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Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 24 By Herbert I. Gross and Richard A. Medeiros next
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Problem #1 © 2007 Herbert I. Gross next John and Bill together have 500 marbles and John has 50 marbles more than Bill. How many marbles does Bill have? Answer: 225 next
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Answer: 225 Solution for Problem #1: Method 1 A common strategy for solving problems of this type is to focus on what we are trying to find. In this case, we want to know how many marbles Bill has; so we let B denote this amount. next © 2007 Herbert I. Gross next Since John has 50 marbles more than Bill, the number of marbles he has is B + 50.
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Solution for Problem #1: The total number of marbles they have is the number of marbles Bill has plus the number of marbles John has. In the language of algebra, this becomes the equation… next © 2007 Herbert I. Gross next B + (B + 50) = 500 …which by the associative property of addition can be rewritten as… (B + B ) + 50 = 500
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Solution for Problem #1: The equation (B + B) + 50 = 500 can be rewritten in the equivalent form… next © 2007 Herbert I. Gross next 2B + 50 = 500 If we then subtract 50 from both sides of the above equation, we obtain the equivalent equation… B = 225 …and if we then divide both sides of the equation 2B = 450 by 2, we obtain the result that… 2B = 450 next
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Notice the convenience of using a letter such as B rather than x to represent the number of marbles Bill has. Namely, if we use x to denote the number of marbles Bill has, after we determine that x = 225, we might not remember that x represented the number of marbles Bill had. © 2007 Herbert I. Gross On the other hand, the letter B is a clear reminder that B = 225 implies that Bill has 225 marbles. Notes on Method 1 next
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In the above context, notice also the importance of reading correctly. © 2007 Herbert I. Gross If we had written the answer as 275, it would have been the correct answer to the question… “How many marbles does John have?” However, this is the incorrect answer to the given question. Notes on Method 1 next
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Also notice how the algebraic solution gives us a hint as to how we might solve the problem by using arithmetic. For example, when we subtracted 50 from both sides of 2B + 50 = 500 to obtain 2B = 450, it was equivalent to saying that if John had 50 fewer marbles, they would both have the same number of the 450 marbles that remained. © 2007 Herbert I. Gross Notes on Method 1 next Then when we divided both sides of the equation 2B = 450 to obtain B = 225, we were saying that if they had equal amounts, they would have each had 225 marbles.
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It is understandable that if we have “algebra anxiety” that we might prefer a trial-and-error solution to a problem rather than an algebraic solution. © 2007 Herbert I. Gross Important Note next However, while it might seem more complicated, the algebraic solution has a huge advantage over a trial-and-error solution.
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We have already encountered this type of situation when we looked at an equation such as x + 1 = x + 2 © 2007 Herbert I. Gross Important Note next Namely, using algebra we simply subtracted x from both sides of the equation to obtain the false statement that 1 = 2.
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In other words, if it was possible that there was a value of x for which x + 1 = x + 2, it would follow inescapably that 1 = 2 was a true statement. © 2007 Herbert I. Gross Important Note next However, since we know that 1 = 2 is a false statement we can say for sure that there is no value of x for which x + 1 = x +2.
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On the other hand, if we were to rely solely on trial-and-error, we would be able show that no number we tested belonged to the solution set of the equation x + 1 = x + 2 © 2007 Herbert I. Gross Important Note next However, since it is impossible to test every number for membership, our task would never end.
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Applying this discussion to Problem #1, in Lesson 23, we showed that J = 275 and B = 225 belonged to the solution set of the system of equations J + B = 500 and J = B + 50. © 2007 Herbert I. Gross Important Note next However, we did not show that J = 275 and B = 225 was the only pair of numbers that belonged to the solution set of this system of equations.
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However, using algebra, we have now shown that if J + B =500 and if J = B + 50; John must have 275 marbles, and Bill must have 225 marbles. © 2007 Herbert I. Gross Important Note next In other words, using algebra we know that J = 275 and B = 225 are the only pair of numbers that can belong to the solution set of the given system of equations.
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In still other words, by using algebra, we have now shown that unless J = 275 and B =225, the conditions stated in Problem #1 cannot be met. © 2007 Herbert I. Gross Important Note next The subtle point is that this doesn’t say that if J = 275 and B = 225 the conditions stated in Problem #1 have been met!
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More specifically, what we are saying is that either J = 275 and B = 225 satisfies the given conditions or else there are no solutions. © 2007 Herbert I. Gross Important Note next So we have to check whether when J = 275 and B = 225 that the conditions J + B = 500 and J = B + 50 are met.
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In this case, it is easy to verify that when J = 275 and B = 225, the conditions J + B = 500 and J = B + 50 are both met. © 2007 Herbert I. Gross Important Note next Once we know this, we can correctly conclude that the two conditions J + B = 500 and J = B + 50 are met if and only if J = 275 and B = 225.
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In summary, the use of algebra leaves us with a list of “candidates” that can possibly belong to the solution set. It doesn’t imply that each of these “candidates” will be elected. © 2007 Herbert I. Gross Important Note next We still have to check each “candidate” separately; and in this sense what is often referred to as “checking the solution” is actually part of the solution.
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Solution for Problem #1: Method 2 Different people may look at the same problem differently and come up with two different strategies, each of which gives a correct solution to the problem. next © 2007 Herbert I. Gross For example, a person could have begun the solution by again letting B represent the number of marbles Bill has. However, this time the person decides to focus on the fact that together they have 500 marbles. next
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Solution for Problem # 1: Method 2 In this case, the number of marbles John has is the difference between 500 and the number of marbles Bill has, or in the language of algebra, the number of marbles John has is 500 – B. next © 2007 Herbert I. Gross Since John has 50 more marbles than Bill, we know that if we added 50 marbles to the number Bill has, it would equal the number of marbles John has. next
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Solution for Problem # 1: Method 2 In the language of algebra this says… 500 – B = B + 50 next © 2007 Herbert I. Gross If we subtract 50 from both sides of our equation, we obtain the equivalent equation… 450 – B = B next
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Solution for Problem # 1: Method 2 …and if we then add B to both sides of the equation 450 – B = B, we obtain… 450 = 2B next © 2007 Herbert I. Gross …and now if we divide both sides by 2, it follows that… B = 225 next
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In solving the problem by using this method, it’s important to distinguish between the commands “subtract” and “subtract from”. © 2007 Herbert I. Gross next For example, it’s important that when we subtract B from 500, we write the difference as … 500 – B and not as B – 500. Notes on Method 2
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next Notice again the algebraic solution tells us how we might have found an arithmetic solution. © 2007 Herbert I. Gross next Namely, the equation 500 – B = B + 50 tells us that if Bill had 50 more marbles, he and John would have had the same number of marbles. Notes on Method 2
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next This means the same thing as saying that if John had 50 fewer marbles, he and Bill would have had the same number of marbles, and that there would then have been only 450 marbles, and this is what the equation 450 – B = B tells us. © 2007 Herbert I. Gross next The equation 450 = 2B then tells us that in this case they would have both had 225 marbles. Notes on Method 2
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Solution for Problem #1: Method 3 It is sometimes easier to use 2 variables so that we don’t have to do as much mental arithmetic. next © 2007 Herbert I. Gross For example, just as we did above, we could let B represent the number of marbles Bill has. But this time we will let J represent the number of marbles John has. next
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Solution for Problem #1: Method 3 Since altogether they have 500 marbles we may write this in the form of the equation… J + B = 500 next © 2007 Herbert I. Gross …and the fact that John has 50 more marbles than Bill can be represented by the equation… J – B = 50 next The above two constraints yield the system of equations… J + B = 500 J – B = 50 next
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Solution for Problem #1: Method 3 If we add the two equations we obtain… next © 2007 Herbert I. Gross Dividing both sides of the equation by 2 we obtain… next J + B = 500 J – B = 50 next + 2 2J = 550 J = 275 Hence, Bill has 225 marbles.
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next The Role of Algebra in Paraphrasing © 2007 Herbert I. Gross next Special Note The statement “…The fact that John has 50 more marbles than Bill can be represented by the equation J – B = 50,” involves paraphrasing. More specifically, the algebraic translation of “John has 50 more marbles than Bill,” is J = B + 50; not J – B = 50.
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next The Role of Algebra in Paraphrasing © 2007 Herbert I. Gross next However, starting with the equation J= B + 50, our rules allow us to subtract B from both sides of this equation to obtain the equivalent equation J – B = 50. Most likely it was not difficult to understand this particular paraphrasing without referring to algebra.
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The Role of Algebra in Paraphrasing © 2007 Herbert I. Gross However, the important thing to remember is that whenever we are in doubt about whether we have paraphrased an arithmetic statement correctly, we can always use our rules of algebra to make sure we are correct.
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Problem #2 © 2007 Herbert I. Gross next Herb has three times as many marbles as Ben. Together they have 120 marbles. How many marbles does Herb have? next Answer: 90
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Solution for Problem #2: In this case, we are asked to find the number of marbles Herb has. next © 2007 Herbert I. Gross Therefore, we may begin by letting H represent the number of marbles Herb has. The statement that Herb has three times as many marbles as Ben means the same thing as saying that Ben has 1 / 3 the number of marbles Herb has. In terms of an algebraic expression, it says that the number of marbles Ben has is 1 / 3 H. next
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Solution for Problem #2: Since they have a total of 120 marbles, it means that… H + 1 / 3 H = 120 next © 2007 Herbert I. Gross Multiplying both sides of the equation by 3 leads to the equivalent equation… 3H + H = 360 next …from which it follows that 4H = 360, or H = 90.
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Again, notice the importance of reading comprehension. © 2007 Herbert I. Gross next Notes on Problem #2 We have to know that the statement “Herb has 3 times as many marbles as Ben” means the same thing as the statement “Ben has 1 / 3 the number of marbles Herb has”. Note: As we mentioned previously, algebra can do the paraphrasing for us. Namely, by dividing both sides of B = 3J by 3, we see that 1 / 3 B = J.
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next © 2007 Herbert I. Gross next Notes on Problem #2 As long as we’re careful in distinguishing H from B, we can eliminate the need for working with fractions if we choose to work with the number of marbles Ben has. That is, if we let B represent the number of marbles Ben has, then 3B will represent the number of marbles Herb has. In this case, the fact that altogether they have 120 marbles leads to the equation B + 3B = 120 from which we see that 4B = 120 or B = 30; then remembering that the number of marbles Herb has is 3B, we see that Herb has 90 marbles.
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next © 2007 Herbert I. Gross next Notes on Problem #2 Notice that the equation 4B = 120 is equivalent to our corn bread model where we let the corn bread represent the 120 marbles, and then use the fact that we slice the corn bread into 4 pieces of equal size,. Corn Bread 120
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next © 2007 Herbert I. Gross next Notes on Problem #2 One of the pieces represents the number of marbles Ben has, 120 and the other three pieces represent the number of marbles Herb has. 30 30 30 30 B 90 next H
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Problem #3 © 2007 Herbert I. Gross next The only money Mary has with her is 30 coins consisting of dimes and quarters. If she has twice as many quarters as dimes, how much money does she have with her? next Answer: $6
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Solution for Problem #3: In this case we may begin by letting d denote the number of dimes Mary has. Knowing that she has twice as many quarters as she has dimes, we know that the number of quarters she has can be represented by 2d. next © 2007 Herbert I. Gross Knowing that she has a total of 30 coins (all of which are dimes and quarters) tells us that… d + 2d = 30 next
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Solution for Problem #3: d + 2d = 30 Now we can see that 3d = 30 or d = 10. In this case, 2d = 20. next © 2007 Herbert I. Gross Thus, she has 10 dimes (which are worth $1) and 20 quarters (which are worth $5), and we see that the total amount of money she has with her is $1 + $5, or $6. next
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© 2007 Herbert I. Gross next Notes on Problem #3 Again reading comprehension and attention to detail are important. The problem does not ask to find the number of dimes or the number of quarters. Rather it asks us to find how much money she has. To do this, we first have to find the number of dimes and the number of quarters she has, but we have to go beyond this to find the answer to the question.
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© 2007 Herbert I. Gross Notes on Problem #3 From a more formal point of view this problem has 3 variables; namely the number of dimes, the number of quarters and the total amount of money she has with her. In order for a unique solution to exist, we need 3 independent constraints. In more detail, we want to find the value of T where T represents the total amount of money (in cents) she has with her. next
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© 2007 Herbert I. Gross Notes on Problem #3 These are the things we know… ► Since each dime (d) is worth 10 cents and each quarter (q) is worth 25 cents, one of the constraints is that… T = 10d + 25q next
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© 2007 Herbert I. Gross Notes on Problem #3 ► and since she has a total of 30 dimes and quarters, another constraint is… d + q = 30 ► and finally since the remaining constraint is that she has twice as many quarters as she has dimes, the third equation is… q = 2d next
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© 2007 Herbert I. Gross Notes on Problem #3 In our solution, we first used the equations d + q = 30 and q = 2d to find that she had 10 dimes and 20 quarters (that is, d = 10 and q = 20), and we then used these values of d and q in the equation T = 10d + 25q to find the value of T. next
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Problem #4 © 2007 Herbert I. Gross next A rectangle’s length is 22 inches more than three times the rectangle’s width. If the perimeter of the rectangle is 180 inches, what is the length of the rectangle? next Answer: 73 inches
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next © 2007 Herbert I. Gross next L W L W L L W W ++ + = 2(L + W) next = 2L + 2W Answer: 73 inches Solution for Problem #4: Recall that the perimeter (P) of a rectangle is the distance around its four sides. That is, the perimeter is equal to 2(L + W), where L represents the length, and W represents the width of the rectangle.
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Answer: 73 inches Solution for Problem #4: If we let W represent the width (in inches) of the rectangle, then 3W represents three times the width and 3W + 22 represents the length that is 22 inches longer than three times the width of the rectangle. next © 2007 Herbert I. Gross
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Solution for Problem #4: In summary… W = the width of the rectangle (in inches). L = 3W + 22 = the length of the rectangle. P = 2(L + W) = the perimeter of the rectangle. next © 2007 Herbert I. Gross From the given information we may replace P in the equation P = 2(L + W) by 180, and we can also replace L in the equation P = 2(L + W) by its value in the equation L = 3W + 22. next
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Solution for Problem #4: If we do this, we obtain the equation … 180 = 2([3W + 22] + W) next © 2007 Herbert I. Gross Our equation may then be paraphrased in the following way… next 180 = 2(3W + 22 + W) 180 = 2(4W + 22) 90 = 4W + 22 68 = 4W 17 = W
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Solution for Problem #4: The equation 17 = W tells us that the width of the rectangle is 17 inches. However, the problem asked us to find the length of the rectangle. next © 2007 Herbert I. Gross To do this, we replace W by 17 in the equation L = 3W + 22 to obtain… next L = 3W + 22 = 3(17) + 22 = 51 + 22 = 73 (inches)
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next © 2007 Herbert I. Gross next Notes on Problem #4 In the step in which we went from… 180 = 2(4W + 22) to 90 = 4W + 22 we divided both sides of the equation by 2. However, we could have elected to use the distributive property and rewrite the equation 180 = 2(4W + 22) in the form… 180 = 8W + 44
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© 2007 Herbert I. Gross Notes on Problem #4 We could then have solved for W by subtracting 44 from both sides of the equation 180 = 8W + 44 to obtain… 136 = 8W …after which we could have divided both sides of the equation by 8 to obtain the same result, namely W = 17. The above note is simply another reminder that there is more than one correct way to solve a problem. next
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Problem #5 © 2007 Herbert I. Gross next 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? next Answer: $8
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Solution for Problem #5: We let J represent the amount of money that Jane has, and we let M represent the amount of money that Mary has. next © 2007 Herbert I. Gross Then… next ► M + 4 represents $4 more than the amount Mary has. ► J – 3 represents $3 less than the amount Jane has.
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Solution for Problem #5: Since Mary has twice as much money as Jane, we may write that… M = 2J next © 2007 Herbert I. Gross The given information tells us that… M + 4 = 4(J – 3) next We may then replace M in the equation M + 4 = 4(J – 3) by its value in the equation M = 2J to obtain… 2J + 4 = 4(J – 3)
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Solution for Problem #5: The following sequence of steps solves our equation… next © 2007 Herbert I. Gross next 2J + 4 = 4(J – 3) 2J + 4 = 4J – 12 4 = 2J – 12 16 = 2J 8 = J So Jane has $8.
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Solution for Problem #5: Check… M = 2J = 2(8) = 16; so Mary has $16; hence, M + 4 = 16 + 4 = 20 and J – 3 = 8 – 3 = 5 next © 2007 Herbert I. Gross next That is… If Mary had $4 more she would have had $20; In that case, Mary would have 4 times as much money as Jane. $20 $5 and if Jane had $3 less she would have had $5.
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Problem #6 © 2007 Herbert I. Gross next 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? next Answer: 7 years old
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Solution for Problem #6: If we let B denote Bill’s present age, the fact that John is now 3 times as old as Bill means that 3B represents John’s present age. next © 2007 Herbert I. Gross next Seven years from now, Bill’s age will be represented by B + 7, and John’s age by 3B + 7. The fact that John's age in 7 years (that is, 3B + 7) will be twice Bill’s age in 7 years (that is, B + 7) yields the equation… 3B + 7 = 2(B + 7)
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Solution for Problem #6: We can then solve the equation by the following sequence of steps. next © 2007 Herbert I. Gross next 3B + 7 = 2(B + 7) 3B + 7 = 2B + 14 B + 7 = 14 B = 7 That is… Bill is now 7 years old.
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Solution for Problem #6: Check… If B = 7, then 3B = 21, B + 7 = 14, and 3B + 7 = 28. next © 2007 Herbert I. Gross next In words, John is now 21 and Bill is 7 (which verifies that John is now 3 times as old as Bill). In 7 years, John will be 28 and Bill will be 14 (which verifies that in 7 years John will be twice as old as Bill).
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Problem #7 © 2007 Herbert I. Gross next A car went from A to B at a constant rate of 30 mph and made the return trip at a constant rate of 20 mph. If the round trip took a total of 30 hours, what is the distance between A and B? next Answer: 360 miles
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Solution for Problem #7: Since the problem asks us to find the distance between A and B, we may begin by letting, say, d, represent the distance. However, the “helpful” information concerns the time it took for the round trip. next © 2007 Herbert I. Gross next Since the speed is constant, we know that if we divide the distance the car traveled by the speed at which it traveled, we will obtain the time it took to make the trip.
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Solution for Problem #7: Therefore… d / 30 = the time it takes the car to go from A to B, and d / 20 = the time it takes the car to return from B to A. next © 2007 Herbert I. Gross next Thus, the total time for the round trip is, on the one hand, 30 hours; and on the other hand, it is also d / 30 + d / 20. Hence, in the form of an equation… d / 30 + d / 20 = 30
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Solution for Problem #7: Since a common multiple of 30 and 20 is 60, we multiply both sides of the equation d / 30 + d / 20 = 30 by 60 to obtain… next © 2007 Herbert I. Gross next or… 60( d / 30 + d / 20 ) = 60(30) next or… 2d + 3d = 1800 d = 360 or… 5d = 1800 next
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© 2007 Herbert I. Gross Notes on Problem #7 Nothing says that we have to start the solution of a problem by working with the variable that is asked for in the problem. next For example, we could work with the time it takes to travel one way; and then use that result to find the distance.
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© 2007 Herbert I. Gross Notes on Problem #7 More specifically, suppose we let t represent the time it takes the car to travel from A to B. Since the total time for the round trip is 30 hours, the time it takes for the return trip is 30 – t. next Since the car is traveling at a constant speed of 30 mph in going from A to B, in t hours, it travels 30t miles. Hence, the distance from A to B is 30t miles. next
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© 2007 Herbert I. Gross Notes on Problem #7 On the other hand, in returning from B to A the car travels at a constant speed of 20 mph for (30 – t) hours. next But the distance from A to B is the same as the distance from B to A. Hence, the distance from B to A is 20(30 –t) miles. next
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© 2007 Herbert I. Gross 30t = 600 – 20t t = 12 50t = 600 Notes on Problem #7 Therefore… 30t = 20(30 – t). Solving the equation, we see that… In other words, in going from A to B the car travels at a speed of 30 mph for 12 hours. Hence, the distance from A to B is 360 miles. next
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© 2007 Herbert I. Gross Notes on Problem #7 We still have to check whether our answer satisfies the given conditions. To this end….. next we have just shown that in going from A to B the car traveled for 12 hours; and in returning for B to A the auto traveled for 18 hours. Hence the time for the round trup was 30 hours; which checks with the given condition.
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Problem #8 © 2007 Herbert I. Gross next John has three times as many marbles as Bill, and Tom has four more marbles than John. How many marbles does Tom have if altogether they have 214 marbles? next Answer: 94 marbles
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Answer: 94 Solution for Problem #8: Although the problem asks us to find the number of marbles Tom has; the equations will be less complicated if we work in terms of the number of marbles Bill has. More specifically… next © 2007 Herbert I. Gross next ► Let B represent the number of marbles Bill has. ► Since John has 3 times as many marbles as Bill has, the number of marbles John has is represented by 3B.
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Solution for Problem #8: ► And since Tom has 4 more marbles than John has, the number of marbles Tom has is represented by 3B + 4. next © 2007 Herbert I. Gross next ► Since the total number of marbles that all three have is 214, the equation that determines the value of B is… B + 3B +(3B + 4) = 214
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Solution for Problem #8: ► The value of B can then be determined from the equation B + 3B +(3B + 4) = 214 by the following sequence of steps… next © 2007 Herbert I. Gross next B + 3B +(3B + 4) = 214 B + 3B +3B + 4 = 214 7B + 4 = 214 7B = 210 B = 30
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Solution for Problem #8: ►And since B = 30, 3B = 90, and 3B + 4 = 94 Therefore, Tom has 94 marbles. next © 2007 Herbert I. Gross next ► Check… Bill has 30 marbles, John has 90 marbles, and Tom has 94 marbles. And since 30 + 90 + 94 = 214, altogether they have 214 marbles.
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© 2007 Herbert I. Gross Notes on Problem #8 We could have solved this exercise by setting up a system of 3 equations in 3 unknowns. Namely, we could have let B represent the number of marbles Bill has; J, the number of marbles John has; and T, the number of marbles Tom has. next In this case, the system of equations would be… B + J + T = 214 J = 3B T = J + 4
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© 2007 Herbert I. Gross Notes on Problem #8 The solution for this system would be very similar to the one we used above. next B + J + T = 214 J = 3B T = J + 4 The reason for using three variables is that it often makes it easier to keep track of the given information.
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© 2007 Herbert I. Gross Notes on Problem #8 Because the problem asks for the number of marbles Tom has, one might be tempted to start the solution by letting T represent the number of marbles Tom has. next There is nothing wrong with this approach, but it does lead to a more cumbersome equation to solve.
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© 2007 Herbert I. Gross Notes on Problem #8 For example, if we let T denote the number of marbles Tom has, then the number of marbles John has is represented by T – 4, and the number of marbles Bill has would be represented by next T – 4 3 In this case the equation that determines T will be… T + (T – 4) + T – 4 = 214 3
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© 2007 Herbert I. Gross Notes on Problem #8 To solve our equation, we could multiply both sides of the equation by 3 to obtain… next T + (T – 4) + T – 4 = 214 3 3T + 3(T – 4) + T – 4 = 642 3T + 3T – 12 + T – 4 = 642 7T – 16 = 642 7T = 658 T = 94
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Problem #9 © 2007 Herbert I. Gross next A piece of string, 65 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? next Answer: 12 inches
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Answer: 8 inches Solution for Problem #9: Since the problem asks for the length of the first piece of string we can begin our solution by letting f denote the length of the first piece of string. next © 2007 Herbert I. Gross next ► Since the second piece of string is twice as long as the first piece, its length is represented by 2f. ► And since the third piece of string is 5 inches longer than the second piece, its length is represented by 2f + 5.
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Solution for Problem #9: ► Finally, since the total length of the three pieces is 65 inches, our equation becomes… next © 2007 Herbert I. Gross next f + 2f +(2f + 5) = 65 f = 12 5f + 5 = 65 5f = 60 Therefore, the length of the first piece is 12 inches.
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Solution for Problem #9: ► Check: Since the length of the first piece is 12 inches, the length of the second piece is 24 inches (2f) and the length of the third piece is 29inches (2f +5). next © 2007 Herbert I. Gross next Hence, the total length of the string is… 12 inches + 24 inches + 29 inches = 65 inches.
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Problem #10 © 2007 Herbert I. Gross next A piece of string, 65 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? next Answer: 15 inches
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Answer: 10 inches Solution for Problem #10: Except for a slight (but important) change in the wording the solution of this problem is very similar to the solution of the previous problem. Namely… next © 2007 Herbert I. Gross next ► Since the problem asks for the length of the first piece of string we can begin our solution by letting f denote the length of the first piece of string.
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Solution for Problem #10: ► Since the second piece of string is twice as long as the first piece, its length is represented by 2f. next © 2007 Herbert I. Gross next ► And since the third piece of string is 5 inches longer than the first piece, its length is represented by f + 5. This is the only difference between this exercise and the previous exercise. In the previous exercise, the third piece of string was 5 inches longer than the second piece.
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Solution for Problem #10: ► Finally since the total length of the three pieces is 65 inches, our equation becomes… next © 2007 Herbert I. Gross next f + 2f +(f + 5) = 65 f = 15 4f + 5 = 65 4f = 60 Therefore, the length of the first piece is 15 inches.
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Solution for Problem #10: ► Check: Since the length of the first piece is 15 inches, the length of the second piece is 30 inches and the length of the third piece is 20 inches. next © 2007 Herbert I. Gross next Hence, the total length of the string is… 15 inches + 30 inches + 20 inches = 65 inches.
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Solution for Problem #10: ► The important point with respect to Problems #9 and #10 is to notice how a very slight change in wording had a profound effect on the answers to the two problems. next © 2007 Herbert I. Gross next ►However, in Problem # 10, the three lengths were 15 inches, 30 inches, and 20 inches respectively. ► More specifically in Problem #9 the three lengths were 12 inches, 24 inches, and 29 inches respectively.
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