Mathematics — ALGEBRA

Subtopic: Linear Equations (for age 13, Kenya)

A linear equation is an equation in which the highest power of the unknown (variable) is 1. We usually write a linear equation in one variable as ax + b = c, where a, b and c are numbers and x is the variable we want to find.

Key steps to solve a linear equation
  1. Simplify both sides (remove brackets and collect like terms).
  2. Get all terms with the variable on one side and numbers on the other side.
  3. Use inverse operations (add/subtract, multiply/divide) to isolate the variable.
  4. Check by substituting the found value into the original equation.

Worked examples

Example 1: Solve 3x + 5 = 20.
Step 1: Subtract 5 from both sides: 3x + 5 − 5 = 20 − 5 → 3x = 15.
Step 2: Divide both sides by 3: x = 15 ÷ 3 → x = 5.
Check: 3×5 + 5 = 15 + 5 = 20 ✓
Example 2 (brackets): Solve 2(x − 3) = 10.
Step 1: Expand brackets: 2x − 6 = 10.
Step 2: Add 6 to both sides: 2x = 16.
Step 3: Divide by 2: x = 8.
Check: 2(8 − 3) = 2×5 = 10 ✓
Example 3 (fraction): Solve x/4 + 3 = 7.
Step 1: Subtract 3: x/4 = 4.
Step 2: Multiply both sides by 4: x = 16.
Check: 16/4 + 3 = 4 + 3 = 7 ✓
Example 4 (variables on both sides): Solve 5x − 2 = 2x + 10.
Step 1: Subtract 2x from both sides: 3x − 2 = 10.
Step 2: Add 2 to both sides: 3x = 12.
Step 3: Divide by 3: x = 4.
Check: 5×4 − 2 = 20 − 2 = 18 and 2×4 + 10 = 8 + 10 = 18 ✓

Simple visual: balance idea

Think of an equation like a balance (both sides must stay equal). Whatever you do to one side, do the same to the other side.

Left: 3x + 5
=
Right: 20

Common types of linear equations

  • ax + b = c (simple one-step or two-step equations)
  • Equations with brackets: a(bx + c) = d
  • Equations with fractions: x/number + ... = ...
  • Variables on both sides: ax + b = cx + d

Common mistakes and tips

  • Forgetting to do the same operation on both sides.
  • Not distributing negative signs correctly in brackets (for example, −(x − 2) = −x + 2).
  • When clearing fractions, multiply every term on both sides by the common denominator.
  • Always check your answer by substituting back into the original equation.

Word problem (Kenyan context)

Jane saves KSh x each week. After 6 weeks she has saved KSh 240. Form an equation and find x.
Equation: 6x = 240 → x = 240 ÷ 6 = 40. Jane saves KSh 40 each week.

Practice exercises

  1. Solve: 4x + 7 = 31.
  2. Solve: 7(x − 2) = 21.
  3. Solve: (x/3) + 5 = 9.
  4. Solve: 6x − 4 = 2x + 12.
  5. Solve: −3x + 9 = 0.
  6. Word problem: A matatu fare for two people is KSh 140. If one person pays KSh x and the other pays KSh (x + 10), find x.

Answers (work shown briefly)

  1. 4x + 7 = 31 → 4x = 24 → x = 6.
  2. 7(x − 2) = 21 → x − 2 = 3 → x = 5.
  3. x/3 + 5 = 9 → x/3 = 4 → x = 12.
  4. 6x − 4 = 2x + 12 → 4x = 16 → x = 4.
  5. −3x + 9 = 0 → −3x = −9 → x = 3.
  6. Let fares be x and x + 10. Then x + (x + 10) = 140 → 2x + 10 = 140 → 2x = 130 → x = 65. So one person paid KSh 65 and the other KSh 75.
Final tip: practise many different forms — with brackets, fractions and variables on both sides. Always write each step clearly and check your answer.

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