Real Number Properties
Real Number Properties
In this assignment we were asked to solve three expressions using the properties of real numbers in order to do so. Each of the real number properties are essential in solving algebraic expressions. Although you may not need to use all of them in the same expression to solve you will need to use at least one. In this paper I will demonstrate the use of the properties and show the steps needed to solve each part of an expression.
Understanding the properties of algebra is important because you must be able to use these properties in order to correctly put expressions and equations into their simplest form by moving terms around to solve or simplify the expression. Distribution is used to multiply across terms inside parentheses, allowing you to remove the parentheses for the problem. Moving terms to different locations is known as the commutative …show more content…
2a^2-10a+4a-20 Using the distributive property I removed the parentheses. 2a^2-6a-20 I added the coefficients here in order to add the like terms.
2a^2-6a-20 This is the simplest form of this expression.
2w-3+3(w-4)-5(w-6)
2w-3+3(w-4)-5(w-6) This is the given expression for problem number two.
2w-3+3w-12-5w-30 Using the distributive property I removed the parentheses.
2w+3w-5w-3-12-30 I used the commutative property to arrange the like terms.
5w-5w-15-30 Two of the variable and two of constant terms are added here.
0-30 The coefficients are added together.
30 The expression is fully simplified. 0.05(0.3m+35n)-0.8(-0.09n-22m)
0.05(0.3m+35n)-0.8(-0.09n-22m) This is the given expression for problem number three.
0.015m+1.75n-0.72n-17.6m Using the distribution I removed the parentheses.
0.015m-17.6m+1.75n-0.72n I used the commutative property to arrange the like terms.
-17.58m+1.03n The coefficients are added together.
-17.58m+1.03n The expression is fully
In this assignment we were asked to solve three expressions using the properties of real numbers in order to do so. Each of the real number properties are essential in solving algebraic expressions. Although you may not need to use all of them in the same expression to solve you will need to use at least one. In this paper I will demonstrate the use of the properties and show the steps needed to solve each part of an expression.
Understanding the properties of algebra is important because you must be able to use these properties in order to correctly put expressions and equations into their simplest form by moving terms around to solve or simplify the expression. Distribution is used to multiply across terms inside parentheses, allowing you to remove the parentheses for the problem. Moving terms to different locations is known as the commutative …show more content…
2a^2-10a+4a-20 Using the distributive property I removed the parentheses. 2a^2-6a-20 I added the coefficients here in order to add the like terms.
2a^2-6a-20 This is the simplest form of this expression.
2w-3+3(w-4)-5(w-6)
2w-3+3(w-4)-5(w-6) This is the given expression for problem number two.
2w-3+3w-12-5w-30 Using the distributive property I removed the parentheses.
2w+3w-5w-3-12-30 I used the commutative property to arrange the like terms.
5w-5w-15-30 Two of the variable and two of constant terms are added here.
0-30 The coefficients are added together.
30 The expression is fully simplified. 0.05(0.3m+35n)-0.8(-0.09n-22m)
0.05(0.3m+35n)-0.8(-0.09n-22m) This is the given expression for problem number three.
0.015m+1.75n-0.72n-17.6m Using the distribution I removed the parentheses.
0.015m-17.6m+1.75n-0.72n I used the commutative property to arrange the like terms.
-17.58m+1.03n The coefficients are added together.
-17.58m+1.03n The expression is fully