FORM 2 • TERM 1 • LESSON 10

In Form 1 you solved equations such as 3x + 5 = 20 in algebraic equations and word problems. This lesson takes the next steps: equations with the letter on both sides, equations with brackets, and word problems that you turn into equations yourself.

Learning outcomes

By the end of this lesson you should be able to:

• 2.6.3.1 – Tell the difference between an expression and an equation.
• 2.6.3.2 – Solve linear equations with the variable on both sides.
• 2.6.3.3 – Solve linear equations that need the distributive law, using different representations.
• 2.6.3.4 – Rewrite word statements as algebraic equations.
• 2.6.3.5 – Solve word problems using algebraic equations.

Expression or equation?

Expression	Equation
Has no equals sign	Has an equals sign
Can be simplified or evaluated	Can be solved to find the unknown
Example: 5x − 3	Example: 5x − 3 = 2x + 9

An equation says that two expressions have the same value. A linear equation has no x², x³ or other powers of the letter – just x on its own.

The balance method

Think of an equation as a balanced scale. Whatever you do to one side, you must do to the other side, so it stays balanced. The aim is to end with x on its own on one side.

Letters on both sides

1. Collect the letters on one side by subtracting the smaller letter term from both sides (this keeps the coefficient positive).
2. Collect the numbers on the other side by adding or subtracting.
3. Divide both sides by the coefficient of x.
4. Check by putting your answer back into both sides of the original equation.

For 5x − 3 = 2x + 9: subtract 2x from both sides to get 3x − 3 = 9; add 3 to get 3x = 12; divide by 3 to get x = 4. Check: 5 × 4 − 3 = 17 and 2 × 4 + 9 = 17. ✓

Equations with brackets

When an equation has brackets, use the distributive law to expand them first, then solve as before. For 4(2x − 1) = 3(x + 7): expand to get 8x − 4 = 3x + 21; subtract 3x to get 5x − 4 = 21; add 4 to get 5x = 25; so x = 5.

Sometimes you can divide first instead: in 3(x + 4) = 27, divide both sides by 3 to get x + 4 = 9, so x = 5. Both methods give the same answer.

From words to equations

1. Choose a letter for the unknown and say what it stands for.
2. Write the other quantities in terms of that letter.
3. Use the facts in the question to write an equation.
4. Solve it, then answer the question in words and check the answer makes sense.

Worked examples

Example 1: Letters on both sides

Question: Solve 2x + 11 = 5x − 4.

1. Subtract 2x from both sides: 11 = 3x − 4.
2. Add 4 to both sides: 15 = 3x.
3. Divide by 3: x = 5.
4. Check: 2 × 5 + 11 = 21 and 5 × 5 − 4 = 21. ✓

Answer: x = 5.

Example 2: A negative answer

Question: Solve 6x + 5 = 2x − 7.

1. Subtract 2x: 4x + 5 = −7.
2. Subtract 5: 4x = −12.
3. Divide by 4: x = −3.
4. Check: 6 × (−3) + 5 = −13 and 2 × (−3) − 7 = −13. ✓

Answer: x = −3.

Example 3: Brackets on both sides

Question: Solve 2(x − 3) = 5(x − 6).

1. Expand: 2x − 6 = 5x − 30.
2. Subtract 2x: −6 = 3x − 30.
3. Add 30: 24 = 3x, so x = 8.
4. Check: 2 × 5 = 10 and 5 × 2 = 10. ✓

Answer: x = 8.

Example 4: Think of a number

Question: I think of a number, multiply it by 4 and subtract 7. The answer is 21. What is my number?

1. Let n be the number. Then 4n − 7 = 21.
2. Add 7: 4n = 28.
3. Divide by 4: n = 7.

Answer: The number is 7.

Example 5: Ages

Question: Maria is three times as old as her brother. Their ages add up to 48. How old is each?

1. Let b be her brother’s age in years. Maria is 3b.
2. b + 3b = 48, so 4b = 48 and b = 12.
3. Maria is 3 × 12 = 36. Check: 12 + 36 = 48. ✓

Answer: Her brother is 12 and Maria is 36.

Example 6: Comparing two plans

Question: Phone plan A costs $50 plus $2 per GB. Plan B costs $20 plus $5 per GB. For how many GB do the plans cost the same?

1. Let g be the number of GB. Plan A: 50 + 2g. Plan B: 20 + 5g.
2. 50 + 2g = 20 + 5g. Subtract 2g: 50 = 20 + 3g.
3. Subtract 20: 30 = 3g, so g = 10.
4. Check: 50 + 20 = 70 and 20 + 50 = 70. ✓

Answer: 10 GB (both cost $70).

Example 7: A perimeter problem

Question: A rectangle is 5 cm longer than it is wide. Its perimeter is 38 cm. Find its length and width.

1. Let the width be w cm. The length is w + 5.
2. Perimeter: 2(w + w + 5) = 38, so 2(2w + 5) = 38.
3. Expand: 4w + 10 = 38, so 4w = 28 and w = 7.
4. Length = 7 + 5 = 12. Check: 2 × (7 + 12) = 38. ✓

Answer: Width 7 cm, length 12 cm.

Common mistakes

• Doing something to one side only. If you subtract 2x on the left, subtract 2x on the right too.
• Moving a term without changing its sign. From 3x − 3 = 9, adding 3 to both sides gives 3x = 12, not 3x = 6.
• Expanding only the first term in a bracket. 4(2x − 1) is 8x − 4, not 8x − 1.
• Dividing only part of a side. From 3x = 12 + 6, first add to get 18, then divide: x = 6.
• Stopping at the value of the letter. In a word problem, answer the question that was asked, in words and with units.

Practice questions

Try these yourself first, then open each answer to check your working.

Quick check

Previous: Simplifying Algebraic Expressions: Terms, Coefficients, Like Terms and Algebra Tiles

Next: Inequalities: Symbols, Real-Life Meaning and Writing Word Problems as Inequalities

Curriculum link: This post is relevant to the Trinidad and Tobago National Lower Secondary (NCSE) Mathematics Curriculum – Form 2, Term 1, Strand 2.6 Algebra, Topic 2.6.3 Solution of Linear Equations – Learning Outcomes 2.6.3.1 (expressions and equations), 2.6.3.2 (variables on both sides), 2.6.3.3 (using the distributive law), 2.6.3.4 (writing verbal statements as equations) and 2.6.3.5 (solving word problems).

Updates

2026/10/23

– First version of this lesson.

2026/10/25

– Put each lettered part of the questions and answers on its own line.

© 2026 Vedesh Kungebeharry. All rights reserved.