Algebra- find solution using two variables

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To create a 72 ml solution of 12% alcohol using 8% and 20% solutions, the chemist needs to set up two equations: one for the total volume (x + y = 72) and one for the total alcohol content (0.08x + 0.20y = 8.64). The amount of alcohol in the final solution is calculated by multiplying the desired percentage (0.12) by the total volume (72 ml), resulting in 8.64 ml of alcohol. The first equation ensures the total volume of the mixed solutions equals 72 ml, while the second equation accounts for the total alcohol content. Both equations are necessary to solve for the volumes of the 8% and 20% solutions needed.
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Homework Statement


A chemist ran out of a 72 ml solution of 12% alcohol. All that is left in the lab is 8% alcohol and 20% alcohol. How many ml of 8% and 20% would be needed to make a 72 ml solution of 12%?


Homework Equations





The Attempt at a Solution


8x + 20y = 12
 
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What would x+ y represent in this situation?

Of course, if you have x ml of 8% solution then .08x (not "8x") would be the amount of alcohol in it and if you have y ml of 20% solution then .20y would be the amount of alcohol in that. So .08x+ .20y would be the amount of alcohol in the combined soltion. But that is NOT 12 or .12. .12 is the percentage of alcohol in the 72 ml solution. What do you need to multiply the .12 by to get the amount of alcohol? You can then multiply the entire equation by 100 to get rid of the decmals.
 
Would I multiply .12 by 72 ? Which would then be 8.64

This is what I have so far.

.12 X 72=8.64

.08x + .2y=8.64

(100) .08x + .2y=8.64 (100)

8x + 20y= 864
 
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8x+20y=864 that is good for amount of alcohol
but you are missing one more equation, what's the total amount of liquid that should be made ?
 
puma072806 said:

Homework Statement


A chemist ran out of a 72 ml solution of 12% alcohol. All that is left in the lab is 8% alcohol and 20% alcohol. How many ml of 8% and 20% would be needed to make a 72 ml solution of 12%?


Homework Equations





The Attempt at a Solution


8x + 20y = 12

One way to think about such problems is to realize that the figures 12%, 8% and 20% are, essentially, measures of the number of alcohol molecules in a ml of solution. (The actual numbers could be found by going to physical/chemical tables and computing some conversion factors.) When you take a bottle of V1 ml of 8% solution (containing N1 alcohol molecules) and V2 ml of 20% solution (containing N2 alcohol molecules), how much solution do you have altogether? How many alcohol molecules are in the new solution? How does that translate into percentage terms?
 
HallsofIvy said:
What would x+ y represent in this situation?

Of course, if you have x ml of 8% solution then .08x (not "8x") would be the amount of alcohol in it and if you have y ml of 20% solution then .20y would be the amount of alcohol in that. So .08x+ .20y would be the amount of alcohol in the combined soltion. But that is NOT 12 or .12. .12 is the percentage of alcohol in the 72 ml solution. What do you need to multiply the .12 by to get the amount of alcohol? You can then multiply the entire equation by 100 to get rid of the decmals.

puma072806 said:
Would I multiply .12 by 72 ? Which would then be 8.64

This is what I have so far.

.12 X 72=8.64

.08x + .2y=8.64

(100) .08x + .2y=8.64 (100)

8x + 20y= 864

The big hint is, again, what does x+y represent? What exactly does this example chemist want? Your original equation was good, but you need one more equation.

What HallsofIvy is trying to explain is that you can use decimal fractions instead of percentage, so your first equation can be
0.08x+0.20y=0.12
 
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I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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