Before we begin: Solve $\frac{x}{4} = 5$ without a calculator. Then solve $\frac{x}{4} + 3 = 5$. How are the two problems different? What extra step did you need?
Equations with fractions look intimidating, but the strategy is simple: eliminate the fractions first. Multiply every term by the lowest common denominator (LCD). This turns a fraction equation into a regular linear equation that you already know how to solve.
$\frac{x}{3} + 2 = 7$: multiply every term by 3. $x + 6 = 21$.
- Find the lowest common denominator (LCD) of algebraic fractions
- Clear fractions by multiplying every term by the LCD
- Use cross-multiplication for single fractions on each side
- Solve equations with algebraic fractions in the denominator
- Identify and state restrictions on the variable
Wrong: Only multiplying one term by the LCD. In $\frac{x}{3} + 2 = 7$, multiplying only $\frac{x}{3}$ by 3 gives $x + 2 = 7$.
Right: Multiply every term on both sides by the LCD: $x + 6 = 21$.
Wrong: Using cross-multiplication when there are multiple terms on one side, such as $\frac{x}{2} + 3 = \frac{x}{4}$.
Right: Cross-multiplication only works for $\frac{a}{b} = \frac{c}{d}$ (one fraction each side). For multiple terms, use the LCD method.
To solve an equation with fractions, find the lowest common denominator of all fractions, then multiply every term on both sides by that number. This clears all denominators in one step and leaves a simple linear equation.
$\frac{x}{3} + \frac{x}{6} = 5$: LCD is 6. Multiply every term by 6. $2x + x = 30$.
When there is exactly one fraction on each side, cross-multiplication is a fast shortcut. Multiply the numerator of the left fraction by the denominator of the right, and set it equal to the denominator of the left times the numerator of the right.
$\frac{3}{x} = \frac{2}{5}$: cross-multiply to get $3 \times 5 = 2 \times x$.
When the variable appears in the denominator, some values are forbidden. A denominator can never equal zero. Identify these restrictions before you solve, and check your final answer does not violate them.
$\frac{5}{x-2} = 3$: restriction is $x \neq 2$. Then solve.
Not every fraction equation needs the same treatment. Look at the structure first, then choose the most efficient method. Using the wrong method wastes time and invites errors.
One fraction each side → cross-multiply. Multiple terms → use LCD. Variable in denominator → state restriction first.
Watch Me Solve It · 3 examples
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1Identify the LCDDenominators are 3 and 6. LCD = 6.6 is the smallest number that both 3 and 6 divide into evenly.
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2Multiply every term by 6$6 \times \frac{x}{3} + 6 \times \frac{x}{6} = 6 \times 5$Every term on both sides must be multiplied, including the 5.
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3Simplify each term$2x + x = 30$6 divided by 3 is 2, so the first term becomes 2x. 6 divided by 6 is 1.
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4Collect and solve$3x = 30$ so $x = 10$Check: $\frac{10}{3} + \frac{10}{6} = \frac{20}{6} + \frac{10}{6} = \frac{30}{6} = 5$. Correct.
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1Check the form and state restrictionsOne fraction each side. Restriction: $x \neq 0$.x cannot be 0 because it would make the left denominator zero.
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2Cross-multiply$3 \times 5 = 2 \times x$Multiply numerator left by denominator right, and denominator left by numerator right.
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3Solve and check$15 = 2x$ so $x = 7.5$Check: $\frac{3}{7.5} = 0.4$ and $\frac{2}{5} = 0.4$. Both match. x = 7.5 is not restricted.
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1State the restriction$x - 2 \neq 0$ so $x \neq 2$The denominator cannot equal zero, so x cannot be 2.
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2Multiply both sides by (x - 2)$5 = 3(x - 2)$This eliminates the fraction and leaves a simple linear equation.
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3Expand and solve$5 = 3x - 6$ → $11 = 3x$ → $x = \frac{11}{3}$Add 6 to both sides, then divide by 3.
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4Check against restriction and verify$x = \frac{11}{3} \approx 3.67$, which is not 2. Valid.Check: $\frac{5}{11/3 - 2} = \frac{5}{5/3} = 5 \times \frac{3}{5} = 3$. Correct.
Brain Trainer · 4 problems
Four equations covering LCD, cross-multiplication and algebraic fractions. Show each step, then reveal the answer to check.
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1 Solve $\frac{x}{4} + 2 = 7$.
Subtract 2: $\frac{x}{4} = 5$. Multiply by 4: $x = 20$. Check: $\frac{20}{4} + 2 = 7$. -
2 Solve $\frac{2x}{5} = 8$.
Multiply by 5: $2x = 40$. Divide by 2: $x = 20$. Check: $\frac{2(20)}{5} = \frac{40}{5} = 8$. -
3 Solve $\frac{x}{2} + \frac{x}{3} = 10$.
LCD = 6. Multiply all terms by 6: $3x + 2x = 60$. $5x = 60$. $x = 12$. Check: $\frac{12}{2} + \frac{12}{3} = 6 + 4 = 10$. -
4 Solve $\frac{3}{x+1} = 1$.
Restriction: $x \neq -1$. Multiply by (x+1): $3 = x + 1$. $x = 2$. Check: $\frac{3}{2+1} = 1$. Valid.
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Solve $\frac{2x}{3} - \frac{x}{4} = 5$. Show all steps, including finding the LCD and checking your answer.
Q7. Solve $\frac{5}{2x-1} = 3$. Identify any restrictions on $x$, then show all working and verify your solution.
Q8. A recipe requires $\frac{2}{3}$ of a cup of sugar per batch. Jordan has $5\frac{1}{2}$ cups of sugar and wants to make as many full batches as possible.
(a) Write an equation to find the number of batches $n$. (1 mark)
(b) Solve the equation and state how many full batches Jordan can make. (1 mark)
(c) Explain why the answer must be a whole number and calculate how much sugar will be left over. (1 mark)
Quick Check
1. B Multiples of 4: 4, 8, 12, 16... Multiples of 6: 6, 12, 18... LCD = 12.
2. C Subtract 3: $\frac{x}{5} = 4$. Multiply by 5: $x = 20$.
3. B Cross-multiply: $3 \times 5 = 2x$, so $15 = 2x$, giving $x = 7.5$.
4. A Multiple terms on each side means LCD method. LCD of 2 and 3 is 6.
5. B Denominator is (x - 3). Setting x - 3 = 0 gives x = 3, which is not allowed.
Show Your Working Model Answers
Q6 (3 marks): LCD of 3 and 4 is 12 [0.5]. Multiply every term by 12: $12 \times \frac{2x}{3} - 12 \times \frac{x}{4} = 12 \times 5$ [0.5]. Simplify: $8x - 3x = 60$ [0.5]. $5x = 60$ [0.5]. $x = 12$ [0.5]. Check: $\frac{24}{3} - \frac{12}{4} = 8 - 3 = 5$ [0.5].
Q7 (3 marks): Restriction: $2x - 1 \neq 0$, so $x \neq \frac{1}{2}$ [0.5]. Multiply by $(2x - 1)$: $5 = 3(2x - 1)$ [0.5]. Expand: $5 = 6x - 3$ [0.5]. Add 3: $8 = 6x$ [0.5]. Divide by 6: $x = \frac{4}{3}$ [0.5]. Check: $\frac{5}{8/3 - 1} = \frac{5}{5/3} = 3$ [0.5].
Q8 (3 marks): (a) $\frac{2}{3}n = \frac{11}{2}$ (since $5\frac{1}{2} = \frac{11}{2}$) [1]. (b) Multiply by 6: $4n = 33$, so $n = 8.25$. Jordan can make 8 full batches [1]. (c) The answer must be a whole number because you cannot make a fraction of a batch. Leftover sugar: $8 \times \frac{2}{3} = \frac{16}{3}$ cups used. $\frac{11}{2} - \frac{16}{3} = \frac{33}{6} - \frac{32}{6} = \frac{1}{6}$ cup left over [1].
The Triple Fraction
Solve $\frac{x+1}{2} - \frac{x-1}{3} = \frac{x+2}{4}$. Hint: find the LCD of all three denominators, multiply every term by it, then solve the resulting linear equation. Show every step.
Reveal solution
LCD of 2, 3 and 4 is 12.
Multiply every term by 12: $12 \times \frac{x+1}{2} - 12 \times \frac{x-1}{3} = 12 \times \frac{x+2}{4}$.
Simplify: $6(x+1) - 4(x-1) = 3(x+2)$.
Expand: $6x + 6 - 4x + 4 = 3x + 6$.
Collect: $2x + 10 = 3x + 6$.
Subtract 2x: $10 = x + 6$.
Subtract 6: $x = 4$.
Check: LHS = $\frac{5}{2} - \frac{3}{3} = 2.5 - 1 = 1.5$. RHS = $\frac{6}{4} = 1.5$. Both match.
Find the LCD
The smallest number all denominators divide into
Multiply Every Term
Do not skip constants when clearing fractions
Cross-Multiply
Use only when there is one fraction on each side
State Restrictions
Denominator cannot equal zero
One Step Per Line
Clear working earns method marks
Always Check
Substitute back into the original equation
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