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**The Game of Algebra or The Other Side of Arithmetic**

Lesson 24 by Herbert I. Gross & Richard A. Medeiros © Herbert I. Gross next

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**Application to Word Problems**

next © Herbert I. Gross

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**Afterwards we’ll look at some others.**

An Algebraic Approach Perhaps the best way to introduce this Lesson is to begin by “revisiting” the type of examples we did in the previous Lesson. Afterwards we’ll look at some others. next © Herbert I. Gross

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Example 1 Erin and Bill together have 500 marbles. Erin has 50 marbles more than Bill. How many marbles does Bill have? Answer: 225 next next next © Herbert I. Gross

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Algebraic Solution To help keep track of which letter represents Erin and which represents Bill, we let E denote the number of marbles Erin has, and we let B denote the number of marbles Bill has. next © Herbert I. Gross 5

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Algebraic Solution The fact that the total number of marbles that they have is 500 leads to the constraint that… E + B = 500 And the fact that Erin has 50 more marbles than Bill has leads to the constraint that… E – B = 50 next next © Herbert I. Gross

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**The above two constraints yield the system of equations…**

And if we add these two equations… And then divide both sides of the resulting equation by 2 … E + B = 500 + E – B = 50 2E = 550 E = 275 next next next next © Herbert I. Gross

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**In either case, Bill has 225 marbles.**

Thus, Erin has 275 marbles. However, the question asks for the number of marbles Bill has. So we may replace E by 275 in either of the two equations… E + B = 500 E – B = 50 B = 500 275 – B = 50 B = 500 – 275 275 – 50 = B B = 225 225 = B In either case, Bill has 225 marbles. next next next next © Herbert I. Gross

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**Notes Problems such as this are often presented to students**

prior to their being taught simultaneous equations. For example, if we know the number of marbles Bill has we may subtract this amount from 500 to determine how many marbles Erin has. So if B represents the number of marbles Bill has, 500 – B represents the number of marbles Erin has. Notice, however that we could have obtained the same result by subtracting B from both sides of the equation E + B = 500 to obtain that E = 500 – B. Notes next © Herbert I. Gross

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**And solving the above equation, we see that B = 225.**

Rewriting the equation E + B = 500 in the form E = 500 – B allows us to replace E by 500 – B in E – B = 50 to obtain… Notes (500 – B) E – B = 50 500 – 2B = 50 500 – = 2B 450 = 2B 225 = B And solving the above equation, we see that B = 225. next next next © Herbert I. Gross 10

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Notes Using E and B rather than using only either E or B, often helps us to organize our thinking. Namely, the more steps we try to keep track of in our head, the easier it becomes to make a mistake. next next © Herbert I. Gross 11

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**Notes In terms of our study of functions, notice that the**

total number of marbles, T, is a function of the number of marbles Erin has and the number of marbles Bill has. So, we may write that… T = f(E,B). In this form, we see that the domain of f is 2-dimensional. Hence, to find a unique solution we need two constraints that involve E and B. This is precisely what the given system of equations has provided for us. next next © Herbert I. Gross 12

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Example 2 Erin has five times as many marbles as Bill. Together they have 180 marbles. How many marbles does Erin have? Answer: 150 next next next © Herbert I. Gross 13

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Algebraic Solution If we let B denote the number of marbles Bill has, knowing that E equals 5B tells us that 5B denotes the number of marbles Erin has. Since they have a total of 180 marbles, we know that B + 5B = 180, or 6B = 180. Hence, B = 30 which means that Bill has 30 marbles and Erin has 5B or 150 marbles. next next next © Herbert I. Gross 14

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**The phrase “5 times as many” indicates a multiplication problem.**

Notes The constraint is that E = 5B not E = 5 + B. We would have written 5 + B if the problem had said “Erin has 5 more marbles than Bill”. The phrase “5 times as many” indicates a multiplication problem. next © Herbert I. Gross 15

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**Notes In the trial-and-error….. method, we picked any**

number to represent the number of marbles Bill had, whereupon five times this number represented the number of marbles Erin had. Our job was to guess the number that would make the total number of marbles equal 180. In the language of algebra, we let B, rather than a specific number, denote the number of marbles Bill has. The fact that Erin has five times as many marbles as Bill can now be represented by the constraint E = 5B. next next © Herbert I. Gross 16

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**…and to solve for B, in the top equation, we replace E by 5B.**

The basic formula, just as it was in Example 1, is still B + E = T; but this time we were given two different constraints; namely, that T = 180 and E = 5B. That is, this time the system of constraints is… Notes E + B = 180 5B + B = 180 E = 5B …and to solve for B, in the top equation, we replace E by 5B. next next next © Herbert I. Gross 17

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**Example 3 The only money Mary has with her is**

40 coins consisting of dimes and quarters, and she has three times as many quarters as dimes. How much money does she have with her? Answer: $8.50 next next next © Herbert I. Gross 18

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Algebraic Solution To translate this word problem into the language of algebra, we might start by letting d denote the number of dimes she has, q denote the number of quarters she has, and t denote the total number of coins she has. d = # of dimes q = # of quarters t = total number of coins next next © Herbert I. Gross 19

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Algebraic Solution She has three times as many quarters as dimes. That is, to find the number of quarters, we multiply the number of dimes by 3. In the language of algebra, this is written as… q = 3d We also know that since the total number of coins is 40… q + d = 40 next next next © Herbert I. Gross 20

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**Therefore, our system of equations is…**

Algebraic Solution Therefore, our system of equations is… q + d = 40 3d + d = 40 q = 3d We replace q in the top equation by its value in the bottom equation … Hence, 4d = 40 and d = 10; and 3d = 30. So she has 10 dimes (which is $1) and 30 quarters (which is $7.50) for a total of $8.50. next next next © Herbert I. Gross 21

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A strategy that is often used in problem solving is to see which quantity we are asked to find and then choosing that to be the variable. Notes In this problem, we are asked to find V (the total value of the money Mary has). This will require that we know the number of dimes (d) and the number of quarters (q). Thus, our problem contains the 3 “unknowns”… d, q, and V. next next next © Herbert I. Gross 22

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**Notes Since each dime is….. worth 10 cents, the value,**

in cents, of d dimes is 10d; and since each quarter is worth 25 cents, the value of in cents of q quarters is 25q. To find the total value (V) of the coins we add the value of the dimes and the value of the quarters. Thus… V = 10d + 25q next next © Herbert I. Gross 23

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**Therefore, our three constraints are…**

Notes Therefore, our three constraints are… V = 10d + 25q q + d = 40 q = 3d …and we already know from our previous solution that d = 10 and q = 30. next next © Herbert I. Gross 24

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Notes V = 10d + 25q q + d = 40 q = 3d So in the top equation in our system we replace d by 10 and q by 30 to obtain… V = 10(10) + 25(30) = = 850 (cents) next next © Herbert I. Gross 25

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**Notes Notice the importance of our adjective/noun theme.**

For example, d modifies the number of dimes, but 10d modifies the value of the dimes. In order to help ensure that you don’t confuse these two adjectives, notice that in any equation we assume the nouns are the same on both sides. 10 dimes = number of coins 10 dimes = $1.00 in value next next © Herbert I. Gross 26

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Notes Do not confuse t (the number of coins) with V (the value of the coins). For example, d + q represents the number of coins (t) while 10d + 25q represents the value of the coins (V). Thus, it would not be correct to write such things as d + q = 850 because the left side of the equation is modifying the number of coins but the right hand side is modifying the value of the coins in cents. next © Herbert I. Gross 27

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Notes So even before beginning to solve an equation, it’s a good strategy to check to make sure that the numbers (adjectives) on both sides of the equation are modifying the same noun. next © Herbert I. Gross 28

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Notes V = 10d + 25q There is sometimes a tendency to write the equation in the form V = 0.10d q because each dime is worth $0.10 and each quarter is worth $0.25. There is nothing wrong with this as long as we remember that if one side of the equation modifies dollars the other side must also modify dollars. In other words, if we replace the top equation by V = 0.10d q, the answer we get represents V dollars. next next next © Herbert I. Gross 29

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Notes A common strategy when we have an equation such as V = 0.10d q is to eliminate the decimals by multiplying both sides of the equation by 100 to obtain the equivalent equation 100V = 10d + 25q. In this case, the value of V is 8.50 and it modifies dollars. next next © Herbert I. Gross 30

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**“10d = the value of d dimes (in cents)”.**

Notes In any event it would be incorrect to simply write either V = 850 or V = 8.5 because in this form the answer doesn’t tell us which noun the adjective is modifying. In a similar way, it is “sloppy” to write something like “d = dimes”. Rather we should write “d = the number of dimes”, and if we are talking about the value of the dimes, we should write something like “10d = the value of d dimes (in cents)”. next next © Herbert I. Gross 31

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Of course we are not obligated to solve explicitly for the quantity we are asked to find. Consider, for example, the following problem… Example 4 Erin has four times as many marbles as Bill and Tom has seven more marbles than Erin. Altogether they have 367 marbles. How many marbles does Tom have? Answer: 167 next next next next © Herbert I. Gross 32

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**Preface to the Solution**

In doing the problem the traditional way, it might seem natural to start by letting T represent the number of marbles Tom has and then expressing the number of marbles Erin and Bill have in terms of T. However, this would lead to a lot of paraphrasing because in the form the problem is worded it’s simpler to start by letting B denote the number of marbles Bill has and then expressing the other two amounts in terms of B. next next © Herbert I. Gross 33

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**Preface to the Solution**

For example, suppose we let T, B, and E represent respectively the number of marbles Tom, Bill and Erin have. If we try to write these amounts in terms of T, we see that… …since T = E + 7 E = T – 7 since E = 4B and E = T – 7 we see that 4B = T – 7, and therefore… B = 1/4 (T – 7) Hence, the fact that T +E + B = 367 would lead to the equation… T + T – 7 + 1/4 (T – 7) = 367 next next next next © Herbert I. Gross 34

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**On the other hand, if we express the other amounts in terms of B, **

we see that… E = 4B and since E = 4B and since T = E + 7, we see that… T = 4B + 7 In this case, T + E + B = 367 is replaced by the simpler equation… (4B + 7) + 4B + B = 367 Using the above equation, it is relatively simple to find the value of B; and once we know the value of B, it is easy to compute 4B and 4B + 7. next next next next © Herbert I. Gross 35

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**And then dividing both sides of the equation by 9, we obtain…**

Algebraic Solution In the equation (4B + 7) + 4B + B = 367, we add the B’s and see that… 9B + 7 = 367 Subtracting 7 from both sides of the equation we get… 9B = 360 And then dividing both sides of the equation by 9, we obtain… B = 40 next next next © Herbert I. Gross 36

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**Algebraic Solution If B = 40, then 4B = 160 and 4B + 7 = 167.**

Therefore, Bill has 40 marbles, Erin has 160 marbles and Tom has167 marbles. As a check, we see that… = 367 next next next © Herbert I. Gross 37

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**Notes In terms of function notation, if we let N denote the total**

number of marbles then N = f(B,E,T). We thus have 3 variables in the domain of f; which means that we need to have 3 constraints in order to have a unique solution. Thus, in the language of functions we might write… B + E + T = 367 N = f(B,E,T) where… E = 4B T = E + 7 next next © Herbert I. Gross 38

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**Notes While we picked N to stand for the total number of marbles,**

the important thing is that we could have used any letter except T. The reason is that we’ve already used T to stand for Tom’s amount, and it would be very confusing if the same letter was used to name two different variables. Some mathematicians will use the same letter twice if one is capitalized and the other isn’t. For example, we could let T stand for Tom’s amount and we could let t stand for the total amount; but this can cause confusion for the beginning student. next next © Herbert I. Gross 39

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**Notes Some people get confused and don’t know whether to**

write E = 4B or B = 4E. If this happens to you, a good strategy is to use specific numbers. That is, pick any number to represent Bill’s amount, and then four times this number will be Erin’s amount. For example, if Bill has 2 marbles, Erin must have 8. That is: to find the amount Erin had, we multiplied the amount Bill had by 4, or more symbolically, E = 4B. next next © Herbert I. Gross 40

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**This will be illustrated in our solution for the next example.**

Notes This is another reason why Lesson 23 is important. Namely, if you get confused when you try to figure out the relationships between variables, replace the variables by specific numbers (just as we do when we use trial and error methods) and see what happens then. Once you get a feeling for the pattern, it’s usually much easier to work with the letters themselves. This will be illustrated in our solution for the next example. next next © Herbert I. Gross 41

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Example 5 An antique automobile travels from A to B at a constant rate of 10 miles per hour and makes the return trip at a constant rate of 15 miles per hour. If the round trip took 20 hours what is the distance between A and B? Answer: 120 miles next next next © Herbert I. Gross 42

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A Solution One way to begin a problem like this is to generalize what we did in the trial-and-error method. Namely, we can pick a number of miles that is a multiple of both 10 and 15. One such number is 30. If the distance between A and B had been 30 miles, it would have taken 2 hours to travel this distance at 15 mph and 3 hours at 10 mph. next © Herbert I. Gross 43

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A Solution Thus, the antique car would have traveled a round trip of 60 miles in 5 hours. In other words, if the distance between A and B had been 30 miles, the round trip would have only taken 5 hours. Since there are four 5 hour periods in a 20 hour period, the distance between A and B must have been 4 times 30 miles (or 120 miles). next next © Herbert I. Gross 44

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An Algebraic Solution To use algebra, we stop guessing what the distance is, and instead, we denote it by a letter of the alphabet. For example, we let d stand for the distance between A and B. Since the time it takes is found by dividing the distance (d) by the speed, we know that it takes… d/10 hours for the object to get from A to B, and that it takes… d/15 hours for the object to get from B to A. next next next © Herbert I. Gross 45

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An Algebraic Solution Since we also know that the round trip took 20 hours, the equation we have to solve is… d/10 + d/15 = 20 To clear the equation of fractions we multiply both sides of our equation by 30 (or any other common multiple of 10 and 15) to obtain… 30(d/10 + d/15) = 30(20) or 3d + 2d = 600 Hence… 5d = 600 and d = 120 next next next © Herbert I. Gross 46

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Notes The point is that the trial-and-error approach helped us to construct the formula; and once we understood the formula, it was not too hard to translate it into an equation that we were able to solve. next © Herbert I. Gross 47

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**Notes Keep in mind that this wasn’t the only algebraic way to solve**

this problem. One other way is to use an indirect approach and find the time it took to go one way. For example, if we knew the time it took to go one way, we could multiply this by the speed of the automobile, and thus find the distance between A and B. So, for example, we might let t denote the time it took to get from A to B at 10 miles per hour. next next © Herbert I. Gross 48

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**Notes At a speed of 10 miles per hour, in t hours the automobile will**

have traveled … 10t miles Since the round trip took 20 hours and one way took t hours, the remainder of the trip took (20 – t) hours. Therefore, at a speed of 15 miles per hour the automobile will have traveled… 15(20 – t) miles next next © Herbert I. Gross 49

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**Notes Since the distance from A to B is the same as the**

distance from B to A, we want to find the value of t for which… Notes 10t = 15(20 – t) That is, 10t = 300 – 15t; or 25t = 300; or t = 12 (and 20 – t = 8). Since t denotes the time the car went at 10 miles per hour, we see that… when t = 12, 10t = 120. Hence, the distance between A and B is 120 miles. next next next next © Herbert I. Gross 50

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**illustrate several methods for solving each problem.**

Notes Keep in mind that we look for the easiest way to solve problems. Sometimes the easiest way to solve a word problem is by trial and error. Other times, the easiest way is to use the algebraic method. The best thing is to understand both ways so that if you have trouble using one method you can still use the other. As you work out the rest of the examples in this Lesson, use whichever method you find easiest for you. In our solutions we’ll illustrate several methods for solving each problem. next next © Herbert I. Gross 51

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**Example 6 John is now 3 times as old as Bill.**

Seven years from now he’ll only be twice as old as Bill. How old is Bill now? Answer: 7 years old next next next © Herbert I. Gross 52

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An Algebraic Solution The problem asks us to find how old Bill is now. So we’ll let B stand for Bill’s present age and try to paraphrase the problem into an equation in which B is the “unknown”. We are told that John is now 3 times as old as Bill. Therefore, if we let J stand for John’s present age, we know that… J = 3B next next © Herbert I. Gross 53

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An Algebraic Solution To find your age seven years from now, all you have to do is add 7 years to your present age. So if Bill’s present age is B, in seven years it will be B + 7, and if John’s age now is J, in seven years it will be J + 7. Thus, the constraint that in 7 years John will be twice as old as Bill translates into the equation… J + 7 = 2(B + 7) next next © Herbert I. Gross 54

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**An Algebraic Solution We may then replace J in the equation**

J + 7 = 2(B + 7) by its value in the equation J = 3B to obtain… J + 7 = 2(B + 7) 3B + 7 = 2(B + 7) 3B + 7 = 2B + 14 3B = 2B + 7 next next © Herbert I. Gross 55

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An Algebraic Solution Subtracting 2B from both sides of the equation 3B = 2B + 7 tells us that B = 7; and since B represents Bill’s present age we see that Bill is now 7 years old. And since John is 3 times as old, he is 21 As a check, if Bill is now 7 and John is 21, John is 3 times as old as Bill (i.e., J = 3B); and in 7 years Bill will be 14 and John will be 28. Thus, John will then be twice as old as Bill (i.e., J + 7 = 2(B + 7)). next next © Herbert I. Gross 56

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**Applying the Corn Bread Solution**

Though it might seem a bit advanced, the fact is that the corn bread can be used early in the elementary grades to solve problems that are usually considered to be algebraic. next © Herbert I. Gross 57

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**Cornbread Solution J = 3B**

The above algebraic solution translates into the following corn bread solution. B = Bill’s Present Age B B B = John’s Present Age next next next © Herbert I. Gross 58

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**Cornbread Solution B 7 B B B 7 B 7 B 7 = Bill’s Age in 7 years**

= John’s Age in 7 years B 7 B 7 = Twice Bill’s Age in 7 years And since in 7 years John will be twice as old as Bill, we may equate the sizes of the bottom two corn breads. That is… 3B + 7 = 2B +14 next next next next © Herbert I. Gross 59

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**This model is quite similar to the corn bread model.**

The Pan Balance Model It is the custom in some texts to introduce algebraic equations in the less threatening form of a pan balance, with one pan on each side of the equal sign. One size weight would represent “x” and another size weight would represent the number of units. This model is quite similar to the corn bread model. next © Herbert I. Gross 60

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**The Pan Balance Model 3x + 7 = 2x + 14 █ █ █ | | | | | | | **

More specifically, if we used, say, █ to represent x and | to represent one, an equation such as 3x + 7 = 2x + 14 would appear as… █ █ █ | | | | | | | 3x = x + 14 █ █ | | | | | | | | | | | | | | Students would then be encouraged to remove equal weights from both sides of the scale so that the scale would remain balanced. next next next next © Herbert I. Gross 61

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The Pan Balance Model Thus, they might begin by removing █ █ from both sides of the scale to obtain… █ █ █ | | | | | | | = █ █ | | | | | | | | | | | | | | Then they would remove | | | | | | | from each side of the scale to obtain… | | | | | | | = █ | | | | | | | | | | | | | | next next next next © Herbert I. Gross 62

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Note Notice that we obtain the pan balance model from the corn bread model simply by replacing….. B by █ and 7 by | | | | | | | next next next © Herbert I. Gross 63

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**Trial and Error Solution**

Whatever age we pick for Bill, we multiply it by 3 to find John’s present age. We then add 7 to each of their ages to find out how old they are in seven years. If we are extremely lucky, our first guess will satisfy the conditions in the problem. Since we don’t expect to be that lucky, we may approach the problem more systematically by assuming that Bill is now 1 year old. We can then add 1 year to his present age every time our guess is incorrect. next next next © Herbert I. Gross 64

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**Trial and Error Solution**

In terms of a chart… Bill’s Age Now John’s Age Now Bill’s Age in 7 Years John’s Age 1 3 8 10 2 6 9 13 3 9 10 16 4 12 11 19 5 15 12 22 6 18 13 25 7 21 14 28 8 24 15 31 next next © Herbert I. Gross 65

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**Trial and Error Solution**

The highlighted row shows John’s age 7 years from now is twice Bill’s age. From the chart, we see this happens in the row for which Bill is now 7 and John is 21. Bill’s Age Now John’s Age Now Bill’s Age in 7 Years John’s Age 5 15 12 22 6 18 13 25 7 21 14 28 8 24 31 7 21 14 28 7 21 14 28 next next © Herbert I. Gross 66

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Example 7 The length of a rectangle is 3 inches longer than twice the width of the rectangle. If the perimeter of the rectangle is 54 inches, what is the length of the rectangle? Answer: 19 inches next next next © Herbert I. Gross 67

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An Algebraic Solution Let W stand for the width of the rectangle, and let L stand for the length. Then the relationship between L and W is… L = 2W + 3 next next © Herbert I. Gross 68

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**For example, suppose the width had been**

Note If you have trouble getting the formula L = 2W + 3, use actual numbers, just as you would in the trial-and-error method. For example, suppose the width had been 5 inches. Then twice the width would be 10 inches, and 3 inches more than that would be 13 inches. If the width had been 15 inches, twice the width would have been 30 inches; and 3 inches more than that would have been 33 inches. next next © Herbert I. Gross 69

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P = 2(L + W) An Algebraic Solution So how did we get from the width to the length each time? We multiplied the width (that is, W) by 2 (that is, 2W) and then added 3 inches (that is, 2W + 3). Next we have to remember what the perimeter of a rectangle is. Recall that it’s the distance around the rectangle. To find the distance around a rectangle, we add the length and the width (that represents half the distance around the rectangle), and then we multiply by 2. next next © Herbert I. Gross 70

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P = 2(L + W) An Algebraic Solution If we let P stand for the perimeter of the rectangle, the formula is… P = 2(L + W) We are told that the perimeter is 54 inches. Hence, we may replace P by 54 in formula 54 P = 2(L + W) From the equation L = 2W + 3, we see that we may replace L by 2W + 3 wherever L appears. 54 = 2( L W) [2W + 3] next next next © Herbert I. Gross 71

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**An Algebraic Solution In this way, formula P = 2(L + W) becomes...**

54 = 2([2W + 3] + W) 54 = 4W W 54 = 6W + 6 48 = 6W 8 = W next next next next next © Herbert I. Gross 72

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P = 2(L + W) An Algebraic Solution Since W denotes the width of the rectangle, we see that the width of the rectangle is 8 inches. The problem, however, asks us to find the length of the rectangle. To do this we simply replace W by 8 in the equation L = 2W + 3, and we see that L = 2( 8 ) + 3 = 19 (inches). As a check, 19 inches + 8 inches = 27 inches; and twice 27 inches is 54 inches. next next © Herbert I. Gross 73

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**Pan Balance Solution Again, if you preferred, you could let**

P = 2(L + W) Pan Balance Solution Again, if you preferred, you could let W stand for the width and | stand for 1 inch. Then 3 inches more than twice the width would look like , and the width would be W | | | That is, the length would be… . W W | | | The length plus the width would be… W | | | or… W | | | next next next © Herbert I. Gross 74

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**Pan Balance Solution Twice the sum of the length and width would be…**

P = 2(L + W) Pan Balance Solution Twice the sum of the length and width would be… W | | | or… W | | | Hence, our equation would now look like… W | | | = 54 next next next © Herbert I. Gross 75

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P = 2(L + W) Pan Balance Solution W | | | = 54 In terms of our equation above, it is not tedious to write six tally marks. However, there are times when we deal with large numbers of tally marks. In such a case, rather than write, say, 75 tally marks we might write . Thus, the above equation might be written as… 75 W = 54 6 next next © Herbert I. Gross 76

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**…and, finally, since 6W = 48, we see that W = 8**

P = 2(L + W) Pan Balance Solution Since we can take away the same amount from both sides of our equation without changing the equality, we may think of 54 as being In this way the equation may be written as… W = 6 48 We may then take away 6 from both sides to get… W = 48 …and, finally, since 6W = 48, we see that W = 8 next next © Herbert I. Gross 77

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**Trial-and-Error Solution**

P = 2(L + W) Trial-and-Error Solution In general, we could guess any number of inches as the width; then 3 more than twice this guess would be the length. We could then add the length and the width, and finally, we would double this to find the perimeter. We’d keep making guesses until the perimeter was 54. next © Herbert I. Gross 78

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**Trial-and-Error Solution**

P = 2(L + W) Trial-and-Error Solution For example, we might start with 5 as the value of the width (W) of the rectangle. In this case the length (2W + 3) is 13. Hence, the length plus the width is 18; and therefore the perimeter is 36. Since we are told that the perimeter is 54 inches, we know that W must be greater than 5 next next © Herbert I. Gross 79

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**Trial and Error Solution**

P = 2(L + W) Trial and Error Solution Thus, we may increase W by 1 (inch) each time as illustrated in the chart below… W 2W L L + W P = 2(L + W) 5 10 13 18 36 6 12 15 21 42 7 14 17 24 48 8 16 19 27 54 8 16 19 27 54 9 18 21 30 60 …and the shaded row indicates the row in which P = 54. This row tells us that W = 8 and L = 19. next next next © Herbert I. Gross 80

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Example 8 Mary has twice as much money as Jane. If Mary had $4 more and Jane had $3 less, Mary would then have 4 times as much money as Jane. How much money does Jane have? Answer: $8 next next next © Herbert I. Gross 81

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An Algebraic Solution This time we let J stand for the amount of money Jane has, and we let M stand for the amount of money Mary has. Since Mary has twice as much money as Jane, we know that… M = 2J $4 more than Mary has would be written as M + 4 and $3 less than Jane has would be written as J – 3. Since Mary would then have four times as much money as Jane, we know that… M + 4 = 4(J – 3) next next © Herbert I. Gross 82

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An Algebraic Solution We now can replace M by 2J whenever we want in this problem. Hence, the equation M + 4 = 4(J – 3) becomes… M + 4 = 4(J – 3) 2J + 4 = 4(J – 3) 2J + 4 = 4J – 12 4 = 2J – 12 16 = 2J 8 = J next next © Herbert I. Gross 83

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**Since $20 is 4 times as much as $5, we see that our answer is correct.**

An Algebraic Solution J = 8 tells us that Jane has $8; and, hence, that Mary has $16. As a check, if Mary had $4 more, she would have $20; and if Jane had $3 less, she would have $5. Since $20 is 4 times as much as $5, we see that our answer is correct. next next © Herbert I. Gross 84

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**Trial and Error Solution**

The wording of the problem suggests that Jane has at least $4 (otherwise she couldn't have $3 less). And since Mary has twice as much money as Jane, if Jane has $4 Mary has $8. J M M + 4 J – 3 4 8 12 1 Thus, we can make a chart similar to the following one. 5 10 14 2 6 12 16 3 7 14 18 4 8 16 20 5 next next © Herbert I. Gross 85

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**We see that M + 4 is four times**

Trial and Error Solution J M M + 4 J – 3 4 8 12 1 5 10 14 2 6 16 3 7 18 20 8 16 20 5 We see that M + 4 is four times as large as J – 3; and that in this row, J = 8 and M = 16. next next next © Herbert I. Gross 86

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Important! It is very important to read the problem carefully so that we are sure that our translation from English into Algebra is accurate. In particular, a small change in wording can greatly change the meaning of a problem. This is illustrated in Examples 9 and 10. next next © Herbert I. Gross 87

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Example 9 A piece of string, 45 inches long, is cut into 3 pieces. The second piece is twice as long as the first piece and the third piece is 5 inches longer than the second piece. How long is the first piece? Answer: 8 inches next next next © Herbert I. Gross 88

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**5 inches longer than the first piece.**

Example 10 A piece of string, 45 inches long, is cut into 3 pieces. The second piece is twice as long as the first piece and the third piece is 5 inches longer than the first piece. How long is the first piece? Answer: 10 inches next next next © Herbert I. Gross 89

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**5 inches longer than the second piece.**

Notice that Examples 9 and 10 are word for word the same except for the two words in italics. A piece of string, 45 inches long, is cut into 3 pieces. The second piece is twice as long as the first piece and the third piece is 5 inches longer than the second piece. Example 9 A piece of string, 45 inches long, is cut into 3 pieces. The second piece is twice as long as the first piece and the third piece is 5 inches longer than the first piece. Example 10 next next © Herbert I. Gross 90

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**…while in Example 10 the three relationships are…**

Yet the change of just one word completely changes the problem. Namely, if we let f denote the length of the first piece; s, the length of the second piece; and t, the length of the third piece; we see that the three relationships in Example 9 are… The Solution f + s + t = 45 s = 2f t = s + 5 …while in Example 10 the three relationships are… f + s + t = 45 s = 2f t = f + 5 next next © Herbert I. Gross 91

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The Solution except for the fact that s and f have been interchanged in the third equation of the two systems. Both systems are identical f + s + t = 45 s = 2f t = s + 5 f + s + t = 45 s = 2f t = f + 5 t = s + 5 t = f + 5 next next © Herbert I. Gross 92

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**The Solution Notice that to solve Example 10, we need**

only to replace s by 2f and t by f + 5 in the equation f + s + t = 45 to obtain… The Solution f + s t = 45 f + (2f) + (f + 5) = 45 1f + 2f + 1f + 5 = 45 4f + 5 = 45 4f = 40 f = 10 next next © Herbert I. Gross 93

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**The Solution But to solve Example 9, if we replace s and t in**

f + s + t = 45 by their values in s = 2f and t = s + 5, we get… The Solution f + s t = 45 f + (2f) + (s + 5) = 45 (s 1f + 2f + (2f + 5) = 45 1f + 2f + 2f + 5 = 45 5f + 5 = 45 5f = 40 f = 8 next next © Herbert I. Gross 94

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**Additional practice is left for the exercise set.**

In summary, the difference in just one word in Examples 9 and 10 is the difference between the lengths of the three pieces being 8 inches, 16 inches and 21 inches rather than 10 inches, 20 inches and 15 inches. Additional practice is left for the exercise set. next next © Herbert I. Gross 95

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