Hi, and welcome. My name is Anthony Varela. and today I'm going to be solving a mixture problem using a system of equations. So we're going to review mixture problems, then we're going to be using information from a mixture problem scenario to setup a system of equations. And we're going to be solving our system using substitution. So there are many different ways to solve a system of equations, but we're going to be using the substitution method in our example.
So let's talk about mixture problems. Now when we think about the classic mixture problem that you're most likely to see in some math or science textbooks, you might see something that looks like this. We have a 1 liter of a 50% concentrated solution, so 50% of some chemical. And we're going to add that to 1 liter with a 20% concentration. So what happens when we add these two together is, of course, we're going to get 2 liters because we're adding 1 liter plus 1 liter, but the percent concentration is not going to be 50, it's not going to be 20. It's going to be a mix of the two.
Now in this case, because I was adding equal amounts, I could just take a simple average of 50 and 20. So that's going to be 35%. And we can see that 1 times 50 plus 1 times 20 equals 2 times 35. So we can use a system of equations to help us solve some mixture problems if you wanted to calculate a certain quantity, like liters, or if you wanted to calculate a certain percent concentration.
So we're going to have a bit of fun here with our scenario, and this is going to involve chocolate chip cookies. And we're going to deal with a percent concentration of cocoa. So here's our situation. I'd like to make some chocolate chip cookies, and the recipe says use two cups of semi-sweet chocolate chips. And semi-sweet chocolate chips are 30% cocoa.
The problem is I don't have semi-sweet chocolate chips. What I do have, I have dark chocolate chips-- that's 55% cocoa-- and I have milk chocolate chips. That's 15% cocoa. So I need to then measure out a specific amount of dark chocolate and a specific amount of milk chocolate, combine them together, and I'm going to get two cups of 30% cocoa.
So let's go ahead and define our variables that will help us set up a system of equations. So what do I not know? I don't know how much dark chocolate chips to use and how much milk chocolate chips to use. So those are going to be my variables. I'm going to say that x equals the cups of dark chocolate chips, and y equals the cups of milk chocolate chips.
So now I already know that I need two cups total. So x plus y is going to equal 2. That's going to be one of my equations in my system. Now I need to set up an equation that relates these two cups of 30% and my x amount of 55% and y amount of 15%.
So here's how I'm going to set up this equation. I'm going to multiply the quantity by its concentration. So then I'll take 55%, write this as a decimal, and multiply that by x because 55% cocoa is dark chocolate chips. X represents the quantity of dark chocolate chips. Now I'm going to add to that, then, the concentration of milk chocolate chips multiplied by that quantity. So this would be plus 0.15y because 15% as a decimal is 0.15, and y represents our quantity of milk chocolate chips.
Now what is this going to equal? This is going to equal the 30% concentration that we're looking for, and this is a combination of dark and milk chocolate chips, so x plus y. So there is our second equation. This makes up our system because x and y in both equations mean the same thing-- cups of dark and milk chocolate chips.
Well we already know that x plus y equals 2. So I'm going to plug in 2 in my first equation. And I can just evaluate that since there's no variables on the right side there. So I know that that's going to equal 0.6.
So here's my system of equations. Now how am I going to then solve for x and solve for y? Well, I'm going to solve this by using substitution. So we're going to be rewriting one of our equations in terms of one of our variables and then substitute it into our other equation.
So I'm going to take x plus y equals 2. And it doesn't matter if you would choose to isolate x or y, so I'm going to take away x from both sides, and now I have an expression for y. I know that y equals 2 minus x. So I can go ahead and take my other equation, and I'm going to substitute then 2 minus x in for y. So here I see y. Let's go ahead and replace that then with 2 minus x.
All right. Well now what I'm going to do, I have a single variable equation here. I don't see any y's in this equation. I just see x's. So I can solve for x. I'm going to then distribute the 0.15 into the 2 and the minus x. So all of this stays the same so far, and then we have 0.15 times 2. That's right here. And then we have 0.15 times negative x, so it's negative 0.15x. And this all still equals 0.6.
Well I notice I have two x terms, so I can go ahead and combine those terms. So 0.55x minus 0.15x, that's going to give me a 0.4x. I still have my plus 0.3, and that still equals 0.6.
Well I'd like to isolate that x term, so I'm going to subtract 0.3 from both sides of this equation. So 0.4x equals 0.3. I'll divide by that coefficient there, and I know then that x equals 0.75. So now then you know that x equals 0.75, I can substitute that into one of my equations from our original system and solve for y.
So x equals 0.75. This is the cups of dark chocolate chips that I'm going to use. Let's go ahead and bring out another equation, x plus y equals 2. So now I'm just going to substitute then 0.75 in for x. So 0.75 plus y equals 2. And then I can easily solve for y, so subtract that 0.75, and y then equals 1.25.
Now what does y stand for, remember? That stands for our cups of milk chocolate chips. So I have a value for x, and I have a value for y. I know that in my recipe, I'm going to be using 3/4 of a cup of dark chocolate chips and 1 and 1/4 cups of milk chocolate chips. That will give me 2 cups of essentially semi-sweet chocolate chips-- 30% cocoa.
So let's review solving mixture problem using a system of equations. First, we want to define our variables. Give them a very specific definition, so we can make sense of the context. Then we can develop equations for the system. And remember, in both equations, those variables have to have the same definitions.
Then we can rewrite an equation so that we may substitute that into another equation. We can solve then for one of our variables, and then bring that back to the original equation and substitute again to solve for another variable. And that's, in general, steps for solving a system of equations using substitution. So thanks for watching this video and solving a mixture problem using a system of equations. Hope to see you next time.
