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Syllabus Objective: Use the four operations and rules of precedence (BODMAS) to manipulate directed numbers.
The questions related to this objective are normally found both in paper one and two as question number one. They simply test your ability to use the rules of precedence (BODMAS) to manipulate directed numbers. Typically you are given three terms, either in fraction form or decimal form, interlinked by two operations. In that particular case it’s up to you to determine which of the given operations you will give precedence to.
Consider the extract from the November 2012 Mathematics Paper 2 Question Paper (4028/2):
1 (a) Simplify , giving your answer as a fraction in its lowest terms. [2]
Solution:
Here given are three fractions interlinked by two operations . The general reasoning with most people is that when there is a plus and a minus they give precedence to the plus operation.
As a result most of the students attempting to solve the above problem would tend to add the last two fractions before subtracting the result from the first fraction. However this reasoning is VERY faulty!! Attempt the permutation below using two different approaches;
Evaluate
BO(DM)(AS)
Even pair operations;
Operations belonging to the same bracket as illustrated above, (AS) implying (Addition-Subtraction) are even in nature i.e they are equal and none presides over the other. As a result an expression involving subtraction and addition would rather be simplified sequentially than bracketing off the last terms and giving precedence to the plus operation.
The steps below illustrate the correct approach to solving the problem;
STEP 1
Convert the individual terms from mixed fraction form to improper fractions;
STEP 2
Identify the common denominator
(the common denominator is the LCM of the three given denominators).
The common denominator=6.
STEP 3
Evaluate each individual term to a denominator of six;
Consider the term. If you are to evaluate this term to a denominator of 6, you are simply establishing its equivalence under a denominator of 6.
By a mere comparison of the denominators, you observe that the denominator 2 increases by a factor of 3 to match the denominator 6. Likewise the numerators should increase by the same factor to maintain the balance. Hence under the denominator 6 the numerator is .
Do the same for the other two terms. Did you get the expression below?
STEP 4
Collect all the terms under one denominator
( Did you observe that after evaluating the individual terms to a denominator of 6 in STEP 3, the fractions now have something in common; they now share the same denominator. Fractions that share the same denominator are called like fractions.).
STEP 5
Simplify the numerator
The numerator also is an expression of three terms interlinked by two operations. As mentioned above the operations involved are an even pair, as a result none of them has precedence over the other. Hence it is advisable to simply evaluate the numerator by simplifying sequentially as was shown in the Sequential approach above.
STEP 6
Simplify the fraction by cancelling out any common factors.
In this case the numerator and the denominator share a common factor of 2. Hence both terms can be reduced by a factor of 2 and the fraction reduces to
W Ngwenya and T Bosha are tutors at the Adult school of Mathematics. They can be reached on the facebook page: Bulawayo Adult School of Mathematics.
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