Working with algebra
Working with algebra you must understand why the properties of real numbers are so important. I will demonstrate my solutions to three problems and I will include my mathematical work. Also I will use five vocabulary words that help me find solutions to the problems. Properties of real numbers are useful for simplifying algebraic expression because a lot of thing we do in life are equations. We use a lot of mathematical terms in the real world. Lastly I will show every step I took to simplify and identify each property of real numbers. The properties of algebra are important to know and understand.
“Algebra is useful because it can be used to solve problems. Since problems are often communicated verbally, we must be able to translate verbal expressions into algebraic expressions and translate algebraic expressions into verbal expressions. ” (Dugopolski, 2012, Chapter 1. 6, ) Each expression has properties that must be simplified and solving methods. You have to know how to follow the order of operation and simplify the equations, variables and like terms in order to complete the mathematical work. Simplify is very important in all expression, must be in simplest form when completely an equations.
Working with algebra Essay Example
Also you have to move and combine like terms. The coefficient of a number is in front of a variable. The coefficient is a factor that can produce a result. The distributive property is a step that multiples a term to be followed in order for the equation to be complete. Lastly removing parentheses is another step you should follow in any equations or expression. You must remove the parentheses in order to solve the equation. I will now demonstrate how the properties of real numbers are used while I simplify the following expressions. 2a (a – 5) + 4(a – 5) 2a2-10a+4a-20 2a2-6a-20
The given expression I multiply by distributive property and that allowed me to remove the parentheses. Then I combined the like terms by adding the coefficients. The distributive property removes the parentheses. I did not have change any order of operations the like term was already together in the middle of the expression. 2w – 3 + 3(w – 4) – 5(w – 6) 2w – 3 + 3w -12 – 5w +30 2w +3w -5w -3 -12 +30 w +15 With this given expression I had to distribute the property in order to remove the parentheses. I move the like terms so they can be arrange together using the commutative property to switch their places.
Also two of the variable terms are added and two of the constant terms are also added. I then added the like terms and now this expression is simplified. 0. 05(0. 3m + 35n) – 0. 8(-0. 09n – 22m) 0. 015m + 1. 75n + 0. 072n + 17. 6m 0. 015m + 17. 6m + 1. 75n + 0. 072n 17. 615m +1. 822n With this given expression I had to distribute the properties and remove the parentheses. I arrange the like terms and then combined them together using the commutative property. The like terms was combined by adding the coefficients. I had to be careful with this expression because of the decimals.
I still followed the same steps as before in the other expression. “Algebraic expressions are used to describe or model real-life situations. ”( Dugopolski, 2012, Chapter 1. 6, ) I learned that algebraic expressions can be used to solve a real life problem. It’s just like going to the store and you need to sum, or the difference, or percentage of an item. I realize you use algebraic expression very often in everyday life. I was able to describe step by step of what I had to do in order to solve the expression. I also touch base with my vocabulary words by using them in my explanation of how to solve the expression. Reference