The Simplifier

Rule 1: Understand the Mathematical  principles/rules/theories

There are a few general rules that must be observed in other to become better in mathematics.

1. Patience: Be patient with yourself and try things over and over again.

2. Note that there are only 10 numbers (digits) in the whole world, which are:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

3. Make Enquiries: you have to learn to use the right books, ask the right tutor(s) and don't be ashamed to make attempts.

4. Believe it is possible: Don't conclude hastily that a problem is not solvable. In fact, every problem is solvable provided the right steps and procedures are known.

5. Be ready to study different mathematical background rules because every mathematical aspect (topic) has rules associated to it. Be it Simple Algebra, Logarithm, Indices, Bearing and Construction, Geometry, Trigonometry, Longitude and Latitude, etc there is a rule guiding it – so don't be in a haste to start solving rather take your time to understand the rules.

6. Do not skip topics: this is a very important rule to note because mathematics is a process and every new topic has a link somehow to an older topic, and you would not understand the next if you skipped the previous.

Keynote
You wouldn’t understand the present and the next, if you skipped the previous – therefore be patient! 

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The Simplifier

Example Three 3. Solve: {1 – [1 – 1 x (1 + 1) – 1 ] of 1} ÷ 1 + 1 This has nested brackets (i.e. bracket inside another bracket and we must start from the innermost bracket). Starting from innermost bracket: {1 – [1 – 1 x (1 + 1)   – 1] of 1} ÷ 1 + 1 = {1 – [1 – 1 x 1 – 1] of 1} ÷ 1 + 1 Apply the Precedence Rule also in the Innermost Bracket:  {1 – [1 – 1 x 1 – 1] of 1} ÷ 1 + 1 = {1 – [1 – 1 – 1] of 1} ÷ 1 + 1 Applying Precedent rule from left to right inside the bracket we have: {1 – [1 – 1 – 1] of 1} ÷ 1 + 1 = {1 – [0 – 1] of 1} ÷ 1 + 1 Furthermore we have: {1 – [0 – 1] of 1} ÷ 1 + 1 = {1 – [–1] of 1} ÷ 1 + 1 Next is to consider the of in the Innermost Bracket (if it exists): {1 – (–1) of 1 } ÷ 1 + 1 = {1 – (–1) } ÷ 1 + 1 Notice that we do not write two mathematical operators together without a bracket to separate them: e.g. instead of 1  –  –  1 we write 1 – ( - 1) . Next Step finalize the operations in bracket...
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The Simplifier

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The Simplifier

Rule 2: Know the Mathematical Operators and their Operations Mathematical operators are simply tools that determine the behaviour of a mathematical process. The operators tell us to join, separate, etc. What we want or are given determine the tools to be used. The basic mathematical tools are: + Plus [Addition (Summation)] - Minus [(Subtraction (Difference)] x Times [(Multiplication (Product)] ÷ Divide [Division (Quotient)] There are lots of other operators, such as roots, power, trigonometric functions, etc, but knowing the basic ones is a great step forward. Keynote Start by knowing the basic operators and how they behave.
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The Simplifier

Example One 1. Simplify: 1 + 1 – 1 x 1 ÷ (1 + 1) of 1 Step 1: Recall the precedence Rule as: • Starting from left to right and • Following the BODMAS sequence  We start the part with bracket : 1 + 1 – 1 x 1 ÷ (1 + 1) of 1 = 1 + 1 – 1 x 1 ÷ 1 of 1 Next we move to the part with the of : 1 + 1 – 1 x 1 ÷ 1 of 1 = 1 + 1 – 1 x 1 ÷ 1 (Note: 1 of 1 = 1 * 1 = 1) Next we move to where we have a division : 1 + 1 – 1 x 1 ÷ 1 = 1 + 1 – 1 x 1 Next is to handle the Multiplication : 1 + 1 – 1 x 1 = 1 + 1 – 1 (Note the negative sign in front of 1 i.e. – 1 x 1) Next to be considered is Addition and Finally Subtraction : 1 + 1 – 1 = 2 – 1 = 1 Keynote This sequence is maintained even if more than one type of the same operator exists at different locations - finish with all additions  before subtractions. Applicable to all others operators!
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The Simplifier

Example Two 2. Evaluate: 1 of 1 – (1 – 1) + 1 ÷ 1 x 1 Start with the bracket: 1 of 1 – (1 – 1) + 1 ÷ 1 x 1 = 1 of 1 – 0 + 1 ÷ 1 x 1 Next is to handle the of ( if it exists ): 1 of 1 – 0 + 1 ÷ 1 x 1 = 1 – 0 + 1 ÷ 1 x 1 Next is to deal with Division ( if it exists ): 1 – 0 + 1 ÷ 1 x 1 = 1 – 0 + 1  x 1 Next is to handle Multiplication ( if it exists ) 1 – 0 + 1 x 1 = 1 – 0 + 1 Next is to consider Addition & Subtraction ( if it exists ) 1– 0 + 1 = 1 + 1 = 2 Keynote It is not a must that all the operators must exist in one problem but the sequence is always followed while the operator not in the operators are skipped to the next available one!
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