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Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 13 By Herbert I. Gross and Richard A. Medeiros next.

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Presentation on theme: "Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 13 By Herbert I. Gross and Richard A. Medeiros next."— Presentation transcript:

1 Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 13 By Herbert I. Gross and Richard A. Medeiros next

2 Problem #1 © 2007 Herbert I. Gross Answer: y = 5x next Write… y = 3x – 2(4 – x) in the form y = mx + b.

3 Answer: y = 5x Solution: We may begin by using the add-the-opposite rule to rewrite y = 3x – 2(4 – x) as… next © 2007 Herbert I. Gross next We then use the distributive property to rewrite our above equation as… y = 3x + - 2(4 + - x) y = 3x + [ - 2(4) + - 2( - x)]

4 Solution: By the rules for the arithmetic of signed numbers, y = 3x + [ - 2(4) + - 2( - x)] may be rewritten as… next © 2007 Herbert I. Gross next Then by the associative and commutative properties of addition, we may rewrite the above equation in the form… y = 3x + [ x] y = (3x + 2x) or y = 5x next

5 Because of the way the question was worded (y = mx + b), the answer had to be written as y = 5x Notes on #1 © 2007 Herbert I. Gross That is, the answer could also be written as y = 5x – 8. However, in that case the answer would have the form y = mx – b. next

6 ASIDE © 2007 Herbert I. Gross We have been stressing the linear form y = mx + b. However, since b does not affect the rate of change of y with respect to x, m still gives us the rate of change even if the form is y = mx – b. next For example, if you are buying items, the cost per item is the same whether you have, say, a previous credit or a previous debit.

7 The point is that just by looking at the equation y = 5x + - 8, we see it immediately as y = mx + b, with m = 5 and b = - 8. Then, since m represents the change of y with respect to x; and, since our first equation is equivalent to the second equation; we now know that in the equation y = 3x – 2(4 – x), the rate of change of y with respect to x is 5. Notes on #1 © 2007 Herbert I. Gross What we have shown is that our original equation, y = 3x – 2(4 – x), and our second equation, y = 5x + - 8, are equivalent. next

8 Notes on #1 © 2007 Herbert I. Gross The relationship between the equations y = 3x – 2(4 – x) and y = 5x – 8 may be easier to visualize if we return to our “computer program” analogy...

9 next Notes on #1 © 2007 Herbert I. Gross Namely, in a more verbal form… next y = 3x – 2(4 – x) is expressed in Program 1 Program 1 (1) Start with input x. (2) Multiply it by 3. (3) Store the result. (4) Take the input and subtract it from 4. (5) Multiply this result by 2. (6) Subtract this from the result stored in (3). (7) Write the answer (y). x 3x3x 3x 4 – x 2(4 – x) 3x – 2(4 – x) next

10 Notes on #1 © 2007 Herbert I. Gross next y = 5x – 8, on the other hand, represents Program 2 Program 2 (1) Start with input x. (2) Multiply it by 5. (3) Subtract 8 (i.e., add - 8). (4) Write the answer (y). x 5x5x 5x – 8 5x – 8 (i.e., 5x + - 8) next What we showed algebraically was that Program 1 and Program 2 are equivalent. That is, we proved that 3x – 2(4 – x) = 5x – 8.

11 Key Point © 2007 Herbert I. Gross next Thus, whenever we wish, we may replace Program 1 by the simpler but equivalent Program 2. The advantage of being able to do this will become clearer when we do the next exercise.

12 Problem #2 © 2007 Herbert I. Gross Answer: x = 31 next For what value of x is it true that… 3x – 2 (4 – x) = 147

13 Answer: x = 31 Solution: In Problem #1, we showed that 3 x – 2 (4 – x) was equivalent to 5x – 8. next © 2007 Herbert I. Gross Hence, we may replace the equation 3x – 2 (4 – x) = 147 by the equivalent equation 5x – 8 = 147. next

14 Solution: In solving the equation 5x – 8 = 147, we may add 8 to both sides… next © 2007 Herbert I. Gross next …and if we then divide both sides of our equation by 5 we see that… + 8 next 5x – 8 = 147 5x = x = 31

15 Solution: To complete our solution, we check our answer by replacing x by 31 in 5x – 8 = 147 and verify whether… next © 2007 Herbert I. Gross next 5 x – 8 = 147(31) …is a true statement. Since 155 – 8 = 147 is a true statement, our solution is complete.

16 next Notes on #2 © 2007 Herbert I. Gross In terms of our “plain English” version, this exercise asks us to find the value of the input for which it’s true that… next Program 1 (1) Start with the input. (2) Multiply it by 3. (3) Store the result. (4) Take the input and subtract it from 4. (5) Multiply this result by 2. (6) Subtract this from the result stored in (3). (7) Write the answer. ? 147 next

17 Notes on #2 © 2007 Herbert I. Gross As a check, we can replace x by 31 in Program 1 to show that… next Program 1 (1) Start with the input. (2) Multiply it by 3. (3) Store the result. (4) Take the input and subtract it from 4. (5) Multiply this result by 2. (6) Subtract this from the result stored in (3). (7) Write the answer. 31 3(31) = – 31= ( - 27) = – next

18 Limitations of the Undoing Method the Undoing Method Based on what we have done in the past, we might have been tempted in Program 1 to start with 147 (Step 7) and then, one-by-one, undo the previous steps. However, notice that Step 6 asks us to subtract 147 from the number that is stored in Step 3. Unfortunately, at this point in the undoing process, we have no idea what number that is. Thus, it would have been necessary to rewrite Program 1 in the form of, say, Program 2 if we had wanted to be able to use the undoing method to answer the question. © 2007 Herbert I. Gross next

19 Notes on #2 © 2007 Herbert I. Gross Thus, the main aim of this exercise was to illustrate the power of paraphrasing.

20 Problem #3 © 2007 Herbert I. Gross Answer: - 8 next If… - y = 8x – 5, what is the rate of change of y with respect to x?

21 Answer: - 8 Solution: To use the fact that m is the rate of change of y with respect to x, the equation has to be in the form y = mx + b. To convert - y = 8x – 5 into the y = mx + b form, we may multiply both sides of the equation by - 1. In this case we obtain… next © 2007 Herbert I. Gross next or… - 1( - y) = - 1(8 x 5) = - 1(8x + - 5) = - 1(8x) + - 1( - 5) y = - 8x + 5

22 Solution: Looking at y = - 8x + 5 we see that m = - 8. next © 2007 Herbert I. Gross Therefore, the rate of change of y with respect to x is - 8. That is: whenever x increases by 1 unit, y decreases by 8 units. next

23 Notes on #3 © 2007 Herbert I. Gross We have to read equations carefully! In this problem for example, we should not confuse the equation - y = 8x – 5 with the equation y = 8x – 5. If you look at - y = 8x – 5 too quickly, you might mistakenly conclude that m = 8 (rather than m = - 8). next

24 Notes on #3 © 2007 Herbert I. Gross We can always check the correctness of our answer by making a chart such as... And we see that each time x increases by 1, y decreases by 8. next x8x 8x - 5 -y-y-y-yy

25 next Remember that in going from - 3 to - 11, y decreases by 8. Reminder © 2007 Herbert I. Gross More generally for negative numbers, the greater the magnitude the less the value of the number. From a more colloquial point of view, the more money we owe, the worse off we are. next

26 Problem #4 © 2007 Herbert I. Gross Answer: -2 / 3 next If… 3y = 12 – 2x, what is the rate of change of y with respect to x?

27 Answer: -2 / 3 Solution: To convert the equation 3y = 12 – 2x into the y = mx – b form, we divide both sides of it by 3 (or, equivalently, multiply both sides by 1 / 3 ) to obtain… next © 2007 Herbert I. Gross next or… 1 / 3 (3y) = 1 / 3 ( x) = / 3 x = -2 / 3 x + 4 y = -2 / 3 x + 4 …and in this form m = -2 / 3, which is the rate of change of y with respect to x. next

28 Notes on #4 © 2007 Herbert I. Gross The form y = mx + b is nice but not necessary. For example, we could have made a chart similar to the one shown here. From this chart we see that when x increases by 3, y decreases by 2. next x - 2x x 3yy

29 next Use of Fractions © 2007 Herbert I. Gross With respect to the use of fractions, notice the psychological difference between saying that y decreases by 2 / 3 of a unit whenever x increases by 1 unit, and saying that y decreases by 2 units whenever x increases by 3 units.

30 next Use of Fractions © 2007 Herbert I. Gross More specifically, the rate 2 / 3 of a “y unit” per 1 “x unit” can be written in the form. 2 / 3 (y unit /x unit) = 2 (y unit) / 3 (x unit) = 2 “y units” per 3 “x units” next

31 Problem #5a © 2007 Herbert I. Gross Answer: 30 next If a 20 pound slab of cheese is sliced into packages, each of which contains 2 / 3 of a pound of cheese: how many packages will there be?

32 Answer: 30 Solution: This is really an arithmetic problem that will be revisited in part (b) of this problem. Since each package weighs 2 / 3 of a pound and we are packaging 20 pounds of cheese, we want to solve the indirect equation… next © 2007 Herbert I. Gross 2 / 3 (pounds per package) × ___ (packages) = 20 (pounds) next

33 Solution: 2 / 3 (pounds per package) × ___ (packages) = 20 (pounds) is equivalent to the following direct computation… next © 2007 Herbert I. Gross 20 (pounds) ÷ 2 / 3 (pounds per package) = ___ (packages) 20 ÷ 2 / 3 = 20 × 3 / 2 = 30 next

34 Notes on #5a © 2007 Herbert I. Gross We already know that 2 / 3 of a pound per package is equivalent to 2 pounds per 3 packages. next And since 20 pounds is the tenth multiple of 2 pounds, the number of packages must be the tenth multiple of 3 packages; that is 30 packages.

35 Notes on 5a © 2007 Herbert I. Gross Stated as a proportion, the equation becomes… next 2 pounds / 3 packages = 20 pounds / x packages … or more briefly, 2 / 3 = 20 / x To solve for x, one then usually “cross multiplies” to obtain… 2x = 60 or… x = 30

36 Problem #5b © 2007 Herbert I. Gross Answer: y = 20 – 2 / 3 x next With respect problem #5a: write the linear equation that relates the number of pounds of cheese (y) that are left after x packages have been sold.

37 y = 20 – 2 / 3 x Solution: We start with a slab of 20 pounds of cheese and every time we sell a package, we are left with 2 / 3 of a pound of cheese less. next © 2007 Herbert I. Gross So if we sell x packages, we will have sold 2 / 3 x pounds of cheese. If we now subtract this amount from the original 20 pounds, the amount of cheese that is left is… 20 – 2 / 3 x. In other words, y = 20 – 2 / 3 x next

38 Notes on #5b © 2007 Herbert I. Gross y = 20 – 2 / 3 x is another example of a linear relationship. next It indicates that we reduce our amount of cheese by 2 pounds every time 3 of the packages of cheese are sold.

39 Notes on #5b © 2007 Herbert I. Gross Notice that there are a couple of constraints in this particular problem. next y = 20 – 2 / 3 x where x is a whole number such that… 0 ≤ x ≤ 30. For example, we cannot sell a fractional part of a package. Hence, x must be a whole number. Secondly, we cannot sell more cheese than we have, and as shown in part (a) of this problem, there are only 30 packages. So a more precise version of the answer to this part of the problem would be…

40 Concluding Note © 2007 Herbert I. Gross There is a tendency to identify linearity in terms of the form y = mx +b. Thus, a person might believe that we have to rewrite y = 20 – 2 / 3 x in the form y = -2 / 3 x Other people, preferring to avoid fractions, might multiply both sides of y = 20 – 2/3x by 3 to obtain the equivalent form 3y = 60 – 2x. next

41 Concluding Note © 2007 Herbert I. Gross However, keep in mind that the definition of linear is that the rate of change of y with respect to x is constant. Thus, it does not matter which form of the equation we use to express this. next


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