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The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard A. Medeiros next Lesson 24

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Application to Word Problems Application to Word Problems © 2007 Herbert I. Gross next

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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 well look at some others. © 2007 Herbert I. Gross next

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Erin and Bill together have 500 marbles. Erin has 50 marbles more than Bill. Answer: 225 © 2007 Herbert I. Gross next How many marbles does Bill have? Example 1 next

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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. © 2007 Herbert I. Gross next Algebraic Solution

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© 2007 Herbert I. Gross next 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

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The above two constraints yield the system of equations… And if we add these two equations… © 2007 Herbert I. Gross next E + B = 500 E – B = 50+ 2E = 550 And then divide both sides of the resulting equation by 2 … 2 E = 275 next

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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… © 2007 Herbert I. Gross next E + B = 500E – B = B = 500 B = 500 – 275 B = – B = – 50 = B 225 = B next In either case, Bill has 225 marbles. next

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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 © 2007 Herbert I. Gross next

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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 © 2007 Herbert I. Gross next E – B = 50(500 – B) And solving the above equation, we see that B = 225. next 500 – 2B = – 50 = 2B 450 = 2B 225 = B

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Using E and B rather than using only either E or B, often helps us to organize our thinking. Notes © 2007 Herbert I. Gross next Namely, the more steps we try to keep track of in our head, the easier it becomes to make a mistake.

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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). Notes © 2007 Herbert I. Gross next 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.

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Erin has five times as many marbles as Bill. Together they have 180 marbles. Answer: 150 © 2007 Herbert I. Gross next How many marbles does Erin have? Example 2 next

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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. © 2007 Herbert I. Gross next Algebraic Solution Hence, B = 30 which means that Bill has 30 marbles and Erin has 5B or 150 marbles. next Since they have a total of 180 marbles, we know that B + 5B = 180, or 6B = 180.

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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. Notes © 2007 Herbert I. Gross next

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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. Notes © 2007 Herbert I. Gross next 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.

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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 © 2007 Herbert I. Gross next …and to solve for B, in the top equation, we replace E by 5B. E + B = 180 E = 5B 5B + B = 180

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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. Answer: $8.50 © 2007 Herbert I. Gross next How much money does she have with her? Example 3 next

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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. © 2007 Herbert I. Gross next Algebraic Solution d = # of dimesq = # of quarters t = total number of coins next

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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. © 2007 Herbert I. Gross next Algebraic Solution 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

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Therefore, our system of equations is… © 2007 Herbert I. Gross next We replace q in the top equation by its value in the bottom equation … q + d = 40 q = 3d 3d + d = 40 Algebraic Solution 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.

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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. © 2007 Herbert I. Gross next In this problem, we are asked to find V (the total value of the money Mary has). Notes 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

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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. © 2007 Herbert I. Gross next To find the total value (V) of the coins we add the value of the dimes and the value of the quarters. Thus… Notes V = 10d + 25q

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Therefore, our three constraints are… © 2007 Herbert I. Gross next …and we already know from our previous solution that d = 10 and q = 30. Notes V = 10d + 25q q + d = 40 q = 3d

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So in the top equation in our system we replace d by 10 and q by 30 to obtain… © 2007 Herbert I. Gross next V = 10(10) + 25(30) next Notes V = 10d + 25q q + d = 40 q = 3d = = 850 (cents)

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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 dont confuse these two adjectives, notice that in any equation we assume the nouns are the same on both sides. © 2007 Herbert I. Gross next Notes 10 dimes = number of coins 10 dimes = $1.00 in value

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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. © 2007 Herbert I. Gross next Notes

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So even before beginning to solve an equation, its a good strategy to check to make sure that the numbers (adjectives) on both sides of the equation are modifying the same noun. © 2007 Herbert I. Gross next Notes

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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. © 2007 Herbert I. Gross next Notes V = 10d + 25q In other words, if we replace the top equation by V = 0.10d q, the answer we get represents V dollars. next 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.

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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. © 2007 Herbert I. Gross next Notes In this case, the value of V is 8.50 and it modifies dollars.

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In any event it would be incorrect to simply write either V = 850 or V = 8.5 because in this form the answer doesnt tell us which noun the adjective is modifying. © 2007 Herbert I. Gross next Notes 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).

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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… © 2007 Herbert I. Gross next 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

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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. © 2007 Herbert I. Gross next Preface to the Solution However, this would lead to a lot of paraphrasing because in the form the problem is worded its 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

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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… © 2007 Herbert I. Gross next …since T = E + 7 next since E = 4B and E = T – 7 we see that 4B = T – 7, and therefore… Preface to the Solution E = T – 7 B = 1 / 4 (T – 7) Hence, the fact that T +E + B = 367 would lead to the equation… T + T – / 4 (T – 7) = 367 next

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On the other hand, if we express the other amounts in terms of B, we see that… © 2007 Herbert I. Gross next and since E = 4B and since T = E + 7, we see that… E = 4B T = 4B + 7 In this case, T + E + B = 367 is replaced by the simpler equation… (4B + 7) + 4B + B = 367 next 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.

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In the equation (4B + 7) + 4B + B = 367, we add the Bs and see that… © 2007 Herbert I. Gross next Subtracting 7 from both sides of the equation we get… 9B + 7 = 367 Algebraic Solution And then dividing both sides of the equation by 9, we obtain… 9B = 360 B = 40

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If B = 40, then 4B = 160 and 4B + 7 = 167. © 2007 Herbert I. Gross next Therefore, Bill has 40 marbles, Erin has 160 marbles and Tom has167 marbles. Algebraic Solution As a check, we see that… = 367

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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… © 2007 Herbert I. Gross next N = f(B,E,T) where… next Notes B + E + T = 367 E = 4B T = E + 7

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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 weve already used T to stand for Toms amount, and it would be very confusing if the same letter was used to name two different variables. © 2007 Herbert I. Gross next Some mathematicians will use the same letter twice if one is capitalized and the other isnt. For example, we could let T stand for Toms amount and we could let t stand for the total amount; but this can cause confusion for the beginning student. next Notes

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Some people get confused and dont 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 Bills amount, and then four times this number will be Erins amount. © 2007 Herbert I. Gross next 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. Notes

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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, its usually much easier to work with the letters themselves. © 2007 Herbert I. Gross next This will be illustrated in our solution for the next example. Notes

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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. Answer: 120 miles © 2007 Herbert I. Gross next If the round trip took 20 hours what is the distance between A and B? Example 5 next

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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. © 2007 Herbert I. Gross next A Solution

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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. © 2007 Herbert I. Gross next A Solution 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

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To use algebra, we stop guessing what the distance is, and instead, we denote it by a letter of the alphabet. © 2007 Herbert I. Gross next An Algebraic Solution 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… next 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.

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© 2007 Herbert I. Gross next An Algebraic Solution Since we also know that the round trip took 20 hours, the equation we have to solve is… next 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

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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. © 2007 Herbert I. Gross next Notes

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Keep in mind that this wasnt 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. © 2007 Herbert I. Gross next Notes 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.

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At a speed of 10 miles per hour, in t hours the automobile will have traveled … © 2007 Herbert I. Gross next Notes 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… 10t miles 15(20 – t) miles next

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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… © 2007 Herbert I. Gross next Notes 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… 10t = 15(20 – t) 10t = 120. next Hence, the distance between A and B is 120 miles. when t = 12,

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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. © 2007 Herbert I. Gross next Notes As you work out the rest of the examples in this Lesson, use whichever method you find easiest for you. In our solutions well illustrate several methods for solving each problem. next

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John is now 3 times as old as Bill. Seven years from now hell only be twice as old as Bill. Answer: 7 years old © 2007 Herbert I. Gross next How old is Bill now? Example 6 next

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The problem asks us to find how old Bill is now. So well let B stand for Bills present age and try to paraphrase the problem into an equation in which B is the unknown. © 2007 Herbert I. Gross next An Algebraic Solution We are told that John is now 3 times as old as Bill. Therefore, if we let J stand for Johns present age, we know that… J = 3B next

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To find your age seven years from now, all you have to do is add 7 years to your present age. So if Bills present age is B, in seven years it will be B + 7, and if Johns age now is J, in seven years it will be J + 7. © 2007 Herbert I. Gross next An Algebraic Solution 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

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We may then replace J in the equation J + 7 = 2(B + 7) by its value in the equation J = 3B to obtain… © 2007 Herbert I. Gross next An Algebraic Solution 3B + 7 = 2B + 14 J + 7 = 2(B + 7) next 3B + 7 = 2(B + 7) 3B = 2B + 7

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Subtracting 2B from both sides of the equation 3B = 2B + 7 tells us that B = 7; and since B represents Bills present age we see that Bill is now 7 years old. And since John is 3 times as old, he is 21 © 2007 Herbert I. Gross next An Algebraic Solution 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

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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. Applying the Corn Bread Solution © 2007 Herbert I. Gross next

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The above algebraic solution translates into the following corn bread solution. © 2007 Herbert I. Gross next Cornbread Solution next B B = Bills Present Age = Johns Present Age BB J = 3B next

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© 2007 Herbert I. Gross next Cornbread Solution next B B = Bills Age in 7 years = Johns Age in 7 years BB next B7 = Twice Bills 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 7 7 B7

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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. The Pan Balance Model © 2007 Herbert I. Gross next

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More specifically, if we used, say, to represent x and | to represent one, an equation such as 3x + 7 = 2x + 14 would appear as… The Pan Balance Model © 2007 Herbert I. Gross next Students would then be encouraged to remove equal weights from both sides of the scale so that the scale would remain balanced. 3x + 7 = 2x + 14 | | | | | | | | | | | | | | | | | | | | | next

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| | | | | | | = The Pan Balance Model © 2007 Herbert I. Gross next 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

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Notice that we obtain the pan balance model from the corn bread model simply by replacing….. © 2007 Herbert I. Gross next Note B by and by | | | | | | | 7 next

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Since we dont 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. © 2007 Herbert I. Gross next Trial and Error Solution Whatever age we pick for Bill, we multiply it by 3 to find Johns present age. next 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.

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In terms of a chart… © 2007 Herbert I. Gross next Trial and Error Solution next Bills Age NowJohns Age Now Bills Age in 7 Years Johns Age in 7 Years

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The highlighted row shows Johns age 7 years from now is twice Bills age. © 2007 Herbert I. Gross next Trial and Error Solution next Bills Age NowJohns Age Now Bills Age in 7 Years Johns Age in 7 Years From the chart, we see this happens in the row for which Bill is now 7 and John is

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The length of a rectangle is 3 inches longer than twice the width of the rectangle. Answer: 19 inches © 2007 Herbert I. Gross next If the perimeter of the rectangle is 54 inches, what is the length of the rectangle? Example 7 next

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Let W stand for the width of the rectangle, and let L stand for the length. © 2007 Herbert I. Gross next An Algebraic Solution L = 2W + 3 next Then the relationship between L and W is…

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If you have trouble getting the formula L = 2W + 3, use actual numbers, just as you would in the trial-and-error method. © 2007 Herbert I. Gross next Note 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.

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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). © 2007 Herbert I. Gross next An Algebraic Solution next Next we have to remember what the perimeter of a rectangle is. Recall that its 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. P = 2(L + W)

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If we let P stand for the perimeter of the rectangle, the formula is… © 2007 Herbert I. Gross next An Algebraic Solution next P = 2(L + W) We are told that the perimeter is 54 inches. Hence, we may replace P by 54 in formula 54 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 P = 2(L + W)

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In this way, formula P = 2(L + W) becomes... © 2007 Herbert I. Gross next An Algebraic Solution next 54 = 2([2W + 3] + W) 54 = 4W W 54 = 6W = 6W 8 = W next P = 2(L + W)

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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). © 2007 Herbert I. Gross next An Algebraic Solution next As a check, 19 inches + 8 inches = 27 inches; and twice 27 inches is 54 inches. P = 2(L + W)

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© 2007 Herbert I. Gross next Pan Balance Solution next Again, if you preferred, you could let stand for the width and | stand for 1 inch. Then 3 inches more than twice the width would look like That is, the length would be… P = 2(L + W) W WW | | |, and the width would be W. The length plus the width would be… WW W WW or… W WW

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© 2007 Herbert I. Gross next Pan Balance Solution next Twice the sum of the length and width would be… P = 2(L + W) Hence, our equation would now look like… or… W WW | | | W WW W WW WWW W WW WWW = 54

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© 2007 Herbert I. Gross next Pan Balance Solution 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 next P = 2(L + W). Thus, the above equation might be written as… W WW | | | WWW = W WWWWW = 54 6

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© 2007 Herbert I. Gross next 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… next P = 2(L + W) We may then take away 6 from both sides to get… W WWWWW = 6486 W WWWWW = …and, finally, since 6W = 48, we see that W = 8

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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. Wed keep making guesses until the perimeter was 54. © 2007 Herbert I. Gross next Trial-and-Error Solution P = 2(L + W)

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Since we are told that the perimeter is 54 inches, we know that W must be greater than 5 © 2007 Herbert I. Gross next Trial-and-Error Solution P = 2(L + W) 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. next

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Thus, we may increase W by 1 (inch) each time as illustrated in the chart below… © 2007 Herbert I. Gross next Trial and Error Solution next W2WLL + WP = 2(L + W) …and the shaded row indicates the row in which P = 54. This row tells us that W = 8 and L = next P = 2(L + W)

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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. Answer: $8 © 2007 Herbert I. Gross next How much money does Jane have? Example 8 next

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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… © 2007 Herbert I. Gross next An Algebraic Solution 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

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We now can replace M by 2J whenever we want in this problem. Hence, the equation M + 4 = 4(J – 3) becomes… © 2007 Herbert I. Gross next An Algebraic Solution M + 4 = 4(J – 3) next 2J + 4 = 4(J – 3) 4 = 2J – = 2J 8 = J 2J + 4 = 4J – 12

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J = 8 tells us that Jane has $8; and, hence, that Mary has $16. © 2007 Herbert I. Gross next An Algebraic Solution next 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.

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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. © 2007 Herbert I. Gross next Trial and Error Solution next JMM + 4J – Thus, we can make a chart similar to the following one.

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We see that M + 4 is four times as large as J – 3; © 2007 Herbert I. Gross next Trial and Error Solution next JMM + 4J – and that in this row, J = 8 and M = 16. next

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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. © 2007 Herbert I. Gross next Important! This is illustrated in Examples 9 and 10.

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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. Answer: 8 inches © 2007 Herbert I. Gross next How long is the first piece? Example 9 next

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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. Answer: 10 inches © 2007 Herbert I. Gross next How long is the first piece? Example 10 next

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© 2007 Herbert I. Gross next 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 next 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 Notice that Examples 9 and 10 are word for word the same except for the two words in italics.

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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… © 2007 Herbert I. Gross next The Solution f + s + t = 45 next s = 2f t = s + 5 …while in Example 10 the three relationships are… f + s + t = 45 s = 2f t = f + 5

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Both systems are identical © 2007 Herbert I. Gross next The Solution next f + s + t = 45 s = 2f t = s + 5 f + s + t = 45 s = 2f t = f + 5 except for the fact that s and f have been interchanged in the third equation of the two systems. t = s + 5t = f + 5

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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… © 2007 Herbert I. Gross next The Solution f + s + t = 45 next f + (2f) + (f + 5) = 45 1f + 2f + 1f + 5 = 45 4f + 5 = 45 4f = 40 f = 10

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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… © 2007 Herbert I. Gross next The Solution f + s + t = 45 next f + (2f) + (s + 5) = 45 1f + 2f + (2f + 5) = 45 5f + 5 = 45 5f = 40 f = 8 1f + 2f + 2f + 5 = 45 (s(s

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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. © 2007 Herbert I. Gross next Additional practice is left for the exercise set.

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