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Patterns, Functions, and Algebra
Session 6 Part A Part B Part C Homework
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Session 6 Materials:



Solutions for Session 6, Part B

See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6 | B7 | B8 | B9 | B10

Problem B1

First, choose a convenient answer. Because we are to multiply the answer by 1/7, use 7: 1 1/7 times 7 is 8. The number to multiply by 8 to get 32 is 4. Then multiply 7 by 4 to get the correct answer, 28. By way of checking, 1/7 of 28 is 4, and 28 + 4 = 32, which is the answer we wanted.

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Problem B2

We could use false position. We could guess-and-check. We could begin multiplying and dividing from the inside out to produce a more easily solvable equation. We could build an operations flowchart for the equation, then follow it backwards to see what input value would lead to 8 as the output. We could begin dividing and multiplying, from the outside in, in the hope that cancellation will lead to a more easily solvable equation. We could build a table of values with a spreadsheet in the hope that a pattern will emerge. We could use a graphing calculator to graph the left and right sides (right side: y = 8) of the equation, then look for the intersections.

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Problem B3


The flowchart steps are multiply by 2, subtract 4, multiply by 3, divide by 6, and multiply by 4.


Working backwards from 8, we divide by 4 to get 2, multiply by 6 to get 12, divide by 3 to get 4, add 4 to get 8, and divide by 2 to get 4 as the correct value of n. Testing n = 4 in the equation shows that this is correct.

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Problem B4


In the original order, the steps are divide by 2, subtract 3, and multiply by 5. So, take 20, divide by 5 to get 4, add 3 to get 7, and multiply by 2 to get 14 as the answer. Testing b = 14 confirms it as correct.


In the original order, the steps are add 1, multiply by 7, and divide by 2. So, take 14, multiply by 2 to get 28, divide by 7 to get 4, and subtract 1 to get 3 as the answer. Test it out!

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Problem B5

Sure, such equations abound, so long as you can find an operation that can't be "undone." An equation like b2 - 5b = 6 does the trick nicely -- most equations that involve squares cannot be undone in a straightforward manner. Also, any equation that has a variable on both sides of the equal sign, like those in Part C, cannot be easily solved by backtracking.

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Problem B6

Undo the steps. Take 16 and multiply by 3 (undoing the last step first) to get 48. Then, divide 48 by 8 to get 6. Finally, add 3 to get 9, the original number. Test 9 to check that it is correct.

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Problem B7

Each of these has two things in common: the variable being solved for only occurs once, and steps in the equation that can be reversed. Since backtracking is so similar to "undoing" the effect of several function machines, any step in the equation must be reversible for backtracking to be successful.

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Problem B8

The first stage has 4 toothpicks, and each stage beyond the first needs 2 more toothpicks. One way of backtracking is to use the fact that we need to add 108 toothpicks from the initial formation, and dividing 108 by 2 (= 54) will tell us how many stages to add. Because there are 4 toothpicks in the first stage, 54 more stages will bring us to the 55th stage, which has 112 toothpicks.

Alternately, we can find a formula for the number of toothpicks, in terms of the stage number. One such formula is T = 2n + 2 (another is T = 2(n - 1) + 4). Then use the techniques of Problem B4 to find n, which is 55.

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Problem B9


Cover up everything inside the brackets to get 4(covered) = 8. So 3(2n - 4) / 6 = 2. Cover up 3(2n - 4) to get (covered)/ 6 = 2. Now 3(2n - 4) = 12. Cover up (2n - 4) to get 3(covered) = 12. So 2n - 4 = 4. Then cover up 2n to get (covered) - 4 = 4. So 2n = 8, and n = 4 is the solution.


Cover up 3(12 / [x - 5]) so the equation reads (covered) + 1 = 13.
Then 3(12 / [x - 5]) = 12. Then cover up (12 / [x - 5]) to get 3(covered) = 12. So 12 / [x - 5]= 4. Finally, cover up (x - 5) to get the
equation 12 / (covered) = 4. From here we get the equation x - 5 = 3, for which x = 8 is the solution.

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Problem B10

One of the best ways to negotiate this problem is to use a variable for the "unknown," the number of heads of lettuce sold each day. Call this number n. Now we can follow the problem either forward or backward.

Following it forward:
Start of first day: 80.
End of first day: 80 - n.
Start of second day: 2(80 - n) = 160 - 2n.
End of second day: 160 - 2n - n = 160 - 3n.
Start of third day: 3(160 - 3n) = 480 - 9n.
End of third day: 480 - 9n - n = 480 - 10n. Since this equals zero, 480 - 10n = 0, and this means 480 = 10n, and 48 = n.

Following it backwards:
End of third day: 0
Start of third day: n
End of second day: 1/3 n [the stock was tripled!]
Start of second day: 1/3 n + n = 1 1/3 n
End of first day: 2/3 n [half of 1 1/3]
Start of first day: 2/3 n + n = 1 2/3 n = 80 heads. So n = 80 divided by 1 2/3 = 48.

Notice that backtracking seems to be a little easier here. A solution obtained by covering up is possible, but it would be more difficult because it would require you to write an equation for the entire problem in which there is only one variable.

You could also use false position and decide whether the number of heads sold in a day was too small (if some were left at the end), too large (if not enough were left to be sold on the third day), or just right (48). In this problem, though, false position amounts to guess-and-check, because there is no easy way to adjust a wrong answer to make a right one.

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