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ACI2606 Assignment THREE 2024
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UNISA ACI2606 Teaching Patterns, Functions and Algebra in the Intermediate Phase Mathematics Assignment THREE of 2024 solutions.
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Question 11.1 The equal sign symbol, =, is basically used to indicate that two quantities have the
same numerical value in Mathematics. Primary school Mathematics teachers need to be
aware of and alert to how/where the equal sign will be used in higher grades.
The equal sign in the Senior phase is not only used to indicate that expressions with
constants are the same, but is also used in equations with unknowns. In the
Intermediate phase it may be used in the following
10 − 4 + 2 = 8 where there are no variables. In the Senior phase, variables are
2
introduced for instance in 𝑥 − 16 = 0. On the previous equation the equal sign means
that the equality (sameness) only holds for certain values of 𝑥. It must not be confused
with meaning always the same for any 𝑥 value.
If we have a situation in which the equality always holds, we use equivalent to, with
symbol ≡. An example of where an equivalence is used is
2 2
(𝑥 + 3) ≡ 𝑥 + 6𝑥 + 9. The previous relation is always true no matter the value of 𝑥.
The equal sign must be introduced properly to avoid the misconception of it being used
at the beginning of a step instead of using therefore or implies that.
Example of = sign inappropriate use on an inequality:
2𝑥 + 4 > 4(𝑥 + 3)
= 2𝑥 + 4 > 4𝑥 + 12
= 2𝑥 − 4𝑥 > 12 − 4 and so on.
On the above steps, it is inappropriate to use the = sign on successive steps. An implies
that sign or therefore should have been used.
1.2 𝑥, 2𝑥, 𝑥 + 12, 𝑥 − 3 Assume that all amounts of money are given in rand (𝑅).
1.2.1 For the total amount of money, we will add the four expressions:
Total amount of money
= (𝑥) + (2𝑥) + (𝑥 + 12) + ( 𝑥 − 3)
= 𝑥 + 2𝑥 + 𝑥 + 12 + 𝑥 − 3
We will now collect like terms by adding or subtracting
= 𝑥 + 2𝑥 + 𝑥 + 𝑥 + 12 − 3
= 5𝑥 + 9
, 1.2.2 This be the total amount of money available divided by the number of people
sharing:
5𝑥+9
4
1.2.3 This will be the total money available in 1.2.1 minus the amount spent.
If four people spend 𝑅15 each then 𝑅15 × 4 = 𝑅60 was spent.
Amount remaining
= 5𝑥 + 9 − 60
= 5𝑥 − 51
1.2.4 The total amount in 1.2.1 is equal to 119
5𝑥 + 9 = 119
We now solve for 𝑥 by isolating it:
5𝑥 = 119 − 9
5𝑥 = 110
110
𝑥= 5
𝑥 = 22
The first person’s share was 𝑥, to substitute 𝑥 = 22 on the previous expression.
This person had 𝑅22.
The second person’s share was 2𝑥, to substitute 𝑥 = 22 on the previous expression.
2 × 22 = 44
This person had 𝑅44.
The third person’s share was 𝑥 + 12, to substitute 𝑥 = 22 on the previous expression.
22 + 12 = 34
This person had 𝑅34.
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