The Simplifier

SIMPLE EXAMPLES

1. Simplify: 1 + 1 – 1 x 1 ÷ (1 + 1) of 1
2. Evaluate:1 of 1 – (1 – 1) + 1 ÷ 1 x 1
3. Solve: {1 – [1 – 1 x (1 + 1) – 1] of 1} ÷ 1 + 1

SOLUTION

Rule1: Don’t be scared
Rule2: Follow the principles of operation (precedence rule).

Note: do not be confused by the different terms: simplicity, evaluate or solve. For now we can assume them to all mean the same thing. Simply see them as saying “manipulate” the digits using the right principles until the simplest form is reached!

Comments

Post a Comment

Popular posts from this blog

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...
Read more

The Simplifier

Attempt The Following 1. Simplify:   1 + 2 – 3 x 4 ÷ (5 + 6) of 7 2. Evaluate:  7 of 6 – (5 – 4) + 3 ÷ 2 x 1 3. Solve:  {8 – [7 – 6 x (5 + 1) – 2] of 3} ÷ 4 + 1 Solve Manually without a calculator and take fractions not decimals . In case you aren’t familiar with those two then let us know. PS : Please make your comments here or send them directly to my mail by using the side bar. You can also make more research on each topic by using the side bar! Ask your questions, make your comments and help us improve. Digging Deep is aimed at turning our fears to strength through simplification of complex processes, therefore we intend covering more on not only mathematics but also other courses!
Read more

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.
Read more

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!
Read more

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!
Read more