6 Proportion: Fractions, Direct and Inverse Variation, and Percent

Transcription

1 6 Proportion: Fractions, Direct and Inverse Variation, and Percent 6.1 Fractions Every rational number can be written as a fraction, that is a quotient of two integers, where the divisor of course cannot be zero. However, this representation is not unique. We need to know how to tell when two fractions represent the same rational number (which we think of as being a unique point on the real number line). We also need to know how to perform the arithmetic operations of addition, subtraction, multiplication, and division on fractions. Subsequently we will deal with fractional expressions where the numerator and/or denominator are made up of algebraic expressions instead of integers, but the same rules for determining when two are equal, and for performing arithmetic operations on them, will apply. The general method for telling when two fractions are equal is the rule for cross-multiplying: Let m, n, r, s be integers and assume that n 0, s 0. Then m = r if and only if ms = rn. Let s look at why this makes sense: we n s know that m = mp for any integer p. If we think about this in terms of the n np number line representation, this is clear. We can think of m, where m, n are n positive, as dividing each unit segment up into n subunits, each of length 1/n, and now we take m of those subunits, to have something of length m/n. If we further subdivide each of those subunits p times, now each little segment will have length 1/(np), and in taking m of the old subunits, we will have taken mp of the new ones, so the length is also mp. It should be the case np that m = r if and only if there is some way to subdivide each of them so n s that we get a fraction mp rt and a fraction where the numerator of each and np st the denominator of each are exactly the same, that is, mp = rt, np = st. But then, mp st = rt np, or mspt = nrpt, and canceling pt, we have ms = nr. The approach just given justifies the criterion for when two fractions are equal, but we can check it more simply if we accept some other facts about fractions. We have that ms = rn if and only if ms sn = rn sn. Now if we accept that dividing by an integer is the same as multiplying by its reciprocal, we have this equivalent to ms = rn, and then if we accept as sn sn above that we can cancel common factors of numerators and denominators, we have m = r. n s We say that a fraction m is represented in lowest terms if the only comn 22

2 mon divisor of m and n is 1. (In a separate section we discuss the use of the Euclidean algorithm to determine the greatest common divisor of two integers.) Recall from 1 the discussion of how fractions are added, subtracted, and multiplied. There we also stated without proof that a = c if and only b d a if. We show this simple fact now, using the remarks above on = c a+b c+d determining when two fractions are equal. We have a = c if and only if a+b c+d a (c + d) = (a + b) c. Applying the distributive law, we see this is equivalent to ac + ad = ac + bc, and subtracting ac from each side of the equation shows us this is the same as ad = bc. Then since we are assuming neither b nor d is 0, we have that this is equivalent to a = c. b d We also revisit the idea of dividing with fractions. There are many ways to look at this. We can accept as obvious the fact that if we take any number and divide it by itself, the quotient will be 1. Thus for rational numbers, m m = 1. In terms of our properties of real numbers, and the n n idea that multiplying any number by its multiplicative inverse gives you 1, we see that dividing by m has the same effect as multiplying by ( m n n ) 1. But what is ( m n ) 1 as a rational number, that is as a quotient of two integers? It is a number which when multiplied together with m gives us a fraction n that is equivalent to 1. Now observe that m n = mn = 1, so we have that n m mn ( m n ) 1 = n. Thus dividing by a fraction is the same as multiplying by its m reciprocal. This also makes sense when we think of division as the inverse operation to multiplication. Working with integers, if we are trying to solve a b = x, we are asking to find the number x such that b times x will give us a. Working with fractions, if we ar trying to solve p m = x, we are looking for q n that number x such that p = m x. Using our understanding of equivalent q n fractions, it is easy to check that x = pn works, thus verifying that dividing qm by m is the same as multiplying by n. n m When our fractions involve algebraic expressions, exactly the same rules will apply. For example, to add x+1 + x+2 we will first have to find a common x 1 x+3 denominator, which here will be the product of the two individual denominators. We need to convert each of the summands into an equivalent fraction have the common denominator (x 1)(x + 3). This will be done by multiplying the first term by x+3 x+3 and the second term by x 1 x 1. This gives (x + 1)(x + 3) 1)(x + 2) +(x (x 1)(x + 3) (x 1)(x + 3) = (x2 + 4x + 3) + (x 2 + x 2) (x 1)(x + 3) 23 = 2x2 + 5x 1 x 2 + 2x 3.

3 We summarize the rules for fractions below. Make sure you understand why each one is true! And never forget that we cannot divide by 0, so the denominator of a fraction can never be Equivalent Fractions: a b = ax bx x 0 2. Cross-Multiplying Rule for Equivalence: a b = c d if and only if ad = bc 3. Multiplication Rule: a b c d = ac bd 4. Division Rule: a b c d = a b d c = ad bc 5. Addition/Subtraction Rule: a b ± c d = ad±bc bd 6. Simplifying Compound Fractions: a/b = a c = ad c/d b d bc 7. Working with Negatives: a = a; a b b b = a b = a b 6.2 Direct and Inverse Variation Problems Two values are said to be directly proportional to each other if one value is a fixed numerical multiple of the other. Thus y is directly proportional to x if there exists a constant c such that y = cx. What this means is that the ratio y is always equal to the constant c. Two values are said to be x inversely proportional to each other if one value is directly proportional to the multiplicative inverse (reciprocal) of the other number. In this case, there would be a constant c such that y = c 1. Equivalently, this would say yx = c. x Direct and inverse proportion problems are easy to solve algebraically and provide a powerful tool. Direct proportion is what is used in order to convert units of one kind to units of another. For example, suppose we know that on a map 1/8 of an inch equals 10miles, and we measure on the map that two towns are 3.5 inches apart. How far are they apart in real life? We know that the ratio of miles to inches is a constant, and we know that constant is equal to 12. Then we also know that the ratio of the distance the two towns 1/8 are apart to their distance on the map is also equal to this constant. To solve the problem, set the two ratios equal to each other. Then we have two equal fractions, so we know if we cross-multiply, they must be equal. We can cross multiply and solve for the unknown distance. In this case we have 12 miles 1/8 inches = x miles 3.5 inches 24

4 which gives = 1 x, or = x, so x = 336 miles. 8 The general method for solving proportion problems is to set up the correct ratios, cross-multiply, and solve. Often, thinking about the units involved will help you set up the ratio in the correct order. Let s look at a few more examples: A man is in a field with a charging bull, which is 363 feet away from him. If the man can run 12 feet per second, and the bull can run 45 feet per second, how long until the bull catches up to the man? If there is a sturdy fence the man can climb over and be safe, which is 150 feet away, will the man make it? We can figure this out as follows: Every second, the bull runs = 33 feet farther than the man. So we can set up the ratio 33 feet / 1 second = 363 feet / x seconds, and solve for x. If we cross-muultiply, we have 33x = 363, so x = 11. Now in 11 seconds, the man will travel = 132 feet, so unfortunately he will not reach the fence. Ratios and direct or inverse proportions come up in many real world problems. The distance an object travels is equal to its speed times the time it travels. Thus distance is directly proportional to speed, and distance is directly proportional to time, and time and speed are inversely proportional to each other. Likewise, the amount of work done by a person who works at a certain rate is the product of the rate at which the person works, multiplied by the amount of time the person works. Let s start with some simple examples and move up to more complicated ones. Example 1: Susie walks at a rate of 3.5 miles per hour. She is walking around a lake path that is 2.25 miles long. If she starts around the lake at 9:30, what time (to the nearest minute) will she get back to where she started? If we let t stand for how long Susie walks, we have 3.5 miles / 1 hour = 2.25 miles / t hours. We cross multiply to get 3.5t = 2.25, or t = 2.25/3.5 = 9/14, so it will take her 9/14 of an hour. Now we know we have 60 minutes / 1 hour, so we have 60 = m, or m = 9 60/14 = 270/7 minutes, 1 9/14 which is minutes, so to the nearest minute it will take her 39 minutes, and she will be back at 10:09. Example 2: It took John two hours to ride his bike to his grandmother s house. On the way home he was tired, and it took three hours. If John was biking 5 mph slower on the way home than he was on the way there, what speed was he biking on the way to his grandmother s, and how far away did she live? Let s let s be the speed (in miles per hour) John biked on the way 25

5 to his grandmother s, and let s let d be the distance he traveled. Then we know d = 2s, because distance is rate time. On the way back, the distance he traveled is still d, but the rate is s 5, and the time is 3, so we have d = 3(s 5). Thus 2s = 3(s 5), so 2s = 3s 15, and s = 15. He biked at a rate of 15 miles per hour on the way there and 10 miles per hour on the way home, and his grandmother lived 30 miles away. Example 3: Bill can paint four houses in 28 days. How long will it take him to paint five houses? John can paint nine houses in 45 days. How long will it take him to paint seven houses? If the two of them work together, how long will it take them to paint two houses? To answer this question, we have to determine the rate at which each of them paint houses. Let b denote the rate at which Bill paints houses, measured in houses per day, and let j denote the rate at which John paints houses, also measured in houses per day. Then b = 4/28 = 1/7, that is, Bill can paint 1/7 house per day. Or, taking reciprocals, it takes him 7 days per house. To paint five houses will take 5 7 = 35 days. And j = 9/45 = 1/5, that is, John can paint 1/5 house per day, or it takes him 5 days per house. To paint seven houses will also take him 35 days. If they both work together, the first thing we need to figure out is the rate at which they paint. In one day, they can paint 1/7 + 1/5 houses working together, so they paint 5/35 + 7/35 = 12/35 houses per day. That means it takes them 35/12 days to paint one house, and 70/12 = 35/6 days to paint two houses. 6.3 Solving Problems Involving Percents There are basically three types of problems involving percents: You may be asked to find the percent, you may be asked to find the part, or you may be asked to find the whole. We ll look at three examples that illustrate this: a) 60 is what percent of 75? (Find the percent.) b) What is 80% of 75? (Find the part.) c) 60 is 80% of what number? (Find the whole.) All three of these problems can be solved by setting up the following ratio: percent 100 = part whole 26

6 In each problem, you are given two of the three values percent, part, and whole. By setting up the ratio and cross-multiplying, you can solve x for the remaining value. Thus (a) becomes = 60 80, (b) becomes = x, and (c) becomes 80 = 60. The task in solving each problem then becomes 100 x identifying which of the three, percent, part, or whole, you need to find to solve the problem, and which have been given. Percent problems may also involve finding the percent increase, or the percent decrease. This is given by solving for the percent where the whole is the original amount, and the part is the amount by which the original value went up (in the case of increase) or went down (in the case of decrease). Suppose the price of gas one day is $1.60, and the next day it has jumped to $1.80. The following day it is back at $1.60. What is the percent increase from the first day to the second? What is the percent decrease from the second day to the third? Notice that although the price goes up $.20 and then back down $.20, the percent decrease will be less than the percent increase, because $.20 is a smaller percent of $1.80 than it is of $1.60. Let s do the calculations: To find the percent increase, we solve x =.20 to obtain x = 12.5, so the price increased 12.5%. To find the percent decrease, we solve x = to obtain x = 11. 1, so the percent decrease is approximately 11%. Summarizing, percent problems are not that difficult if you remember to set up the ratio percent/100 = part/whole, and cross-multiply to solve. 27