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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. Decimal and Percentages I | Week 9
    2 Topics
    |
    1 Quiz
  10. Decimal and Percentages | Week 10
    3 Topics
    |
    1 Quiz



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An equation is basically saying that two things are EQUAL since in a balanced situation the two sides of the balance hold equal weight, we can model simple equations with a balance.

In the picture below, each circle represents, “one” (1), and the block represents the unknown “x“. To find out what the block weighs, you can do the following:

1. Add the same amount (circles or blocks) to BOTH sides.
2. Take away the same amount from BOTH sides.  
3. Multiply BOTH sides by the same amount.
4. Divide BOTH sides by the same amount.

That way both sides will maintain the balance or “equality”

Example 1:

The diagram shows a balanced lever. The objects on both sides of the fulcrum (triangle) are equal.

Represent the unknown with the letter x

lever

A block \( \scriptsize \boxed {?} \) Represents x which is unknown

A circle \( \Large\circ \) Represents 1 

∴  \( \scriptsize x \: + \: 3 = 5 \)

The lever will remain balanced if we removed 3 circles from each side.

lever 4

We took away three circles from BOTH sides to make the lever balanced.

lever 2

This reads as one block = two circles.

i.e.  x  = 2

If x = 2 this can also mean that a block is equal to two circles, which means they both have the same weight.

Therefore the diagram can also be drawn like this;

lever 3

Without the scale model, the solving process looks like this:

⇒ \( \scriptsize x \: + \: 3 = 5 \)

⇒ \( \scriptsize x \: + \: 3 \: – \: 3 = 5 \: – \: 3 \)

⇒ \( \scriptsize x = 2 \)

Example 2

lever 6

⇒ \( \scriptsize 3x \: + \: 2 = 2x \: + \: 6 \)

First, take away two blocks (Two x’s) from both sides.

lever 7

Then we have \( \scriptsize x \: + \: 2 = 6 \)

Take away two circles from both sides

The balance will stay balanced.

lever 8

i.e. x  =   4

Without the scale model, the solving process looks like this:

\( \scriptsize 3x \: + \: 2 = 2x \: + \: 6 \)

(Take away 2x from both sides)

\( \scriptsize 3x \: + \: 2 \; – \: 2x = 2x \: + \: 6 \; – \: 2x \)

Collect like terms

\( \scriptsize 3x \; – \: 2x \: + \: 2 = 2x \; – \: 2x \: + \: 6\)

\( \scriptsize x \: + \: 2 = 0 \: + \: 6\)

\( \scriptsize x \: + \: 2 = 6\)

Take away 2 from both sides

\( \scriptsize x \: + \: 2 \; – \: 2 = 6 \; – \: 2\)

\( \scriptsize x \: + \: 0 = 4\)

\( \scriptsize x = 4\)

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