Five System Equation: electricians, nails, screws, tape,

G

Go^3

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Oct 24, 2008
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Hi, I'm a freshman in high school taking Algebra II Honors. I've been lurking this board a lot, and I love math and plan to be an actuary at this time.

Anyway, in my Algebra class we were given a story problem that I believe I need to make a five system equation for.

You are an accountant for an electrical company. Your company has many electricians that worry more about getting their job done than making sure that you receive the proper paperwork for the purchases they make during the day. However, they are able to tell you how many of each item they bought and what their total bill was.

Electrician 1: Four packs of nails, three screws, four rolls of tape, one breaker, and one outlet for $23.00.

Electrician 2: Two screws, one pack of nails, one breaker, two outlets, and a roll of tape for $14.50.

Electrician 3: A roll of tape, a pack of nails, three breakers, three outlets, and three screws for $27.25.

Electrician 4: Two outlets, one breaker, two rolls of tape, one pack of nails, and four screws for $18.00.

Electrician 5: Four outlets, two breakers, one roll of tape, six screws, and three packs of nails for $31.25.

Find how much each unit costs algebraically so that you can itemize each item in your accounting books. Show all of your work.
Click to expand...

I got this assignment today and it won't be due till Thursday, so I've got a little time, but I'm an overachiever that likes to get things done fast, so instead of waiting and trying to get something out of my teacher on Monday, I figured I ought to come here.

I'll try to make my work easy to follow.

n=nails
s=screws
t=tape
b=breaker
u=outlets (I find o to be too close to 0 when I scribble down my work and it confuses me)

So I came up with an equation.

4n+3s+4t+b+u=23
n+2s+t+b+2u=14.5
n+3s+t+3b+3u=27.25
n+4s+2t+b+2u=18
3n+6s+t+2b+4u=31.25

And, using my handy graphing calculator and RREF, I used matrices to find the answer to find out what each answer was. Trouble is, I'm not sure what he means by finding and showing it algebraically.

n=1.75
s=1
t=1.5
b=4.75
u=2.25
Click to expand...

Anyway, from personal experience I know its all smooth sailing once you find one variable. I have about 5 pages of doodling by using substitution and elimination, but I think they are pretty irrelevant to the matter at hand.

If anyone could help me "algebraically" find one variable, it'd be great. I can do everything else, its just starting the problem that I need help with.

Thanks in advance for any and all help! =]
 
Re: Five System Equation

Go^3 said:
... using my handy graphing calculator and [with the single stroke of my finger executed this calculator's built-in command] RREF, I ... [read the answer off of the computer's display screen, and, thus, found out] out what [the] answer [is]. Trouble is, I'm not sure what he means by finding and showing it algebraically.
Click to expand...


If I were him, then I would write that he means for you to solve the system by hand using sheets of paper and something to write with. (Your brain, of course, will also be involved.)

In other words, write out each elementary row operation, and progress in a manner in which he can follow your train of thought, until you arrive at the same result as your "handy" calculator.

You wrote that you "used matrices" to solve this exercise. I concluded that you understand how to move an augmented coefficient matrix into row echelon form (REF) (followed by backsolving) OR proceed all the way through to reduced row echelon form (RREF).

Therefore, the algebra of solving those equations simultaneously will be much easier for you, as you understand how to use the abbreviated system known as matrix arithmetic and linear algebra, than it would be for somebody who has to write out each equation (with symbols) at each step.

Enjoy your weekend ... (just kidding!)

Cheers,

~ Mark :)

 
Re: Five System Equation

Thanks for the quick reply Mark.

Yeah, I did everything you said a few hours ago and I just cant seem to find a variable algebraically. I've gotten for example, t=3.5-2s, but that's about as far as I've gotten.


Two systems are easy, three systems are tedious, but the five system I can't even figure out. I'm pretty sure he wants me to find it all by elimination and substitution, but I honestly can't figure out how.

Thanks again Mark.
 
Re: Five System Equation



Do you know how to use elementary row operations to reduce a coefficient matrix to RREF form?

(I'm searching at the Great God Google for a decent introductory-level example ...)


Oh, sorry, I did not realize that Bill Maher's guest panel is just about to start on cable ... gotta go!

MY EDITS: corrected spelling
 
Re: Five System Equation

I'm assuming you mean 5 columns by 6 rows?

If so, yes. But not much, as we just got into this unit and its all new to me.



Gauss Elimination Method eh? I'll look it up. I'm guessing its different than standard elimination?


Bye Mark, and thanks for your time.


Edit:

I see what you both are getting at, and I have tried both of those ways earlier today to no avail, which is why I came here. I really do not want to seem lazy, but I have tried both ways and can't seem to get a clean equation like

4n-3t=50
-4n+4t=20.
 
Re: Five System Equation

Did you find any examples that give you the steps? There is a straightforward algorithm, which, if followed, will reduce the augmented coefficient matrix to RREF.

I'm not sure, but I'm assuming that you should not have to spend time scanning more than a couple dozen sites to find the algorithm. (This type of research is what honors students do.)

One disadvantage of using this algorithm is that it often results in many additional steps. (But, it works.) After you're more fluent in elementary row operations (by virtue of doing a bunch of them by hand), then you will begin to realize the shortcuts. My point: it's not a disadvantage for you.

Look for a step-by-step list of how to mechanically achieve RREF by brute force.

(Hint: it has something to do with starting by getting a zero in the upper left corner, and working down and to the right until you've got an "echelon" of zeros, toward the lower right, row and column by row and column by row and column, trodding along ...).
 
Re: Five System Equation

Go^3 said:
I've gotten for example, t = 3.5 - 2s, but that's about as far as I've gotten.
Click to expand...
Well, if you got that right off the bat, then you KNOW what you're doing:
first, pick the 2 equations where the most "cancelling" is possible :idea:

Any teacher that asks for a "5 set" to be solved by elimination/substitution
should be shot at sunrise :shock:
 
Re: Five System Equation

Denis said:
... Any teacher that asks for a "5 set" to be solved by elimination/substitution should be shot at sunrise :shock:
Click to expand...


I realize that Denis is joshing, but I do not agree with the implication that the set of "any teacher" includes those elements which are teaching honor-level material to students who have been singled out to be challenged. An analogy is allowing elementary-grade students who show promise to skip the rote memorization drills for permanently stamping the complete multiplication table from 0X0 through 12X12 into some neural sub-network within the trillions and trillions of connections (more connections than the total number of stars in the universe, amazingly) of their brain, for ever and ever -- always available for instant recall.



Start today taking better care of your brain, all ye who read these words, because the current actuarial data show that, out of all of us who live to be at least 85 years old, half of us will be suffering from Alzeimers. For most people, it's easy to avoid winding up in that 50 percent whose life ends up fading away in a not-so-graceful manner, if you start today.
 
Re: Five System Equation

In a way, I agree with you both. I do agree that teachers who asks student to solve 5x5 system by hand - should solve a 10x10 system by hand in the class (fate worse than being shot).

I do agree that students should be stretched - but doing 5x5 set by hand is not it (heck, pka is even against partial fractions). I think, more can be achieved by say discussing "ill-conditioned" matrices - or error propagation in these types of calculations - than doing 5x5 by hand (where book-keeping is more important than mathematics).

But that will take too much thinking - it is much easier to tell students to do every problem at the end-of the chapter or solve 5x5 by hand.
 
Re: Five System Equation

Yeah, I'm not a fan of it either, but it needs to be done.

So which variable would be easiest to solve for? t?

If so, what should I do beyond t=3.5-2s?
 
Re: Five System Equation

Your equations my way :

1a + 2b + 1c + 1d + 2e = 1450 [1]
1a + 3b + 1c + 3d + 3e = 2725 [2]
1a + 4b + 2c + 1d + 2e = 1800 [3]
3a + 6b + 1c + 2d + 4e = 3125 [4]
4a + 3b + 4c + 1d + 1e = 2300 [5]

1a + 4b + 2c + 1d + 2e = 1800 [3]
1a + 2b + 1c + 1d + 2e = 1450 [1] : 2b + 1c = 350 [6]

Can't eliminate 3 in 1 shot no more, so try for 2:

1a + 3b + 1c + 3d + 3e = 2725 [2]
1a + 2b + 1c + 1d + 2e = 1450 [1] : 1b + 2d + 1e = 1275 [7]

4a +12b + 4c +12d +12e =10900 [2]*4
4a + 3b + 4c + 1d + 1e = 2300 [5] : 9b + 11d + 11e = 8600 [8]

4a + 8b + 4c + 4d + 8e = 5800 [1]*4
4a + 3b + 4c + 1d + 1e = 2300 [5] : 5b + 3d + 7e = 3500 [9]

We're in luck! [7][8][9] : 3 equations, 3 variables
You said you're ok with 3 variables...so go ahead!
Then get c from [6], then a : game over.
 
Re: Five System Equation

Thanks for the effort you put in to solving those equations!

But I have a new problem. I can't seem to solve for a variable because I end up getting 0=0 instead of b=4.8 or anything.

Here's when I tried to solve for b:

45b+90d+45e=57375
45b+55d+55e=43000
45b+27d+63e=31500


35d-10e=14375
28d-8e=11500

140d-40e=57500
140d-40e=57500

0=0
Click to expand...

Hopefully you can follow that.

I tried the same method for d and e, and got the same thing. If it would help, I can type it out too if I need to.

So what do I do from there?
 
Re: Five System Equation

Inthese problems it is hard to follow my own work - let alone somebody else's work. So I'll put my work - Matrices involved through Gauss-Jordan elimination scheme(shortened).
Code: 
1.00	2.00	1.00	1.00	2.00	1450.00
1.00	3.00	1.00	3.00	3.00	2725.00
1.00	4.00	2.00	1.00	2.00	1800.00
3.00	6.00	1.00	2.00	4.00	3125.00
4.00	3.00	4.00	1.00	1.00	2300.00

1.00	2.00	1.00	1.00	2.00	1450.00
0.00	1.00	0.00	2.00	1.00	1275.00
0.00	2.00	1.00	0.00	0.00	350.00
0.00	0.00	-2.00	-1.00	-2.00	-1225.00
0.00	-5.00	0.00	-3.00	-7.00	-3500.00

1.00	0.00	1.00	-3.00	0.00	-1100.00
0.00	1.00	0.00	2.00	1.00	1275.00
0.00	0.00	1.00	-4.00	-2.00	-2200.00
0.00	0.00	-2.00	-1.00	-2.00	-1225.00
0.00	0.00	0.00	7.00	-2.00	2875.00

1.00	0.00	0.00	1.00	2.00	1100.00
0.00	1.00	0.00	2.00	1.00	1275.00
0.00	0.00	1.00	-4.00	-2.00	-2200.00
0.00	0.00	0.00	-9.00	-6.00	-5625.00
0.00	0.00	0.00	7.00	-2.00	2875.00

1.00	0.00	0.00	0.00	1.33	475.00
0.00	1.00	0.00	0.00	-0.33	25.00
0.00	0.00	1.00	0.00	0.67	300.00
0.00	0.00	0.00	1.00	0.67	625.00
0.00	0.00	0.00	0.00	-6.67	-1500.00

1.00	0.00	0.00	0.00	0.00	175.00
0.00	1.00	0.00	0.00	0.00	100.00
0.00	0.00	1.00	0.00	0.00	150.00
0.00	0.00	0.00	1.00	0.00	475.00
0.00	0.00	0.00	0.00	1.00	225.00
I did through algebra - but I took assitance of excel to keep track of book-keeping (and fast calculation).
 
Re: Five System Equation

Thanks a ton!

I can kinda figure that out, I'm just trying to figure out what you did to go from the original set to the second set.

Like, for example, you got rid of all the first variables in all of the equations but the first.How did you go about doing that?

1.00 2.00 1.00 1.00 2.00 1450.00
0.00 1.00 0.00 2.00 1.00 1275.00
0.00 2.00 1.00 0.00 0.00 350.00
0.00 0.00 -2.00 -1.00 -2.00 -1225.00
0.00 -5.00 0.00 -3.00 -7.00 -3500.00
Click to expand...

Just trying to figure that stuff out so I can make sense of it all. Again, thanks a BUNCH.
 
Re: Five System Equation

I guess you didn't notice that equation [4] had not been used in obtaining [7],[8],[9]; it MUST be used somehow.

First, I'll change my [6]'s format to c = 350 - 2b [A]

Only one relationship is possible from [7],[8],[9]; I did it this way:
5b + 10d + 5e = 6375 [7]*5
5b + 3d + 7e = 3500 [9] : 7d - 2e = 2875 ; e = (7d - 2875) / 2

Now we bring in [4]; I did it this way:
2a + 8b + 4c + 2d + 4e = 3600 [3]*2
3a + 6b + 1c + 2d + 4e = 3125 [4] : -a + 2b + 3c = 475
Substitute [A], simplify: a = 575 - 4b [C]

Now we're ready to wrap up by substituting [A],,[C] in 2 of original equations; I did it this way:

1a + 2b + 1c + 1d + 2e = 1450 [1] ; substitute [A],,[C]:
2d - b = 850 [D]

1a + 3b + 1c + 3d + 3e = 2725 [2] ; substitute [A],,[C]:
9d - 2b = 4075 [E]

9d - 2b = 4075 [E]
4d - 2b = 1700 [D]*2 : 5d = 2375 : d = 475 : YEA!!

: e = 225

[D] : b = 100

[C] : a = 175

[A] : c = 150
 
Hurrah!

Awesome!I haven't tried it personally yet, but I'm 99% sure it's gonna work and I'm going to understand it, and I wanted to thank you before you got off. Geez, you guys should get paid for this. :p Thanks so much Denis and everybody that helped!
 
Re: Five System Equation

from row (2) subtracted row (1)

from row (3) subtracted row (1)

from row (4) subtracted row [(1)*3]

from row (5) subtracted row [(1)*4]
 
Re: Five System Equation

Awesome! I got it and it all made sense! Thanks so much! Now nobody has to waste their time trying to help me with this anymore :p
 
Re: Five System Equation

Go^3 said:
... Like, for example, you got rid of all the first variables in all of the equations but the first.How did you go about doing that?
Click to expand...


(Edit: Excuse me; I did not realize that there is a page 2 when I posted, so I had not read all of the posts up to this point.)

The suggestions are all good. I'm pointing out that the algorithm always gets you there, but is often more work than necessary. As you gain experience, you begin to see shortcuts.



Looks like you did not have the time to peruse up to a couple dozen references at Google to find the algorithm that gives you the steps to reduce an augmented coefficient matrix into RREF.

Here's the first two steps in that algorithm.

(1) If the element in [1, 1] (i.e., the leading coefficient in the first equation) is not one, then divide the first row by that element to make it so.

(2) Subtract a multiple of the first row [equation] from each of the remaining rows [equations] in order to zero out the leading coefficent in all rows but the first (i.e., the first column will have all zeros except for the first row, which will have a one).

The leading coefficient in your augmented matrix is already one.

Subtract row one from row two.

Subtract row one from row three.

Subtract three times row one from row four.

Subtract four times row one from row five.

Cheers,

~ Mark :)



MY EDITS: Added comments in red
 
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