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Simple Equation – JSS1 Mathematics Lesson Note (Third Term)

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Lesson note at a glance

Subject
Mathematics
Class
JSS1
Term
Third Term
Topic
Simple Equation
Order
Topic 1 of 7 this term
Reading time
2 min

A simple equation is an algebraic sentence that says two expressions are equal, such as 3 × p = 18, which means three times an unknown number is equal to eighteen. To solve a linear equation with mixed operations, first remove the unwanted terms by adding or subtracting, then remove the coefficient of the unknown by multiplying or dividing.

Behavioural objectives

By the end of the lesson, students should be able to:

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  • translate simple equations into words;
  • state whether an equation is true or false for a given value of the unknown;
  • collect like terms in a linear equation;
  • solve linear equations with mixed operations.

What is a simple equation?

The expression 3 × p = 18 is an algebraic sentence. It means three times an unknown number is equal to eighteen.

How do you translate equations into words?

Translate the following equations into words.

  • 5x + 7 = 37 means if 7 is added to five times a number, the result is 37.
  • 11 = 3x – 1 means 11 is equal to 3 times a certain number minus one.

How do you tell whether an equation is true or false?

Put the given value of the unknown into the equation. If both sides are equal, the statement is true. If they are not equal, the statement is false.

Examples

State whether each of the following is true or false.

  1. x + 8 = 15 (when x is 7)
  2. 3x – 4 = 11 (when x is 6)

Solution to example 1

x + 8 = 15
7 + 8 = 15
Therefore the statement is true.

Solution to example 2

3x – 4 = 11
3 × 6 – 4 = 11
18 – 4 = 14
Since 14 is not equal to 11, the statement is false.

How do you solve a linear equation with mixed operations and like terms?

When there is more than one operation, follow these steps:

  1. Eliminate the unwanted terms by adding or subtracting.
  2. Eliminate the coefficient of the unknown by multiplying or dividing.

Example 1: 4y – 5 = 7

Solution

Add 5 to both sides.
4y – 5 + 5 = 7 + 5
4y = 12

Divide both sides by the coefficient of the unknown.
4y/4 = 12/4
y = 3

Example 2: 5x – 10 = 15x – 60

Solution

5x – 10 = 15x – 60

Collect the like terms.
5x – 15x = -60 + 10
-10x = -50

Divide both sides by -10 (the coefficient of the unknown).
-10x/-10 = -50/-10
x = 5

Evaluation

Solve the following equations.

  1. 6m + 2 = 20 + 5m
  2. 4c – 8 = 10 – 5c
  3. 5x + 5 = 35 + 2x
  4. 56 + 7y = 5y + 16

This term's Mathematics topics

JSS1 Third Term · 7 topics

  1. Simple Equation
  2. Geometry: Plane Shapes & Their Properties
  3. Area & Perimeter of Plane Shapes
  4. Three Dimensional Shapes
  5. Statistics: Mean, Median & Mode
  6. Identification & Properties of Angles
  7. Construction
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