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The Properties of Real Numbers

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Mathematics

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Homework

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Running Head: THE PROPERTIES OF REAL NUMBERS
1
The Properties of Real Numbers
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Running Head: THE PROPERTIES OF REAL NUMBERS
2
Algebraic equations usually result from modeling real life problems. In order to find solutions to
these problems, they have to be manipulated. This is achieved by applying the distributive,
commutative, associative, inverse and identity property of numbers and their operators. These
properties will result in algebraic equations attain its simplest form.
The distributive and associative properties help in eliminating the parentheses in the equation
and group the like terms, respectively, so that the equation is rewritten in its simplest form.
The commutative property helps in rearranging the terms without losing the equality property
across the equations. The inverse and zero property help in separating the like terms on either
side of the equations helping in solving the problem.
The following examples provide a demonstration of applying these properties in solving
algebraic equations.
2a(a-5)+4(a-5) The given expression to be simplified
2a . a -2 x 5 + 4 . a - 4 x 5 Distribute the numbers or terms outside of the
parentheses into the numbers inside of them. This is
equivalent to multiplying the terms.
2a
2
-10 + 4a - 20 Simplify the expressions by carrying out the
operations.
2a
2
+ 4a - 20 - 10 Regroup the like terms in the expression. This will
help align the numbers for simplification.
2a
2
+ 4a - 30 The given expression simplified to the lowest possible
state. Like terms are combined by adding the
coefficients.
There was one operation in this expression which required (step 3) regrouping in order to
simplify the expression. There was no need to add up terms using the coefficients.
2w – 3 + 3(w – 4) – 5(w – 6) The given expression.
2w – 3 + 3w – 3 x 4 – 5w –5 x(- 6) Distribute in the expression to get rid of the
parentheses.

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Running Head: THE PROPERTIES OF REAL NUMBERS
3
2w + 3w -5w - 3 -12 + 30 Use associative property to group like terms.
0w + 15 = 15 The terms with variable (w) is added by adding the
coefficients. Simplification leads to a situation where
there is no variable. Similarly adding all the constants
yields an expression which is a constant.
0.05(0.3m + 35n) – 0.8(-0.09n – 22m) The given expression.
0.05 (0.3)m + 0.05(35)n-0.8(-0.09)n-
22m
Using distributive property, parentheses are removed
0.015m + 1.75 n + 0.072n – 22m Simplify terms by carrying out multiplication
-21.985 m + 1. 822 Add the coefficients of like terms to simplify the
expression.
This problem demonstrates that the primitive operations and properties of numbers remain the
same.

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