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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:

  • Substituting Numbers

The value of an expression such as x + 8  can be found by replacing x with a given number. This is called substitution. For example:  

For the expression x + 8 

If  x = 8,  then  8 + 8  = 16

If x = 10, then 10 + 8  =  18

Example 1.3.1:

Find the value of each of the following mentally when x = 3

i. \( \scriptsize x + 10 \)
ii. \( \scriptsize x\: – \: 1 \)
iii. \( \scriptsize x \: \times \: 8 \)
iv. \( \scriptsize x \: \div \: 9 \)
v. \( \scriptsize x \: \times \: 3 + 1\)

Solution

i. \( \scriptsize x + 10 \\ \rightarrow \scriptsize 3 + 10 \\ = \scriptsize 13 \)

ii. \( \scriptsize x\: – \: 1 \\ \rightarrow \scriptsize 3\: – \: 1 \\ = \scriptsize 2 \)

iii. \( \scriptsize x \: \times \: 8 \\ \rightarrow \scriptsize 3 \: \times \: 8 \\ = \scriptsize 24 \)

iv. \( \scriptsize x \: \div \: 9 \\ \rightarrow \scriptsize 3 \: \div\: 9 \\ = \frac{3}{9} \\ = \frac{1}{3} \)

v. \( \scriptsize x \: \times \: 3 + 1 \\ \rightarrow \scriptsize 3 \: \times \: 3 + 1 \\ = \scriptsize 10 \)

Example 1.3.2:

Find the value of each of the following mentally when y = 20

i. \( \scriptsize y + y + 2 \)
ii. \( \scriptsize (y \: – \: y ) \: + \: 5 \)
iii. \( \scriptsize (y \: – \: 10) \)
iv. \( \scriptsize y \: \div \: y \: + \: 25 \)
v. \( \scriptsize y \: – \: y \: + \: 8 \)

i. \( \scriptsize y + y + 2 \\ \rightarrow \scriptsize 20 + 20 + 2 \\ = \scriptsize 42 \)

ii. \( \scriptsize (y \: – \: y ) \: + \: 5 \\ \rightarrow \scriptsize (20 \: – \: 20) \: + \: 5 \\ \scriptsize = 0 \: + \: 5 \\ \scriptsize = 5 \)

iii. \( \scriptsize (y \: – \: 10) \: \times \: 8 \\ \rightarrow \scriptsize (20 \: – \: 10) \: \times \: 8 \\ = \scriptsize 10 \: \times \: 8 \\ \scriptsize = 80 \)

iv. \( \scriptsize y \: \div \: y \: + \: 25 \\ \rightarrow \scriptsize 20 \: \div \: 20 \: + \: 25 \\ = \scriptsize 1 + 25 \\= \scriptsize 26 \)

v. \( \scriptsize y \: – \: y \: + \: 8 \\ \rightarrow \scriptsize 20 \: – \: 20 \: + \: 8 \\ \scriptsize = 20 \: – \: 28 \\ \scriptsize = \; – 8 \)

Remember BODMAS.

B = Brackets \( \scriptsize \left( \right)\)

O = Orders \( \scriptsize x^2 \: \: \sqrt{x} \)

D = Division \( \scriptsize \div \)

M = Multiplication \( \scriptsize \times \)

A = Addition \( \scriptsize + \)

S = Subtraction \( \scriptsize – \)

For question ii, we solved the brackets () first

For question iii, we solved the brackets () first

For question iv, we divided first before adding

For question v, we solved 20 + 8 first which is equal to 28. We then solved the subtraction 20 – 28 = -8.

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