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| Mad Scientists Science and Mathematics discussion-- theories, arguments, citations, proofs and pudding. |
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#1
 2008-09-19, 05:11
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stiletto 
Regular
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Systems of Equations
I'm having a little trouble with setting up word problems of systems of a equations. I can work through them when i see the problems out in front of me, but extracting the coefficients and variables out of the word are a bit tough.
Problem: How many liters of a 20% acid solution and how many liters of a 40% acid solution must be mixed to to produce a 40 liter 25% acid solution.
I'm guessing i can set up one equation as X(liters of 20%)+Y(liters of 40%)=Z(40 liters of 25%)
Would this become .20X+.40Y=Z ? as for the second equations i'm not sure of how to set it up, but i'm guessing with the use of substitution i can solve this.
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#2
 2008-09-19, 23:19
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Shadout Mapes 
Regular
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Re: Systems of Equations
Where did your 40 liters with 25% acid go?
Your problem seems to just be with visualizing - which is admittedly a little difficult in this particular problem. The way I think about it, 20% of the first volume is all acid, 40% of the second volume is all acid, and 25% of the 40L volume (just 10L!) is all acid. Since the amount (volume) of acid should be conserved:
.2x + .4y = 10
Similarly, the total amount of solution (acid + solvent) should also be conserved throughout the experiment - in other words:
x + y = 40
Then the math is trivial. I guess a part of doing these kinds of problems is getting used to the wording of whoever wrote them, and hoping there isn't much implicit information.
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#3
 2008-09-27, 16:53
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Moderator
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The fine line between genius and madness
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Re: Systems of Equations
If X=the volume of the 20% solution, and Y=the volume of the 40% solution, then:
X+Y=40
and
20%X+40%Y=25%(40)
X+Y=40
.2X+.4Y=10
X+Y=40
.8X+1.6Y=40
X+Y=.8X+1.6Y
5X+5Y=4X+8Y
Y=1/3X and X=3Y
so by substituting the values in the X+Y=40, you get
X+1/3X=40 and 3Y+Y=40
4/3X=40 and 4Y=40
X=30 and Y=10
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