Practice Maths

Equations with Fractions

Key Ideas

Key Terms

Simple fraction equations
Multiply both sides by the denominator to clear the fraction.
Multiple fractions
Find the LCD of all denominators, then multiply every term by the LCD.
Cross-multiplication
If a/b = c/d, then ad = bc (one fraction each side only).
Always check your solution
By substituting back into the original equation.
Fraction Equation Methods
Equation formMethod
x/3 = 5Multiply both sides by 3
x/2 + x/3 = 10Find LCD (6), multiply every term by 6
(x+1)/3 = (2x−1)/5Cross-multiply: 5(x+1) = 3(2x−1)
(x+1)/2 + (x+3)/4 = 7Find LCD (4), multiply every term by 4
Hot Tip When multiplying by the LCD, make sure you multiply every term on both sides, not just the fraction terms. Forgetting to multiply whole-number terms is the most common error.

Worked Example — LCD Method

Question: Solve x/3 + x/4 = 7.

Step 1 — Find the LCD of 3 and 4.
LCD = 12

Step 2 — Multiply every term by 12.
12 × x/3 + 12 × x/4 = 12 × 7
4x + 3x = 84

Step 3 — Solve.
7x = 84 → x = 12

Check: 12/3 + 12/4 = 4 + 3 = 7 ✓

Why Fractions Make Equations Harder

Equations containing fractions can look intimidating, but there is a reliable strategy: clear the fractions first by multiplying every term on both sides by the lowest common multiple (LCM) of all the denominators. This turns the equation into a regular linear equation with no fractions.

Clearing Fractions: The Method

To solve x/3 + x/4 = 7: the denominators are 3 and 4. LCM = 12. Multiply every term by 12:

12 × x/3 + 12 × x/4 = 12 × 7 → 4x + 3x = 84 → 7x = 84 → x = 12.

Check: 12/3 + 12/4 = 4 + 3 = 7. Correct.

Another example: (x + 1)/2 − (x − 3)/5 = 1. LCM of 2 and 5 is 10. Multiply every term by 10: 5(x + 1) − 2(x − 3) = 10 → 5x + 5 − 2x + 6 = 10 → 3x + 11 = 10 → 3x = −1 → x = −1/3.

Equations with a Variable in the Denominator

When the variable appears in the denominator (like 6/x = 3), multiply both sides by x to clear it: 6 = 3x → x = 2.

Always check that your solution does not make any denominator equal to zero — this would make the original equation undefined. If x = 0 makes a denominator zero, then x = 0 is not a valid solution.

Finding the LCM of Denominators

The LCM is the smallest number divisible by all denominators. For simple denominators, list multiples until you find a common one. For denominators 6 and 9: multiples of 6 are 6, 12, 18...; multiples of 9 are 9, 18...; LCM = 18.

If one denominator is a multiple of another (like 3 and 6), the LCM is just the larger one (6). This makes clearing fractions easier.

Most common mistake: Multiplying only some terms by the LCM instead of every single term. In x/3 + 2 = x/4, you must multiply all three terms by 12: 4x + 24 = 3x. A common error is to write 4x + 2 = 3x (not multiplying the 2 by 12). The rule is: multiply every term on both sides by the LCM — no exceptions.

Mastery Practice

  1. Solve each simple fraction equation. Fluency

     EquationSolution
    (a)x/2 = 5 
    (b)x/3 + 1 = 4 
    (c)y/4 = −5 
    (d)(a + 1)/2 = 6 
    (e)(m − 3)/5 = 4 
    (f)(2x + 1)/3 = 5 
    (g)n/4 − 3 = 1 
    (h)(4k − 3)/5 = 3 

  2. Multiply through by the LCD and solve. Fluency

     EquationSolution
    (a)x/2 + x/3 = 10 
    (b)y/4 − y/6 = 2 
    (c)a/3 + 2 = a/2 
    (d)m/5 + m/4 = 9 
    (e)x/2 + 3 = x/4 + 7 
    (f)2x/3 − x/4 = 5 
    (g)(x + 1)/2 + (x + 3)/3 = 5 
    (h)(2m − 1)/3 − (m + 2)/6 = 1 

  3. Solve equations with binomial numerators using cross-multiplication or LCD. Fluency

     EquationSolution
    (a)(x + 1)/2 = 3 
    (b)(2x − 3)/5 = 1 
    (c)(x − 2)/3 = (x + 4)/5 
    (d)(3x + 1)/4 = (x + 3)/2 
    (e)(2x + 3)/4 = (x − 1)/2 
    (f)(x + 5)/3 = (2x − 1)/4 
    (g)(4x − 1)/3 = (2x + 5)/2 
    (h)(x − 4)/6 = (x + 2)/9 

  4. True or False? Check each claim by substituting the given value into the equation. Fluency

    1. x = 10 is a solution of x/2 + x/3 = 10
    2. x = 5 is a solution of x/3 + x/6 = 5
    3. x = 11 is a solution of (x − 2)/3 = (x + 4)/5
    4. x = 4.5 is a solution of (4x − 3)/5 = 3
    5. x = 1/2 is a solution of 3/x = 6
    6. The equation (2x + 3)/4 = (x − 1)/2 has no solution.

  5. Solve and verify by substitution. Understanding

    Check your work. For each equation below, solve it fully and then substitute your answer back into both sides to verify.
    1. Solve 5/(x + 1) = 2. Show each step and verify.
    2. Solve 3/x = 6. (Hint: rewrite as 3 = 6x.) Verify your answer.

  6. Forming fraction equations. Understanding

    Number puzzles. Translate each sentence into a fraction equation and solve.
    1. One-third of a number increased by 5 equals half the number. Find the number.
    2. When a number is divided by 4 and then 2 is added, the result equals the number divided by 2 minus 1. Find the number.

  7. Shared amounts. Understanding

    Savings. Aidan has saved x/3 dollars and Sophie has saved x/4 + 10 dollars. They have saved the same amount.
    1. Write an equation and find x.
    2. How much has each person saved?

  8. Identify and fix the error. Understanding

    Spot the mistake. A student solves x/3 + x/6 = 5 and gets x = 5. Their working is: “LCD = 6. x/3 + x/6 = 5 → x + x = 5 → 2x = 5 → x = 2.5.”
    1. Substitute x = 5 into the left-hand side to check if it is correct.
    2. Identify the student’s error in their working.
    3. Solve the equation correctly.

  9. Pipes filling a tank. Problem Solving

    Water tank. Pipe A fills 1/3 of a tank per hour. Pipe B fills 1/4 of the tank per hour. Both pipes run together for t hours until the tank is full.
    1. Write an equation in terms of t for when the tank is exactly full.
    2. Solve for t.
    3. Express your answer as hours and minutes.

  10. Scaled recipe. Problem Solving

    Baking. A recipe uses three-quarters of a cup of sugar for a standard batch. A baker scales up the recipe so that three-quarters of the scaled batch equals 9 cups. Let b represent the number of cups in the full scaled batch.
    1. Write an equation for this situation.
    2. Solve for b.
    3. How many times larger is the scaled batch than the standard one-cup batch?
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