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May 9th, 2012, 08:59 PM   #1
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Variables

I have been helping my daughter with her Jr. High math; it has been years since I have done these types of problems so I don't remember why it works. And it appears to me that this formula doesn't work after all. Here is the problem she was given and her math steps below to solve the problem.

15x + 12y = 141

141 divided by 12 = 11.7
7*15 = 105
105 divided by 15 = 7
x = 7
141- 105 = 36
36 divided by 12 = 3
y = 3

15*7 + 12*3 = 141

It was a story problem and the answer is correct, it was easy enough to do in my head. I hope you can understand how we worked the problem, I am sure it is not written with the correct math lingo.

My question is this: What am I missing? When the problem gets more complex and depending on the situation the variables x and y could be anything.

Example: I want to find out how many bricks are on a pallet, I use an accurate scale to weigh the pallet and determine I have 3000 lbs of bricks. I also determine I have 2 different types of bricks and I weigh each of them, one is 5 lbs and the other is 2 lbs. How many bricks of each type are on the pallet?

This could be written as:
5x + 2y = 3000

There could be 598 - 5 lbs bricks and 5 - 2 lbs bricks
5*598 + 2*5 = 3000
Or
5*596 + 2*10 = 3000
Or
5*30 + 2*1425 = 3000

And so on and so on and so onů
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May 9th, 2012, 10:06 PM   #2
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Re: Variables

In your first equation there are two variables and 1 equation, so unless we are restricting the variables to be positive integers, there will be an infinite number of solutions. We could divide through by 3 to get:



I would look at subtracting multiples of 5 from 47 to see which are multiples of 4:

no

no

yes " />

no

no

no

yes

no

no

Since , once we find the first solution for x, we know the next will be 4 multiples of 5 greater.

In your example, 3000 is divisible by both 2 and 5, so there are many solutions given by:



So, as long as y is a multiple of 5 where , you will get a non-negative integral value for x. Hence, there are 301 ordered pairs that will work.
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May 10th, 2012, 08:38 AM   #3
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Re: Variables

Quote:
Originally Posted by gbutler1492
My question is this: What am I missing? When the problem gets more complex and depending on the situation the variables x and y could be anything.

Example: I want to find out how many bricks are on a pallet, I use an accurate scale to weigh the pallet and determine I have 3000 lbs of bricks. I also determine I have 2 different types of bricks and I weigh each of them, one is 5 lbs and the other is 2 lbs. How many bricks of each type are on the pallet?

This could be written as:
5x + 2y = 3000
Usually, another "clue" is given; like: there are 100 more 2lb bricks than 5lb bricks; then:
x = 5lb bricks
y = 2lb bricks
(important to show above; plus keeps teacher in good mood!)

5x + 2y = 3000 [1]
y - x = 100 [2]

From [2]: y = x + 100 ; substitute in [1]:
5x + 2(x + 100) = 3000

Solve that for x, then go get y ; OK?
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