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Author Topic: How would you solve this problem?
AgingAmigaoid
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Icon 5 posted December 07, 2006 13:39      Profile for AgingAmigaoid     Send New Private Message       Edit/Delete Post   Reply With Quote 
I'm not looking for an answer, I'm looking for a discussion of the process and a discussion of the best way to teach the process.

First Generalized

y+x = z Solve for x

We want to move y to the other side of the equation so substract y from both sides of the equation.

y-y+x = z-y

x = z-y

Next one

3+x = 7 Solve for x

Again, we want to move 3 to the other side of the equation so subtract 3 from both sides of the equation.

3-3+x = 7-3

x = 4


Now! The point of contention!

(-3)+x = 7 Solve for x

We want to move -3 to the other side of the equation so... subtract -3 from both sides of the equation.

(-3)-(-3)+x = 7-(-3)

x = 10


There are those who say, "Since you are moving a negative number then you add the positive value." So

(-3)+3+x = 7 + 3

x = 10

Let me state that I understand that (-3)-(-3) is the same as (-3)+3 what I disagree with is why, just because it's -3, do we want to change the "subtract y from both sides of the equation" rule to say, "subtract y from both sides of the equation unless y is negative in which case you should add the positive value of y to both sides of the equation" It gives the rule a messy "except when...." clause that is completely, mathematically, unnecessary.

So my question to the assembled geeks is, in your mind when you do the problem do you convert (-3) to 3 and then add it or do you subtract (-3)?

I substract (-3) while others say I'm "adding a step." I say others are doing the step "in their head" and calling it a "special case rule."

In the end, for teaching, I like the simple, clean, "subtract y from both sides of the equation," which works for any value of y.

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Black Widow
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Icon 1 posted December 07, 2006 13:56      Profile for Black Widow     Send New Private Message       Edit/Delete Post   Reply With Quote 
So my question to the assembled geeks is, in your mind when you do the problem do you convert (-3) to 3 and then add it or do you subtract (-3)?

Simple answer: convert (-3) to 3 and then add it

A bit longer: when doing that problem in my head, I "take off" the -3 in front of the +x, and +3 to the 7 on the other side. There is no +3 to the -3 in front of the +x. I just outright dispose of it.

Gah I hope that makes sense. It does in my mind. [crazy]

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spungo
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Icon 1 posted December 07, 2006 13:57      Profile for spungo     Send New Private Message       Edit/Delete Post   Reply With Quote 
I would just say

(-3) + x = 7

is the same as

x - 3 = 7

then the whole adding or subtracting thing becomes more logical.

If you try to instruct with rules on such matters, you risk hindering your pupils' ability to think more clearly about problems. I see this all the time with folk trying to solve differential equations - they keep asking about "integrating factors", and understanding their rote instruction, when if you just use a bit of common sense, it's a lot easier.

Anyway, that's my 2 bits... although, no doubt, it's worth much less than that. [Wink]

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Ashitaka

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Icon 1 posted December 07, 2006 14:21      Profile for Ashitaka     Send New Private Message       Edit/Delete Post   Reply With Quote 
In doing such algerbra I would say I first just add all like terms and set the equation equal to what I want to solve for (x) or zero. This is just to get an overview. I then solve and rearrange the equation as I need.

I don't know how to answer your question though. In your example problem I do not do any algerbra. I just see that x must be ten.

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nerdwithnofriends
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Icon 1 posted December 07, 2006 16:24      Profile for nerdwithnofriends     Send New Private Message       Edit/Delete Post   Reply With Quote 
Having just gotten out of an assembly programming final, I think I'll be fecetios on this one:

If we represent -3 as a binary (2's complement) number, we don't care about adding or subtracting; the beauty of twos complement is that it requires only one arithmetic operation for both positive and negative numbers.

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stevenback7
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Icon 1 posted December 07, 2006 16:31      Profile for stevenback7   Author's Homepage     Send New Private Message       Edit/Delete Post   Reply With Quote 
your making such things to hard by putting the negative integers in brackets. That is how you make mistakes. I know if i put all my negative numbers in brackets and stuff then i first of all i would be adding like 10 extra lines to my problem and the chance of me accidently making a mistake because of the signs basically triples.

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Jace Raven

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Icon 1 posted December 07, 2006 17:48      Profile for Jace Raven         Edit/Delete Post   Reply With Quote 
If you know that a negative and a negative make a positive

x - 3 = 7

can be written as

x + (-3) = 7

Essentially (as I understand it and it's been a while) the combination of two numbers, adding or subtracting, can be simplified to the addition of a positive or negative intiger. That's why if you subtract a negative from anything you're adding the difference in value. In this case 3 - (-3) or 3 + (-1)(-3). This is the logic behind distributing the negative to a grouped item and to be honest I prefer to add terms not subtract them.

Please correct me if I'm wrong. I am by no means a math wizard.

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Xanthine

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Icon 1 posted December 07, 2006 18:05      Profile for Xanthine     Send New Private Message       Edit/Delete Post   Reply With Quote 
Looks okay Jace, but a negative multiplied by a negative goes positive.

The way I finally came to understand algebra is this (And believe me, it was an almighty battle): you are trying to isolate x. Everything on the x side of the equation needs to get moved to the other side, so whatever happens to the x side also happens to the other side. If you add on one side, you ad on the other. If you subtract on one side, you subtract on the other. You do whatever you need to do to isolate x, but you have do it to the other side as well.

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Jace Raven

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Icon 1 posted December 07, 2006 18:11      Profile for Jace Raven         Edit/Delete Post   Reply With Quote 
Oh. I know Xan. It was the only way I could express my point
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The-Tech
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Icon 1 posted December 07, 2006 18:42      Profile for The-Tech     Send New Private Message       Edit/Delete Post   Reply With Quote 
Jaces first equation
X-3=7

is probably the best way to teach it ... use the cummutative property of addition and get all the variables on the left. This is easier when you are dealing with

2+x-5=7
x+2-5=7 ... get the x on the left
x-3=7

As one my profs. liked to say "At This point the solution is trivial"

Dave

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Xanthine

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Icon 1 posted December 07, 2006 20:51      Profile for Xanthine     Send New Private Message       Edit/Delete Post   Reply With Quote 
I had a couple profs who used to say that. Occasionally I even agreed with them.

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Richard Wolf VI
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Icon 3 posted January 17, 2007 08:40      Profile for Richard Wolf VI   Author's Homepage     Send New Private Message       Edit/Delete Post   Reply With Quote 
It's quite simple: The only thing you are doing to solve this equation is adding the opposite of y on both sides in order to cancel it in the side where x is. So in the case y is positive, its opposite is negative and viceversa.

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ZER
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Icon 1 posted February 13, 2007 10:35      Profile for ZER   Author's Homepage     Send New Private Message       Edit/Delete Post   Reply With Quote 
I would burn the math book and throw a stick of dynamite in the ashes.

But thats just me.

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