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by: William Jones

Algebra1 Algebra1

William Jones

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This 4 page Class Notes was uploaded by William Jones on Wednesday February 17, 2016. The Class Notes belongs to Algebra1 at University of Kentucky taught by in Winter 2016. Since its upload, it has received 10 views.


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Date Created: 02/17/16
APPLICATIONS OF SOLVING EQUATIONS This chapter presents applications of material learned in the first chapter. It shows how solving equations are useful when dealing with word problems, rate problems, and mean average problems. One of the most important applications of equations is in solving word problems. Thus, the first section outlines the steps to follow when dealing with a word problem. By breaking them down into simple steps, word problems become less daunting. The second section deals with a particular type of word problems: rate problems. These problems deal with the relationship between time spent traveling, rate of travel, and total distance traveled. This section explain how to calculate these numbers for a single traveler or for multiple travelers. This is useful to know not only for textbook problems, but also for real life--using the tools learned in this section, you can calculate your own distances, times, and rate of travel. The third and final section explains how to solve problems dealing with mean averages. It shows, given a set of numbers, how to calculate the number needed to produce a specific average. This is especially useful when dealing with test scores--one can calculate, given one's previous tests, what score one needs on the next test in order to have a certain average. While the last chapter dealt with the theory behind solving equations, this chapter deals with some of the applications of equations to real life. Equations are useful in many situations, from daily ventures such as shopping and driving, to jobs in the business world. Solving word problems, rate problems, and mean average problems is a useful skill to have. The skills learned in this chapter will help you understand mathematical problems as well as problems you encounter in daily life. Mean Average - The mean average of a group of values is the sum of the values divided by the total number of values. Strategies for Solving Word Problems with Variables Often, word problems appear confusing, and it is difficult to know where to begin. Here are some steps that will make solving word problems easier: 1. Read the problem. 2. Determine what is known and what needs to be found (what is unknown). 3. Try a few numbers to get a general idea of what the solution could be. 4. Write an equation. 5. Solve the equation by inverse operations or by plugging in values. 6. Check your solution--does it satisfy the equation? Does it make sense in the context of the problem? (e.g. A length should not be negative.) Example 1: Matt has 12 nickels. All the rest of his coins are dimes. He has just enough money to buy 2 slices of pizza for 95 cents each. How many dimes does he have? 1. Read the problem. 2. What is known? Matt has 12(5) = 60 cents in nickels. Matt has 2(95) = 190 cents total. What needs to be found? The number of dimes that Matt has. 3. Try a few numbers: 5 dimes? 10(5) + 60 = 110. Too low. 10 dimes? 10(10) + 60 = 160. Still too low. 20 dimes? 10(20) + 60 = 260. Too high. So we know the answer is between 10 and 20. 4. Write an equation: 10d + 60 = 190 where d is the number of dimes Matt has. 5. Solve using inverse operations: 10d + 60 - 60 = 190 - 60 10d = 130 = d = 13 6. Check: 10(13) + 60 = 190? Yes. Does 13 dimes make sense in the context of the problem? Yes. Thus, Matt has 13 dimes. Example 2: Jen is shooting free-throws on the basketball court. She makes 85% of her shots. If she makes 51 shots, how many does she miss? 1. Read the problem. 2. What is known? Jen makes 85% -- or -- of her shots. Jen makes 51 shots. What needs to be found? The number of shots that Jen misses. 3. Try a few numbers: 5 shots? = . Not enough misses. 10 shots? = . Too many misses. So we know the answer is between 5 and 10. 4. Write an equation: = where x is the number of misses. 5. Solve using inverse operations: = 51( ) = 51( ) 51 + x = 60 51 + x - 51 = 60 - 51 x = 9 6. Check: = ? Yes. Does 9 shots make sense in the context of the problem? Yes. Thus, Jen misses 9 shots. Diagram of a Square Example 3: The area of this square is 2 times its perimeter. How long is a side? 1. Read the problem. 2. What is known? The area of the square is 2 times its perimeter. The formula for 2 area is A = x and the formula for perimeter is p = 4x . What needs to be found? The length of a side. 3. Try a few numbers: x = 5 ? A = 5 = 25 , p = 4(5) = 20 . Area too small. x = 10 ? A = 10 = 100 , p = 4(10) = 40 . Area too large. So we know the answer is between 5 and 10. 4. Write an equation: x = 2(4x) . x = 8x 2 5. Solve by plugging in values----or by using inverse operations: = x = 8 6. Check: 8 = 8(8) ? Yes. Does 8 make sense in the context of the problem? Yes. Thus, the square has side length 8.


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