1) Review the following information: https://www.freemathhelp.com/systems-inequalities.html
2) Review: http://ltcconline.net/greenl/courses/152b/QuadraticsLineIneq/systems.htm
3) Review: https://www.khanacademy.org/math/algebra2/systems_eq_ineq/systems_inequalities_precalc/e/graphing_systems_of_inequalities_2
MY TRIP TO PURCHASE A DVD RECORDER
I have a large collection of home movies recorded using a video camera. Some of my older home movies were transferred from 8mm to VHS tape a few years back. Have you seen that commercial on television for a DVD recorder where you can transfer all your home movies to DVD? Well, I wanted one of those recorders. Not only that, but most of the movies I want to rent at my local Blockbusters are in DVD and I need a DVD player to watch them--I only had a VHS player. So I went to my local technology store to purchase a nice DVD player/recorder. After making my very expensive purchase, the sales person asked if I wanted to join a club to rent DVD movies. The cost to join the club for a year is $49.95 then I can rent DVD movies for only 39-cents each. At the local Blockbusters, it costs $2.50 each time I rent a new release. What would I have to consider before purchasing the DVD movie club? This problem is a perfect example of a system of linear equations. As you learned from your reading assignment in the textbook, a system of linear equations is when you have two or more related equations (the variables in each equation represents the same type of quantity). You also learned from your reading that a system of two equations may have one solution, no solution, or infinitely many solutions.
In my DVD movie club example, the solution would be the break-even point in the two purchase plans. Let's ignore taxes right now and look at the equations in my DVD problem:
Let's let y = the total amount of money I would spend in a year Let's let x = the number of movies I rented At Blockbusters, the total amount of money I would spend in a year is $2.50 times however many movies I rented: y = 2.50x
In the DVD movie club, the total amount of money I would spend in a year is 39-cents times however many movies I rented plus my initial investment of $49.95: y = 0.39x + 49.95
I have two related equations using the same two variables. This is a system of equations. I could use any of the three methods described in the textbook (addition, substitution, or graphing) to solve my equations. So, a system of equations is used whenever you have two related equations with two variables and you want to find a specific solution.
HOW ELSE ARE SYSTEMS OF EQUATIONS USED?
With the high cost of a DVD recorder, I'm not going to be purchasing another one soon. So is this the only time in my life I'll need to use systems of equations? No. Let me give you a few other times when it is helpful to know systems of equations:
When I have money to invest--I'll need to compare different investment plans. (Of course, after purchasing a DVD recorder, it will be a long time before I have money to invest.)
Companies use systems of equations when doing cost accounting.
Those in marketing use systems of equations to compare various marketing plans.
Any time an individual does comparison shopping, he or she is using systems of equations.
Systems of equations are used to find a solution when you have two variables. Use the two variables to write two equations then solve them. In real life applications, you will usually see a system of equations when you need to compare two items. The first equation will describe the first item; whereas, the second equation will describe the second item in the comparison.
HOW ARE SYSTEMS OF INEQUALITIES USED?
Because I like to stay fit, I am very familiar with situations where there isn't a single answer but a range of answers. For example, when I exercise, I should keep my heart rate between 65% and 75% of my maximum heart…