Subject Tutorial & Difficult Topics

Understanding and Solving Linear Equations: Step-by-Step ECZ Maths Tutorial

ECZ Solutions Editorial Team  |  March 1, 2026  |  13 Min Read
Tutorial Modules

1. What is a Linear Equation?

A linear equation is a mathematical statement where an unknown quantity (represented by a letter like \(x\), \(y\), or \(n\)) has an exponent/power of 1. When graphed on a coordinate plane, a linear equation forms a straight line.

Examples of linear equations:

  • \(2x + 5 = 15\)
  • \(3(y - 2) = 12\)
  • \(\frac{x}{4} + 1 = 6\)

2. The Golden Rule of Equation Balancing

Think of an equation as a balanced seesaw or scale. Whatever operation (addition, subtraction, multiplication, division) you perform on the left side of the equals sign, you MUST perform on the right side.

  • To undo Addition \((+)\), perform Subtraction \((-)\).
  • To undo Subtraction \((-)\), perform Addition \((+)\).
  • To undo Multiplication \((\times)\), perform Division \((\div)\).
  • To undo Division \((\div)\), perform Multiplication \((\times)\).

3. One-Step and Two-Step Linear Equations

Worked Example 1 (Two-Step Equation):
Solve for \(x\): \(3x + 7 = 22\)

Step 1: Subtract 7 from both sides to isolate the term with \(x\):
\(3x + 7 - 7 = 22 - 7\)
\(3x = 15\)

Step 2: Divide both sides by 3:
\(\frac{3x}{3} = \frac{15}{3}\)
\(x = 5\).

4. Solving Equations with Brackets and Fractions

When an equation contains brackets or fractions, expand the brackets or clear the denominators first:

Worked Example 2 (Equation with Fractions):
Solve for \(x\): \(\frac{2x + 1}{3} = 5\)

Step 1: Multiply both sides by the denominator (3):
\(3 \times \left(\frac{2x + 1}{3}\right) = 5 \times 3\)
\(2x + 1 = 15\)

Step 2: Subtract 1 from both sides:
\(2x = 14\)

Step 3: Divide by 2:
\(x = 7\).

5. Translating Word Problems into Linear Equations

ECZ examination papers frequently feature word problems testing your ability to write equations from real-life scenarios:

Word Problem Example:
Question: Mwamba bought 4 exercise books and a pen costing K10. The total cost was K50. What is the cost of one exercise book?

Solution:
1. Let the cost of one exercise book be \(b\).
2. Write the equation: \(4b + 10 = 50\)
3. Subtract 10: \(4b = 40\)
4. Divide by 4: \(b = 10\).
Answer: One exercise book costs K10.

6. Practice Questions & Step-by-Step Answers

Try solving these practice questions on paper before checking the answers below:

Practice Test Questions:
  1. Solve \(5x - 4 = 21\)
  2. Solve \(4(y + 2) = 28\)
  3. Solve \(\frac{x - 3}{2} = 8\)
Solutions:
  1. \(5x = 25 \implies x = 5\)
  2. \(4y + 8 = 28 \implies 4y = 20 \implies y = 5\)
  3. \(x - 3 = 16 \implies x = 19\)