Algebra- find solution using two variables

In summary: This will make the working easier, since then you don't have to use decimal fractions, but it will not change the solution. The equation8x+20y=864will give you the same results as the equation above.In summary, the chemist needs to create a 72 ml solution with 12% alcohol, using only 8% and 20% alcohol solutions. This can be achieved by setting up the equation 8x + 20y = 864, where x represents the amount of 8% solution and y represents the amount of
  • #1
puma072806
7
0

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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  • #2
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.
 
  • #3
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
 
Last edited:
  • #4
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 ?
 
  • #5
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?
 
  • #6
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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FAQ: Algebra- find solution using two variables

What are variables in algebra?

In algebra, variables are symbols or letters that represent unknown quantities or values. They are used to write and solve equations.

How do I find the solution using two variables?

To find the solution using two variables, you will need to have two equations with two different variables. You can then use methods such as substitution or elimination to solve for the values of the variables.

What is the importance of finding solutions using two variables?

Finding solutions using two variables allows us to solve real-world problems and make predictions. It also helps us understand the relationship between two quantities and how they affect each other.

Can I have more than two variables in an algebraic equation?

Yes, it is possible to have more than two variables in an algebraic equation. However, it may be difficult to solve for multiple variables at once, so most equations are limited to two variables for simplicity.

What are some real-life applications of finding solutions using two variables in algebra?

Finding solutions using two variables is useful in many fields, including finance, engineering, and science. It can be used to calculate interest rates, determine optimal solutions for business decisions, and model relationships between different physical quantities.

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