = -\frac{4}{3}$ .

$y - 4 = -\frac{14}{3} - \frac{12}{3} = -\frac{26}{3}$ , and $-\frac{26}{3} \div 2 = -\frac{13}{3}$ .

$\text{LHS} = -\frac{4}{3} - \left(-\frac{13}{3}\right) = \frac{9}{3} = 3 = \text{RHS} \quad \checkmark$

$y = -4\frac{2}{3}$ (or $-\frac{14}{3}$ )

Question 8

Solve $6 = \frac{12}{3t - 4}$ .

ShowHide solution

Restriction: $3t - 4 \neq 0$ , so $t \neq \frac{4}{3}$ .

Multiply both sides by $(3t - 4)$ :

$6(3t - 4) = \frac{12}{3t - 4} \times (3t - 4)$

$6(3t - 4) = 12$

Expand the bracket:

$18t - 24 = 12$

Add $24$ to both sides:

$18t - 24 + 24 = 12 + 24$

$18t = 36$

Divide both sides by $18$ :

$t = 2$

Since $2 \neq \frac{4}{3}$ , the solution is valid.

Check: $\text{LHS} = 6$ ; $\text{RHS} = \frac{12}{3(2) - 4} = \frac{12}{2} = 6 = \text{LHS} \quad \checkmark$

$t = 2$

Question 9

Solve $\frac{2p - 5}{p + 4} = \frac{5}{4}$ .

ShowHide solution

Restriction: $p + 4 \neq 0$ , so $p \neq -4$ .

Both sides are single fractions. Multiply both sides by $4(p + 4)$ , that is, cross-multiply:

$4(2p - 5) = 5(p + 4)$

Expand brackets on both sides:

$8p - 20 = 5p + 20$

Subtract $5p$ from both sides:

$8p - 20 - 5p = 5p + 20 - 5p$

$3p - 20 = 20$

Add $20$ to both sides:

$3p - 20 + 20 = 20 + 20$

$3p = 40$

Divide both sides by $3$ :

$p = \frac{40}{3} = 13\frac{1}{3}$

Since $13\frac{1}{3} \neq -4$ , the solution is valid.

Check: $2p - 5 = \frac{80}{3} - \frac{15}{3} = \frac{65}{3}$ and $p + 4 = \frac{40}{3} + \frac{12}{3} = \frac{52}{3}$ .

$\text{LHS} = \frac{65}{3} \div \frac{52}{3} = \frac{65}{52} = \frac{5}{4} = \text{RHS} \quad \checkmark$

$p = 13\frac{1}{3}$ (or $\frac{40}{3}$ )

Question 10 (Setting: Baking / Bread Making)

A baker uses $\frac{1}{3}$ of a sack of flour for baguettes and $\frac{1}{6}$ of the sack for croissants. If he uses a total of 12 kg of flour for both, form an equation in $w$ (the mass of the sack in kg) and find the mass of the full sack.

ShowHide solution

Let the mass of the full sack be $w$ kg.

Flour for baguettes: $\frac{w}{3}$ kg.

Flour for croissants: $\frac{w}{6}$ kg.

Total flour used: 12 kg.

Form the equation:

$\frac{w}{3} + \frac{w}{6} = 12$

The LCM of $3$ and $6$ is $6$ . Multiply every term by $6$ :

$6\left(\frac{w}{3}\right) + 6\left(\frac{w}{6}\right) = 6(12)$

$2w + w = 72$

$3w = 72$

Divide both sides by $3$ :

$w = \frac{72}{3} = 24$

Check: $\frac{24}{3} + \frac{24}{6} = 8 + 4 = 12$ kg $\quad \checkmark$

The equation is $\frac{w}{3} + \frac{w}{6} = 12$ ; the mass of the full sack is 24 kg.

Question 11 (Setting: Monthly Pocket Money)

A student spent $\frac{2}{5}$ of her monthly pocket money on MRT fares and $\frac{1}{4}$ of it on school stationery. If she spent a total of $39 on these two items, form an equation in $m$ (her monthly pocket money in dollars) and find $m$ .

ShowHide solution

Let her monthly pocket money be $m$ dollars.

Amount spent on MRT fares: $\frac{2}{5}m$ dollars.

Amount spent on stationery: $\frac{1}{4}m$ dollars.

Total spent: 39 dollars.

Form the equation:

$\frac{2}{5}m + \frac{1}{4}m = 39$

The LCM of $5$ and $4$ is $20$ . Multiply every term by $20$ :

$20\left(\frac{2}{5}m\right) + 20\left(\frac{1}{4}m\right) = 20(39)$

$8m + 5m = 780$

$13m = 780$

Divide both sides by $13$ :

$m = \frac{780}{13} = 60$

Check: MRT fares cost $\frac{2}{5}(60) = 24$ dollars; stationery cost $\frac{1}{4}(60) = 15$ dollars; total is $24 + 15 = 39$ dollars $\quad \checkmark$

The equation is $\frac{2}{5}m + \frac{1}{4}m = 39$ ; her monthly pocket money is $60.

Question 12 (Setting: Mr Tan's Age)

One quarter of Mr Tan's age 4 years ago, added to one sixth of his age 8 years from now, is 17. Let Mr Tan be $a$ years old now. Form an equation in $a$ and find his present age.

ShowHide solution

Mr Tan's age 4 years ago: $(a - 4)$ years. One quarter of it: $\frac{a - 4}{4}$ .

His age 8 years from now: $(a + 8)$ years. One sixth of it: $\frac{a + 8}{6}$ .

Form the equation:

$\frac{a - 4}{4} + \frac{a + 8}{6} = 17$

The LCM of $4$ and $6$ is $12$ . Multiply every term by $12$ :

$12\left(\frac{a - 4}{4}\right) + 12\left(\frac{a + 8}{6}\right) = 12(17)$

$3(a - 4) + 2(a + 8) = 204$

Expand brackets:

$3a - 12 + 2a + 16 = 204$

Combine like terms on LHS ( $3a + 2a = 5a$ and $-12 + 16 = 4$ ):

$5a + 4 = 204$

Subtract $4$ from both sides:

$5a + 4 - 4 = 204 - 4$

$5a = 200$

Divide both sides by $5$ :

$a = \frac{200}{5} = 40$

Check: 4 years ago he was 36, and $\frac{36}{4} = 9$ ; in 8 years he will be 48, and $\frac{48}{6} = 8$ ; $9 + 8 = 17$ $\quad \checkmark$

The equation is $\frac{a - 4}{4} + \frac{a + 8}{6} = 17$ ; Mr Tan is 40 years old now.

Key Points

To clear numerical denominators, multiply every term on both sides by the LCM of all denominators.
Place brackets around a numerator with more than one term before multiplying: $-\frac{x - 3}{4} = -\frac{(x - 3)}{4}$ .
For proportions $\frac{A}{B} = \frac{C}{D}$ , use cross-multiplication: $AD = BC$ .
When the unknown is in the denominator, state the restriction that the denominator cannot equal zero, and check that the solution is not the excluded value.
Always verify your answer by substituting back into the original fractional equation.