Thread: Systems of Equations

A chemist has three acid solutions at various concentrations. The first is 10% acid, the second is 20% acid, and the third is 40% acid. How many milliliters of each should he use to make 100 mL of 18% solution, if he has to use four times as much of the 10% solution as the 40% solution?

I have no idea how to go about this. I'm sure I could solve the problem if I could just set it up! So can anyone help me set up a system of equations for this problem?

EDIT: I've been contemplating this system:

x + y + z = 100
.1x + .2y + .4z = .18
x = 4z

But I don't know how I'd go about setting up a matrix for that...

Originally Posted by blacksuzaku

A chemist has three acid solutions at various concentrations. The first is 10% acid, the second is 20% acid, and the third is 40% acid. How many milliliters of each should he use to make 100 mL of 18% solution, if he has to use four times as much of the 10% solution as the 40% solution?

I have no idea how to go about this. I'm sure I could solve the problem if I could just set it up! So can anyone help me set up a system of equations for this problem?

Let x=10% acid y=20% acid z=40% acid.

We know that the total volume of the solution should be 100 mL so

is the first equation

For the next one the percents multiplied by the volume used need to add up so we get



Note: I multiplied the above equation by 10 to eliminate the decimals.

For the last equation we know that he uses 4 times as much of 10% as 40%

note that the equation is not this is a common error.

We now have a system of three equations with three unknowns.

Good luck.

But I don't know how I'd go about setting up a matrix for that...

If you don't mind me asking, how does E2 become x + 3y + 4z = 180? Wouldn't it be x + 2y + 4z = 180?

The second solution is 20% acid, not 30%. ^_^;;

Originally Posted by blacksuzaku

If you don't mind me asking, how does E2 become x + 3y + 4z = 180? Wouldn't it be x + 2y + 4z = 180?

The second solution is 20% acid, not 30%. ^_^;;

Yes you are correct. I will fix my above post.

Aww, don't cry. Your answers are still extremely helpful and I thank you for life for helping me. After that, it was a simple matter to reduce the matrix to find a solution of x=50, y=40, and z=10. ^_^

You have saved me from getting gray hairs at the tender age of 22.