SOLUTION: Write a statement reflecting on your appreciation for algebra and how it can be applied to a real situation. Support your statement with examples of applications.

Question 440990: Write a statement reflecting on your appreciation for algebra and how it can be applied to a real situation. Support your statement with examples of applications.
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
Actually you use algebra all the time. Here are some straight up easy answers.
Successfully using Algebra in is determining which situations call for which formulas and concepts. Luckily, the most common real life problems call for widely applicable and highly recognizable techniques.
You now have enough information about algebra to try out an applied problem. The general plan, when applying math to the real world, is to find a way to match up objects, events and properties in the real world (pieces of rope, lakes, flying, temperature) to mathematical objects, actions and relationships in the mathematical world (lines, shapes, numbers, addition, subtraction, equals), and then use what is known about the mathematical world to come to conclusions about the real world. Let's see how this works by considering a real world scenario:
first watch this movie:
http://www.thefutureschannel.com/dockets/realworld/designing_backpacks/
not only backpacks, everything what we have (clothing, household things, cars, bikes, etc), everything should be designed before production.
some more examples:
1
Use quadratic equations to find the maximum or minimum possible value of something when increasing one aspect of the situation decreases another. For instance, if your restaurant has a capacity of 200 people, buffet tickets currently cost $10, and a 25 cent increase in price loses about four customers, you can figure out your optimum price and maximum revenue. Because revenue equals price times the number of customers, set up an equation that would look something like this:
where "X" represents the number of 25 cent increases in price. Multiply the equation out to get
which, when simplified and written in standard form (), would look like this:
. Then, use the vertex formula () to find the maximum number of price increases you should make, which, in this case, would be -40/(2)(-1) or 20. Multiply the number of increases or decreases by the amount for each and add or subtract this number from the original price to get the optimum price. Here the optimum price for a buffet would be $10.00 + .25 (20) or $15.00.
2
Use linear equations to determine how much of something you can afford when a service involves both a rate and a flat fee. For instance, if you want to know how many months of a gym membership you can afford, write out an equation with the monthly fee times "X" number of months plus the amount the gym charges up front to join and set it equal to your budget. If the gym charges $25/month, there is a $75 flat fee, and you have a budget of $275, your equation would look like this: 25x + 75 = 275. Solving for x tells you that you can afford eight months at that gym.
3
Bring together two linear equations, called a "system," when you need to compare two plans and figure out the turning point that makes one plan better than the other. For example, you could compare a phone plan that charges a flat fee of $60/month and 10 cents per text message with one that charges a flat fee of $75/month but only 3 cents per text. Set the two cost equations equations equal to each other like this:
where x represents the thing that might change from month to month (in this case number of texts). Then, combine like terms and solve for x to get approximately 214 texts. In this case, the higher flat rate plan becomes a better option. In other words, if you tend to send less than 214 texts per month, you're better off with the first plan; however, if you send more than that, you're better off with the second plan.
4
Use exponential equations to represent and solve savings or loan situations. Fill in the formula
when dealing with compound interest and
when dealing with continuously compounded interest. "A" represents the total amount of money with which you will end up or will have to pay back, "P" represents the amount of money put into the account or given in the loan, "r" represents the rate expressed as a decimal (3 percent would be .03), "n" represents the number of times interest is compounded per year, and "t" represents the number of years the money is left in an account or the the number of years taken to pay back a loan. You can calculate any one of these parts by plugging in and solving if you have the values for all of the others. Time is the exception because it is an exponent. Therefore, to solve for the amount of time it will take to amass, or pay back, a certain amount of money, use logarithms to solve for "t."

5. When filling your car up with gas you can use a form of algebra. Lets say you only have $20.00 to spend on gas today and gas is $3.50 a gallon. How many gallons could you buy?
Let x = # of gallons of gas

6. For the last one lets say you are all grown up now and have to move across country for a new job. Lets use Buffalo, NY to Sacramento, CA which is roughly 2500 miles of driving. How much money do you need to save for gas if the national average is $3.23/gallon.
Let x = amount of money you need to save

dollars
There you go, simple examples of using algebra in the real life.
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