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New Member
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Sep 30, 2007, 11:49 AM
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How to set up equations out of word problems.
I am having some trouble thinking of how to set up equations out of word problems most of them I can get but there were a few on my homework that I am not getting.
Here is the Problem:
A mixture of 12 liters of chemical A, 16 liters of chemical B and 26 liters of chemical C is required to kill a destructive crop insect. Commercial spray X contains 1, 2, and 2 parts respectively, of these chemicals. Commercial spray Y contains only chemical C. Commercial spray Z contains only chemicals A and B in equal amounts. How much of each type of commercial spray is needed to get the desired mixture?
I need to set up three equations out of this to solve it because there are three variables I am trying to solve for.
I get to this point:
X = 1/5A + 2/5B + 2/5C
Y = C
Z = 1/2A + 1/2B
but I can not see how to solve for the variables so I am thinking that these equations are wrong.
So my question is:
How can I figure out what is important and what isn't important in the word problem above?
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Senior Member
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Sep 30, 2007, 01:51 PM
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The way I would look at this is:
since you can add chemical C directly from Spray Y that doesn't matter yet.
spray Z only includes A and B in equal amounts, so it doesn't help you get 12 of A and 16 of B so that's no good yet
spray x is how you get the difference between a and b (4 litres difference), to do this you need 20 litres of spray x giving 4A, 8B and 8C
you then add 16 litres of spray Z to bring A up to 12litres and B up to 16litres C remains at 8 litres so you add 18 litres of spray Y to finish.
I'm not sure how you do the equasions for this or even if you can but that is my logic - hope it helps.
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Ultra Member
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Sep 30, 2007, 02:20 PM
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StuMegu did good by reasoning it out.
Here's an algebraic way to set it up using three equations:
Solving, we get:
20 liters of X, 18 liters of Y, and 16 liters of Z
Which equals 54 liters as needed.
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New Member
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Sep 30, 2007, 02:45 PM
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hmm I am a bit lost in StuMegu's reasoning, seems like he/she is guessing, I know that I am suppose to use Gaussian Elimination to solve this. I am not sure how you, galactus, got those equations. If you could explain it might help.
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New Member
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Oct 1, 2007, 02:45 PM
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Ok I got some help from my math teacher today. Basically, you know you need chemicals A, B, and C in the given amounts. And you know of mixtures that can give you those amounts.
So...
Solution X: 1/5A + 2/5B + 2/5C
Solution Y: C
Solution Z: 1/2A + 1/2B
So which solutions have amounts of chemical A?
X and Z
in what amounts?
1/5 and 1/2
so..
1/5X + 1/2Z = 12
the same logic follows the remaining ones
final equations:
1/5X + 1/2Z =12
2/5X + 1/2Z = 16
2/5X + Y = 26
Which are the same equations that galactus posted, just in fraction form and re-arranged.
So this problem is fixed!
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Junior Member
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Oct 1, 2007, 03:12 PM
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Wow you all make me feel really stupied but that's okay... I'm sure ill need your help to figure out a couple of question ( grade 11 pre-cal) it's the same type of stuff but with quadreatics and stuff
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