JSS1: MATHEMATICS - 2ND TERM
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Algebraic Processes | Week 14 Topics|1 Quiz
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Simplification of Algebraic Expressions | Week 24 Topics|1 Quiz
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Simplification of Algebraic Expressions 2 (Use of Brackets) | Week 34 Topics|1 Quiz
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Simple Equations | Week 41 Topic|1 Quiz
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Simple Equations II | Week 53 Topics|1 Quiz
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Plane Shapes I | Week 65 Topics|2 Quizzes
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Plane Shapes II | Week 77 Topics|1 Quiz
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Plane Shapes III | Week 87 Topics|1 Quiz
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Decimals and Percentages I | Week 92 Topics|1 Quiz
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Decimals and Percentages II | Week 103 Topics|1 Quiz
Dividing
Topic Content:
- Dividing
In some situations, you have to divide both sides of the equation by the same number. When is that? It’s in the situation where there are only x’s (blocks) on one side and there is more than one circle on the other side.
Example:

Let \( \scriptsize \boxed {?} = x \\ \scriptsize \; and \; \LARGE \circ \scriptsize (circle) = 1\\ \scriptsize \therefore x \: + \: x = 8 \\ \scriptsize \rightarrow 2x = 8 \)
Note: Because there are 8 circles and 2 boxes
If you take away half of the things on the left side and similarly half of the things on the right side, the balance will stay balanced.

Therefore we have:

Therefore x = 4
Note: Because there are 4 circles and 1 box
Solving the equation without a scale model;
2x = 8
Divide both sides by 2
⇒ \( \frac{2x}{2} = \frac{8}{2} \\ \frac{\not{2}x}{\not{2}} = \frac{\not{8}^4}{\not{2}} \\ \scriptsize x = 4\)
Note: The scale model is a good example of solving and explaining simple equations, but you can solve simple equations using the balancing method without using the scale model.
Example 5.2.1:
Find the value of x from the following equations using the balancing method.
i. 2x + 3 = 5
ii. 2x + 5 = x + 4
iii. x + 7 = 15
iv. 16 = x – 6
v. 6x + 3 = 2x + 5
Solution
i) 2x + 3 = 5
Subtract 3 from both sides
2x + 3 – 3 = 5 – 3
2x = 2
Divide both sides by 2
⇒ \( \frac{2x}{2} = \frac{2}{2} \\ \frac{\not{2}x}{\not{2}} = \frac{\not{2}^1}{\not{2}} \\ \scriptsize x = 1\)
Note: that the equation remains balanced if the operation carried out on the Left-Hand Side (LHS) is the same operation carried out on the Right-Hand Side (RHS).
ii) 2x + 5 = x + 9
Subtract x from both sides
2x + 5 – x = x + 9 – x
Collect like terms
2x – x + 5 = x – x + 9
2x – x + 5 = 0 + 9
x + 5 = 9
Subtract 5 from both sides
x + 5 – 5 = 9 – 5
x = 4
iii) x + 7 = 15
Subtract 7 from both sides
x + 7 – 7 = 15 – 7
x = 8
iv) 16 = x – 6
Add 6 to both sides
16 + 6 = x – 6 + 6
22 = x
x = 22
v) 6x + 3 = 2x + 5
Subtract 2x from both sides
6x + 3 – 2x = 2x + 5 – 2x
Collect like terms
6x – 2x + 3 = 2x – 2x + 5
4x + 3 = 5
Subtract 3 from both sides
4x + 3 – 3 = 5 – 3
4x = 2
Divide both sides by 4
⇒ \( \frac{4x}{4} = \frac{2}{4} \\ \frac{\not{4}x}{\not{4}} = \frac{\not{2}}{\not{4^2}} \\ \scriptsize x = \normalsize \frac{1}{2} \scriptsize \; or \; 0.5\)