
#1
Nov2011, 07:50 PM

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An algebra tutorial on the Mainland High School Algebra Lab stie
(http://www.algebralab.org/lessons/le...eEquations.xml) Solves (example ii): [itex] 2x 3] = x  5 [/itex] and obtains solutions [itex] x = 2 [/itex] and [itex] x = 8/3 [/itex] I've emailed them about this twice and they haven't corrected it yet. Are there algebra texts that actually teach the fallacious method of solution used on that page? 



#2
Nov2011, 10:21 PM

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PF Gold
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2x3 = x5implies x=2 or x=8/3They've just forgotten to finish the problem from there. 



#3
Nov2011, 10:41 PM

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#4
Nov2011, 11:10 PM

P: 1,295

Will Mainland HS ever fix its algebra page?
I agree with D H.
If you notice, they have provided incomplete procedure for solving these problems: 



#5
Nov2111, 12:18 AM

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The appropriate way to solve the problem would be to use the definition of the absolute value function  namely:
By definition [itex] a = a [/itex] if [itex] a \ge 0 [/itex] and [itex] a = a [/itex] if [itex] a < 0 [/itex]. So we have two cases. Case 1) Assume [itex] 2x  3 \ge 0 [/itex] This implies [itex] x \ge 3/2 [/itex] and [itex]  2x  3  = 2x 3 [/itex]. So we solve [itex] 2x 3 = x  5 [/itex] obtaining [itex] x = 2 [/itex] However, we have assumed [itex] x > 3/2 [/itex], so [itex] x = 2 [/itex] is not a solution. Case 2) Assume [itex] 2x 3 < 0 [/itex]. This implies [itex] x < 3/2 [/itex] and [itex] 2x 3= (2x 3) = 2x + 3 [/itex]. So we solve [itex] 2x + 3 = x  5 [/itex] obtaining [itex] x = 8/3 [/itex]. But we have assumed [itex] x < 3/2 [/itex], so this is not a solution. This method has the advantage of teaching the formal definition of absolute value. It trains students to realize that "minus a quantity" need not be a negative number and all the conditions that are required of the solution are deduced. The method on the page uses some mumbojumbo about "opposites" and could be kludged up by telling students that they must "always check your solutions". That's not bad advice with any method, but why use a method that avoids proper deductive reasoning. The correct method can be generalized to solve equations like [itex] 2x  3 = x  5 [/itex] by cases. I don't know how you apply the method of "opposites" to such equations. In view of the thread http://www.physicsforums.com/showthread.php?t=377783, we might get Bruce Tonkin to weigh in on this matter. 



#6
Nov2111, 04:21 AM

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PF Gold
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The approach is perfectly good, and more efficient to boot. (in fact, the "method of opposites" is reversible in this case: x=y is true if and only if x=y or x=y) It is possible there are good arguments that teaching to solve by cases is pedagogically better  but the one you give is simply not one of them. 



#7
Nov2111, 05:20 AM

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P: 14,433

What's important is to teach that (a) sometimes what appears to be a solution is not a solution; one has to check that it is a solution, and (b) students should always check their work.
In this case I would have used the very advice presented in the next problem: "Before jumping into solving this problem, think about it first." (Their problem iii should have been the very first problem.) "Thinking about it first" in the case of problem ii says that to ensure that the left hand side is not negative we must have x≥5. This in turn means that one need not check the alternative 2x3=x5 because 2x3≥8 for all x≥5. So there is at most one solution here, found by solving 2x3=x5. The solution, x=2, is not a solution to the original problem. There are no solutions to this problem! While a bit verbose for this problem, eliminating entire branches from consideration by "thinking about it first" can save a lot of time on more complex problems. 



#8
Nov2111, 11:06 AM

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P: 3,167

The advantages of the alternate technique in this case are that it does not burden students with learning the definition of the absolute value function, it does not require precise deductive thinking, and it reinforces the advice that one should check answers. 



#9
Nov2111, 11:29 AM

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P: 14,433

Stephen: You are arguing with Hurkyl over the correct way to say "tomato." Both his approach and yours will yield the same answers to any and all questions of this form. It's a bit silly to argue over which approach is the "right" approach since the two are ultimately equivalent.
As for why they won't fix it, perhaps you are being a bit too argumentative in your messages or too adamant in telling them that the very approach they are using is wrong. That's just a perhaps; I don't know the history of your communications with them. Do you really care whether they use your approach, or Hurkyl's, or someone else's, so long as the approach they do use does yield the correct answers? 



#10
Nov2111, 12:36 PM

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P: 3,167

A similar debate could be had over whether to teach students to understand the distributive law or just teach them "the FOIL method". On the one hand, given that there are many achievement tests in secondary education nowadays and that teaching precise reasoning is a burdensome task for teachers, there are pragmatic reasons for teaching students how to get the right answers quick at the expense of how to get right answers in some deductive manner. The deductive manner isn't likely to be the subject of a multiple choice question. Suppose a person is writing a proof and must state the consequences of [itex] 3x  \delta < 4 \epsilon + 2 [/itex]. What will the correct approach be? To use the "method of opposites" and then try to "check your answer"? Granted that not all students of secondary mathematics are going to proceed into such a situation. And perhaps as students proceed in mathematics they must periodically suffer disappointments when the way they learned some things in high school is not recognized as valid. The "method of opposites" can be fixed so it uses valid deductions. The procedure on the Mainland site could be solved by introducing a theorem: If a] =b then a = b if and only if a >= 0 and a = b if and only if a < 0. Then the cases to work the example would be Case 1: 2x  3 = x  5 and 2x  3 >= 0 Case 2: 2x 3 = (x5) and 2x 3 < 0 I don't see that's any simpler than the cases: Case 1: 2x  3 >= 0 and 2x3 = x5 Case 2: 2x 3 < 0 and (2x3) = x 5 I don't see that introducting the theorem and avoiding the use of the definition of the absolute value function has any pedagogical advantage. In subsequent math courses, the definition of the absolute value function will be the more important concept. It's silly to argue with a closed mind, but reasonable people can argue profitably. 



#12
Feb2212, 09:32 AM

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Now I know which web sites hold the real power in mathematical affairs! 



#13
Feb2212, 09:51 AM

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#14
Feb2212, 10:03 AM

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P: 3,167




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