Section A — Chapters 1–22

Algebra III: Problem-Solving with Linear Equations

Junior Cycle — 1st Year

  • By the end of this lesson students will be able to translate word problems into algebraic expressions.
  • By the end of this lesson students will be able to formulate linear equations from real-world scenarios.
  • By the end of this lesson students will be able to solve linear equations to find unknown values in word problems.
  • By the end of this lesson students will be able to interpret the solution of a linear equation in the context of the original problem.

Key concepts

Algebraic Expression

An algebraic expression is a mathematical phrase that can contain numbers, variables (letters representing unknown values), and operation symbols (like +, -, ×, ÷). It does not contain an equals sign. For example, 'x + 5' or '3y - 2'.

Linear Equation

A linear equation is a mathematical statement that shows two expressions are equal. It contains an equals sign and usually one or more variables. The highest power of any variable in a linear equation is always 1. For example, 'x + 5 = 12' or '2y - 3 = 7'.

Variable

A variable is a letter, such as x, y, or a, that is used to represent an unknown numerical value in an expression or equation. It allows us to write general mathematical statements.

Keywords for Operations

Certain words in a problem indicate specific mathematical operations. Understanding these keywords is crucial for translating word problems into algebraic form. For example:\n* Addition: sum, total, more than, increased by, plus.\n* Subtraction: difference, less than, decreased by, minus, take away.\n* Multiplication: product, times, multiplied by, 'of' (e.g., half of).\n* Division: quotient, divided by, shared equally.

Key facts to remember

  • 1A variable is a letter used to represent an unknown numerical value.
  • 2An algebraic expression is a mathematical phrase without an equals sign.
  • 3A linear equation is a mathematical statement showing two expressions are equal, with the highest power of the variable being 1.
  • 4To solve an equation, perform the same operation on both sides to maintain balance.
  • 5The primary goal when solving an equation is to isolate the variable.
  • 6Always define your variables clearly at the beginning of a problem.
  • 7Keywords in word problems (e.g., 'sum', 'difference', 'product') indicate specific mathematical operations.
  • 8Always check your solution by substituting it back into the original problem statement to ensure it makes sense.

Worked examples

Example 1

Write an algebraic expression for 'a number decreased by nine'.

ILet the unknown number be represented by the variable 'n'.
IIThe phrase 'decreased by nine' means we subtract 9 from the number.
IIISo, the algebraic expression is n - 9.

Answer

n - 9

Always define your variable clearly at the start.

Example 2

When a number is multiplied by 4 and then 7 is subtracted, the result is 21. Find the number.

ILet the unknown number be 'x'.
IIThe number multiplied by 4 is '4x'.
IIIThen 7 is subtracted: '4x - 7'.
IVThe result is 21, so we set up the equation: 4x - 7 = 21.
VTo solve for x, add 7 to both sides of the equation: 4x - 7 + 7 = 21 + 7.
VIThis simplifies to: 4x = 28.
VIIDivide both sides by 4: 4x / 4 = 28 / 4.
VIIITherefore, x = 7.

Answer

The number is 7.

Check your answer: (4 × 7) - 7 = 28 - 7 = 21. The answer is correct.

Example 3

The perimeter of a rectangular garden is 40 metres. The length of the garden is 6 metres more than its width. Find the dimensions (length and width) of the garden.

ILet the width of the garden be 'w' metres.
IIThe length is 6 metres more than the width, so the length is '(w + 6)' metres.
IIIThe formula for the perimeter of a rectangle is P = 2(length + width).
IVSubstitute the given perimeter and our expressions for length and width into the formula: 40 = 2((w + 6) + w).
VSimplify the expression inside the brackets: 40 = 2(2w + 6).
VIDistribute the 2 on the right side: 40 = 4w + 12.
VIISubtract 12 from both sides of the equation: 40 - 12 = 4w + 12 - 12.
VIIIThis simplifies to: 28 = 4w.
9Divide both sides by 4: 28 / 4 = 4w / 4.
10Therefore, w = 7.
11The width of the garden is 7 metres.
12Now find the length: Length = w + 6 = 7 + 6 = 13 metres.

Answer

The width of the garden is 7 metres and the length is 13 metres.

Remember to answer the question fully by finding both the length and the width. Check: Perimeter = 2(13 + 7) = 2(20) = 40 metres. Correct.

Common mistakes

  • Not defining variables: Students often jump straight to an equation without stating what their variable represents.
  • Confusing expressions with equations: Writing 'x + 5' when an equation 'x + 5 = 12' is required.
  • Incorrectly translating keywords: Forgetting the order for 'less than' (e.g., '5 less than x' is x - 5, not 5 - x).
  • Errors in balancing equations: Performing an operation on one side of the equals sign but not the other.
  • Not answering the full question: Solving for 'x' but forgetting to calculate the actual quantity asked for in the problem.

Exam tips

  • Read the problem carefully multiple times to fully understand what is being asked.
  • Underline or highlight key information and keywords that indicate operations or relationships.
  • Draw a simple diagram if it helps visualise the problem, especially for geometry-related questions.
  • Show all your working steps clearly and logically; marks are often awarded for method even if the final answer is incorrect.
  • Always write down what your chosen variable represents (e.g., 'Let x be the number of apples').
  • Check your final answer by substituting it back into the original word problem to ensure it is logical and satisfies all conditions.

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