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Author Topic: Algebra: System Applications  (Read 6233 times)

Offline -shiNe!

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Algebra: System Applications
« on: September 19, 2011, 01:00:00 PM »
There was a problem on my class work which, I know Is simple but I can't figure out the 3rd equation.

We're dealing with candy here:
Candy 1: Pepper mints (x)= $1.35 per pound
Candy 2: Gum (y)= $1.55 per pound
Candy 3: Jolly Ranchers (z)= $1.85 per pound

A dude buys some candy and the total price Is $19.26

I put:
Equation 1: x+y+z=19.26
Equation 2: 1.35x+1.55y+1.85z=19.26
Equation 3: ?

The last bit of Information which would fill the void for equation 3 says: "twice as many Jolly Ranchers than Pepper mints"

I can't figure out what It would be, I kept getting weird answers In my calculator for each one that I tried.

What do yall think?
I hope I gave yall all the Information I could, I turned It In, because It was due at the end of class, but I still want to know what It would have been.

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Offline jim360

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Re: Algebra: System Applications
« Reply #1 on: September 19, 2011, 01:08:01 PM »
For starters I think Equation 1 has to be wrong since it says something totally different from the correct equation 2. Secondly I would need to know how much the sweets he bought weighed in total. After I got that information I should be able to solve this.
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Offline -shiNe!

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Re: Algebra: System Applications
« Reply #2 on: September 19, 2011, 01:22:04 PM »
For starters I think Equation 1 has to be wrong since it says something totally different from the correct equation 2. Secondly I would need to know how much the sweets he bought weighed in total. After I got that information I should be able to solve this.
I don't think It said how much It all weighed. 1.35 multiplied by what ever the weight Is plus all the other variables =19.26 Is what we got to work with.

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Offline jim360

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Re: Algebra: System Applications
« Reply #3 on: September 19, 2011, 01:38:44 PM »
All I'm saying really is that combining your equation 1 with equation 2 gives x+y+z = 1.35x+1.55y+1.85z which will have no solutions. So I think you made a mistake when copying out the question or something like that, since equation one needs an answer like x+y+z=6.3 or some other number.

Anyway, equation 3 would just be z=2x.
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Offline -shiNe!

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Re: Algebra: System Applications
« Reply #4 on: September 19, 2011, 03:31:46 PM »

Equation 1: x+y+z=19.26
Equation 2: 1.35x+1.55y+1.85z=19.26


I never said x+y+z=1.35x+1.55y...etc.
They're two different equations, not one conjoined. So It does have an answer, the answer Is 19.26

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Offline BFM_Kiwi

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Re: Algebra: System Applications
« Reply #5 on: September 19, 2011, 09:39:38 PM »
If x = weight of Peppermints, y = weight of Gum etc.  then x + y + z = the total weight of the candy.  It would NOT equal 19.26, which seems to be the total PRICE.

Second thing I notice is that the prices all are multiples of 0.05, so any whole numbers are going to result in $$ amounts that are multiples of 0.05.  So to get a value of 19.26 you'd need a fraction of a pound, so you're not going to get any nice simple answers.

Equation 3 would mean you could substitute one of the two variables, but then you have just equation 2, which has two variables, so you can't get an answer without equation 1. 

I think Equation 1 must be x + y + z = <something> where that something is the total weight.  But I doubt it's 19.26 pounds.




Equation 1: x+y+z=19.26
Equation 2: 1.35x+1.55y+1.85z=19.26


I never said x+y+z=1.35x+1.55y...etc.
They're two different equations, not one conjoined. So It does have an answer, the answer Is 19.26


Offline jim360

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Re: Algebra: System Applications
« Reply #6 on: September 20, 2011, 02:46:21 AM »

Equation 1: x+y+z=19.26
Equation 2: 1.35x+1.55y+1.85z=19.26


I never said x+y+z=1.35x+1.55y...etc.
They're two different equations, not one conjoined. So It does have an answer, the answer Is 19.26


Just to be clear about it, they are two different equations to start with. But look at the number on the right hand side of both equations, it's 19.26 both times. So you can combine the two equations simply by writing

x+y+z = 19.26 = 1.35x+1.55y+1.85z

Because both equations express 19.26 in terms of x, y, and z. So now all you do is cross out the 19.26 in the middle and you find that x+y+z = 1.35x+1.55y+1.85z. Which has no solutions.

Remember when you are solving these equations you will be combining them anyway, so they aren't separate. You copied down the question wrongly, or missed out the information about how much the total weight was.
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Offline -shiNe!

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Re: Algebra: System Applications
« Reply #7 on: September 20, 2011, 12:47:06 PM »
I get what yall are saying, but I did not get the total number of pounds In the question.

When I get the assignment back, I'll post It up to full.

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Offline BFM_Rickp9

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Re: Algebra: System Applications
« Reply #8 on: September 20, 2011, 01:46:42 PM »
I'll pass this to my son he is a wiz in algebra.



    


Offline BFM_SüprM@ñ

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Re: Algebra: System Applications
« Reply #9 on: October 02, 2011, 09:55:49 AM »
1.35x+1.55y+1.85z=19.26

z = 2x

1.35x+1.55y+1.85(2)x=19.26

1.35x+1.55y+3.70x=19.26

5.05x+1.55y=19.26

IF you're just looking for the equations then

Equation 1 would be the 1.35x+1.55y+1.85z=19.26

E2 would be the z=2x

and E3 would be the 5.05x+1.55y=19.26, which as they said would give you a fraction of a pound somewhere in there since it's 19.26 doesn't end in a 5 or 0.
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