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JSS1: MATHEMATICS - 2ND TERM

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  1. Algebraic Processes | Week 1
    4 Topics
    |
    1 Quiz
  2. Simplification of Algebraic Expressions | Week 2
    4 Topics
    |
    1 Quiz
  3. Simplification of Algebraic Expressions 2 (Use of Brackets) | Week 3
    4 Topics
    |
    1 Quiz
  4. Simple Equations | Week 4
    1 Topic
    |
    1 Quiz
  5. Simple Equations II | Week 5
    3 Topics
    |
    1 Quiz
  6. Plane Shapes I | Week 6
    5 Topics
    |
    2 Quizzes
  7. Plane Shapes II | Week 7
    7 Topics
    |
    1 Quiz
  8. Plane Shapes III | Week 8
    7 Topics
    |
    1 Quiz
  9. Decimals and Percentages I | Week 9
    2 Topics
    |
    1 Quiz
  10. Decimals and Percentages II | Week 10
    3 Topics
    |
    1 Quiz
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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:

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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.

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Therefore we have:

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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

⇒ \( \frac{4x}{4} = \frac{2}{4} \\ \frac{\not{4}x}{\not{4}} = \frac{\not{2}}{\not{4^2}} \\ \scriptsize x = \normalsize \frac{1}{2} \scriptsize \; or \; 0.5\)