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# Solving Mixture Problems using a System of Equations

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Author: Colleen Atakpu
##### Description:

This lesson will demonstrate how to solve mixture problems using a system of equations.

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Tutorial

## Video Transcription

Today, we're going to talk about solving mixture problems using system of equations.

A mixture problem involves combining two or more things, and then determining a quantity, such as price, percentage, or concentration from that mixture. And we can use a system of equations to develop relationships between our known and unknown variables from the problem, and then solving the system of equations. So let's see how we could do that through an example.

So let's do an example. Let's say that I have two solutions, A and B. Solution A has a concentration of 30% alcohol and Solution B has a concentration of 90% alcohol. I want to mix these two solutions to get an overall concentration of 75% alcohol, and I want 40 milliliters of this mixed solution.

So let's start by defining a couple of variables. I'm going to define a variable x to be the amount of Solution A that I have. And then I'll define a variable y to be the amount of solution B that I have. So I'm going to write two equations to represent this situation.

So the first equation I'm going to write is to show that I want a total of 40 milliliters of my mixture of Solution A and B. So that means my equation would be x, the amount of Solution A I have, plus y, the amount of solution B that I have, should equal 40, or 40 milliliters.

Then, my second equation should show that I want an overall concentration of 75%. So I'm going to multiply my concentration of alcohol of 30% by the amount of Solution A that I have. So that will be 0.30 times x. And I'm going to add to that 90%, or 0.90, times the amount of Solution B that I have, or my y-variable. So I'll have 0.90 times y.

And I want that to have an overall percentage of concentration of 75%. So that's going to be 0.75 times the total amount of A and B mixed together. Which again, is 40 milliliters. Or, it's going to be my variables x plus y.

So now that I have my equation, I can see here that I can combine these two equations by substituting my value for x plus y into my second equation. So this would tell me that this evaluates to be 0.75 times 40. And 0.75 times 40 is just going to give me 30. So my second equation will be this. So now, let's see how we can solve this system of equations to determine the amount of Solution A and the amount of Solution B that I need.

So let's see how we can solve this system of equations for x and y. So I'm going to use the substitution method because I can easily isolate either my x or my y-variable in my first equation, and then substitute it into my second equation.

So I'm going to isolate my x-variable. And I'm going to do that by subtracting y from both sides, which would make this equation x is equal to 40 minus y. So now that I know what x is equal to, I can take that expression for x and substitute it into my second equation for x.

So now my second equation is going to become 0.30 times 40 minus y plus 0.90y equals 30. So now that I have only a single variable in my equation, I can solve for that variable.

So I'm going to start by distributing. 0.30 times 40 will give me 12. 0.30 times minus y will give me minus 0.30y. Bring down my other terms.

I'm going to combine these two terms together. Negative 0.30y plus 0.90y is going to give me a positive 0.60y. Bring down my other terms.

I'm going to subtract 12 from both sides, which will give me 0.60y is equal to 18.

And dividing both sides by 0.60, I find that y is going to be equal to 30. So I know that y, or the amount of Solution B that I need, is going to be 30, or 30 milliliters.

So now, I can use either one of my original equations, or even my third equation, to figure out the value of x, which is the amount of Solution A that I need. So I'm going to go ahead and use my third equation because that's the easiest as x is already isolated.

So that will give me that x is equal to 40 minus my value for y, which I just found is 30. And 40 minus 30 is 10. So x, or the amount of Solution A that I need, is 10, or 10 milliliters.

So let's go over our key points from today. As usual, make sure you get them in your notes if you don't have them already so you can refer to them later.

A mixture problem involves combining two or more things and then calculating a quantity from the mixture, such as price, percent, or concentration.

A system of equations can help solve mixture problems by creating equations to represent relationships between known and unknown quantities.

So I hope that these key points and examples helped you understand a little bit more about solving mixture problems using systems of equations. Keep using your notes and keep on practicing and soon you'll be a pro. Thanks for watching.