Now that we've seen solving a system of equations using substitution and elimination, let's see how we can use those skills to solve some more interesting problems. Let's try one with interest. You invest $10,000 in two funds, one of the funds pays at 7.5% and another fund pays at 8%. You earned a total interest of $787.50. So, how much money did you invest at each rate? Maybe you lost your tax forms or maybe you don't have your bank statements. Either way, you can use algebra to help figure out how much money you invested in your trade. When creating a system of equations, we know that we need two variables, x and y. This question actually indicates what variables we should use. We're looking for how much money was invested at 7.5% and at 8%. So that's going to be our two variables. I'm going to go ahead and let x be the amount we invested at equation could you write involving the amount of money you invested? You want to use an amount plus the amount should equal your total amount. Write that equation here. And remember, it should involve x and y.
We know the total amount of money we invested was $10,000. And if I add up the other amounts together they should total to this. Well I don't know how much I invested at 7.5%, that's x. And I don't know how much I invested at 8%, that's y. But I know that if I add those two unknown amounts together I should get $10,000, so this is my first equation.
Now that we have what equation, we need to set up another one. The other equation needs to involve interest. I know if I add the interest from one account and the interest from the other account, I'll get my total interest of $787.50. But I need to find the interest first. To find the interest, we take the rate and multiply it by the amount of money. For example, one of their rates is 7 point 5 percent. And, the amount of money that I invested at 7 point 5 percent was, well, x. I don't know it. Remember, our equation should involve two variables, x and y. So, we need to make sure that it appears in the terms. We convert the percentage rate, 7 point 5 percent, to a decimal, and multiply it by x. So, this is the first term. So what are the other terms in the equation. What should go here in the second interest. What should go here in the total interest. You can write the terms in these boxes.
I know the second rate must be 8% because that was from our other fund. The second amount was y. That's the amount we invested at 8%. We don't know it, so that goes here. We convert 8% to a decimal and get 0.08y or .08 times y. Our total interest we actually knew. We gained $787.50. So that goes here. This quiz is pretty tough, and there's a lot of algebraic reasoning going on. So don't worry if you didn't get it right on your first try. Stay with me so that we, we can continue to make sense of this word problem.
So, with some logic and some algebra, we can come up with two equations for this problem. We used the individual amounts from each account to equal the total amount. And we used the interest rates and those amounts to get our total interest. Now that we have two equations, we can use them to figure out how much was invested at each rate. I'm going to show you how to solve the system using substitution. I think it's easier than elimination because I can either isolate the x in this first equation or the y. If I subtract x from both sides, I can solve for y. So y would equal 10,000 minus x. Now that y is solved for, I can replace this y with 10,000 minus x, this is my substitution. I've made the substitution in this line, the I distribute the 0.08. Multiplying by 10,000 moves my decimal point right four places, so I get 800. And then I have minus constant terms on the right, and then my x terms on the left. Next, I combine my like terms and I get negative 0.005 times x. Then, I divide by the coefficient of x to get x is equal to 2,500. But we need to make sense of what this means. Let's look back at our problem. X is 2,500. Well, that's the amount of money I invested at 7.5%. Knowing this, I want you to finish out the rest of this problem. How much was invested at 8%? Write your answer here.
Well, we know we invested a total amount of $10,000. So x plus y must be 10,000. If I solve for y here, I know that number must be 7,500. Great thinking if you got that right. We know when we plug in the values for x and y, we get a true statement in this first equation. And it turns out if we plug in x and y in our second equation, we'll also get a true statement. [inaudible] Try working out this math to make sure that we have the right answer. The equations are checked, which means that we have the correct x and y. It works in both equations.
Here's another problem involving interest. This time you invest money at 7% and money did you invest at each rate? To get started solving this problem let's first come up with some equations. We'll let x be the amount of money we invested at 7% and y be the amount of money we invested at 8%. Using these variables write two equations for representing this problem. Use the ideas of amount, and amount, plus the total amount. And interest plus interest should be total interest. And make sure to use those variables. Good luck.
We don't know the first amount, so that's x. And we don't know the second amount, so that's y. If we add the two amounts together, we get a total amount of $25,000, just like in our problem. For our second equation, we know our interest rate for this first amount was 7%. So I have 0.07, 7 hundredths times x. Our second amount gained 8% interest. So we have 8 hundredths times y, and our total interest was 1900, that was written in our problem. If you're able to write both of these equations that's exceptional. This is strong algebraic reasoning.
Here are the two equations that we just found, and I want you to figure out, how much money was invested at each rate? How much did we invest at 7%, and how much did we invest at 8%? Try your best.
After solving, we find that x is $10,000 and y is $15,000. Congrats if you got both of these right. If you didn't, that's okay, too. Follow along with me in the solution, and then pause it when you find a mistake. Then continue solving, and see if you can get these answers. Remember, mistakes are okay. And this is the time that we learn from them. I'm going to rewrite this first equation by subtracting x from both sides to get y is equal to 25,000 minus x. I did this so I could use substitution. I think that's the easier choice here. Now I perform the substitution, then I can distribute my 8 hundredths to both terms. Here's my first term, 2,000, and my second term, negative 8 hundredths times x. I subtract my constant term to move every constant to the right. And then that leaves me with x on the left. I combine my like terms, then I divide by the coefficient of x. I get x is equal to 10,000 or $10,000 which we invested at 7%. And remember this is a system of equations. And so we have two variables, x and y. Well we know if x is $10,000, y must be $15,000, since those two have to add up to $25,000. And as always, we can check our answers. By plugging in x and y into both equations, we'll get true statements.
Now that we've done some interest problems, let's try a mixture one. Mixture problems like this are helpful if you're at home and you want to make lemonade or punch. Usually, you're trying to make a different concentration than what you already have in your fridge. Scientists also use this approach when they're trying to make different concentrations of chemical solutions in the laboratory. Let's see how it works. Let's use the same approach as before with rates and amounts to try and set up equations. The wording is slightly different here because we're working with volumes and percentages, but I think you'll see a pattern. For this problem we have 5 gallons of a punch mix with 65% water. We want to mix it with a punch mix that's 80% water. So that way we get a mix that's 75% water in the end. The question is, how much of this 80% mixture should we use? These are the two equations that we're going to try and set up. The first one takes into the account the gallons of water. For example, the first volume that I have is 5 gallons of punch mix. This punch mix is 65% water. So if I take my percentage of water and multiply it by the gallons of punch, I'll get my gallons of water. I change 65% to its decimal, and multiply by 5, to get 3.25 or 3 and 25 hundredths. Remember the question can help us identify the variable. We don't know how much punch mix we need that's 80%, so I'm going to let that variable be x. Using x, and the information in the problem, I want you to complete our equation. What should go in these boxes? Mixes. Don't worry about volume 3. We actually don't know that quantity yet.
I'm mixing a punch mix that has 65% water with a punch mix that has 80% water. So 80% as a decimal is 0.80. I don't know how much of it I'm using, so that volume is x. Multiplying the terms together I get 0.8 times x. For this third percentage, we know that our punch mix needs to wind up being 75% water so I want to write this as a decimal. Keep in mind that we have 65% of a punch mix that's water, and five gallons of that punch mix. When I multiply these numbers together, I get gallons of water. So I have gallons of water with gallons of water should equal my total gallons of water. We'll check this in the end to make sure it's true.
Notice that we have an equation, but we still have an unknown, this total final volume. I'm going to use a bucket to try and help us figure it out. Let's see if we can get an expression for this volume. We know, for our first volume, we had 5 gallons. So, I'm going to pour those 5 gallons into my bucket. For the second volume, I don't know how much of it I used. We're going to pour an x amount of that. So, we know our final volume is going to be 5 gallons plus x gallons. This is the amount of gallons we wind up in the end, our volume 3. Now, that we know that the total volume is 5 gallons plus x gallons, we have one equation with one variable. Now, we're ready to solve. So, you solve this. How much punch mix should we use that's 80% water to get our 75% water punch mix? Write your answer here. And I've already included the unit of gallons, so don't write that.
We should use 10 gallons of a punch mix that's 80% water to get a punch mix that's 75% water. Here's how we can do that. First, we distribute, next we subtract 3 and 25 hundredths from both sides. This gets my constant on the right, which means I want to move my x to the left. Then I subtract this term from both sides. And finally, I divide by the coefficient in front of x, 5 hundredths. So I get x is equal to 10. 10 gallons.
Try solving this mixture problem. You have 10 liters of a chemical mixture that is 8% water, and you want to add to that a mixture that is 15% water to get a mixture that's 10% water. How much of the 15% mixture should you use? One thing that you should notice in mixture problems is that the percentage of your final mixture must be in between the percentage of the other two mixtures. The because we can only come up with the percentage that would be between these two. I couldn't make a 60% solution from an 8% water solution and a 15% water solution because, well, these two would be too low.
To make a 15% mixture, we would need 4 liters. If you got that one right, that's amazing work. You've really mastered how to solve these system problems. If this one gave you trouble, that's okay. Stay with me to figure out what might have gone wrong. The question asks for how much of the 15% mixture we should use. So that's going to be our variable, x. We start with 10 liters of a chemical mixture that is 8% water. So that's going to be my percent 1 and volume 1. For percent 2 and volume 2, I'm adding 15% of x liters of solution. And for my final percentage, I know the mixture needs to be 10% water. I don't know my final volume. But if I remember back from the bucket, if I pour in one volume, and I pour in another volume, that gives me my total volume. So I poured in 10 liters first, and then I poured in x liters. This must mean that the total volume is 10 plus x. I can use that here. Now we have one equation with one variable, and we can solve for x. I multiply my terms together, and then I distribute the 0.10. Next, I subtract 0.8 from both sides. I'm moving my constant terms to the right. This means that I must move my x terms to the left. I subtract 0.1x from both sides to get 0.05x equals 0.2. And finally I divide by the coefficient of x, to get x is equal to 10 liters.
To start solving this problem we're going to let d equal the number of dimes and q equal the number of quarters. The question indicates the variables that we need to use. In this case we need to know how many of each coin we have. So we need two variables, d and q. We need to set up two equations. One will involve the number of coins And the other would involve the value of the coins. We know the number of dimes plus the number of quarters has to equal a total number of coins. A register only contains dimes and quarters and it has a total number of number of dimes so we use d, and we don't know the number of quarters so we use q. We know the total value of the coins is $2.70. A dime is worth $0.10. So to get the value of dimes we multiply $0.10 by the number of dimes we have. A quarter on the other hand is worth $0.25 cents. So to get the value of the quarters. We multiply 25 cents times our number of quarters, q. And if we add these two values together we should get $2.70. I've rewritten the two equations down here and now we can use either substitution or elimination to solve this problem. I'm going to use substitution since I can easily solve for one of the variables in this first equation. If we subtract q from both sides of this equation, we'll get d is equal to 21 minus q. Now that d is solved for we can perform substitution and plug in 21 minus q, and for d in our second equation. Next I distribute 1 tenth to 21 and negative q. I combine like terms. Nest I subtract 2 and 1 tenth from both sides. And finally I divide both sides by fifteen hundredths. I get Q is equal to 4, which stood for the number of quarters. So we have four quarters. But I still need to find the number of times. My first equation I plug in 4 for Q to find the number of dimes, D. Solving for D, I get D is equal to 17. So we have 4 quarters in a register and sense. We said we had 17 dimes and 4 quarters. Well, the total number of coins should be 21 and, yes, those two add up to 21. So we know this first equation is true. 1 dime is worth 10 cents, so 17 dimes would be worth $1.70. 1 quarter is worth 25 cents, so 4 quarters would be worth $1. And with simple addition we know that $1.70 and $1 add up to $2.70. So the second equation is true as well.
For this problem, a grocer wants to mix pounds of raisins with pounds of nuts to get a trail mix. He's wondering how much of each he should use to make the trail mix. So, we know we're going to need two variables, r will stand for the pounds of raisins he'll use and n will stand for the pounds of nuts he'll use. The trail mix will contain both the raisins and the nuts. So, if we add up the pounds of the raisins and the pounds of the nuts, we'll get our total pounds of trail mix. I can use these variables to write my first equation. The pounds of raisins, r, plus the pounds of nuts, n, should equal 15, our total amount of trail mix. For the second equation, we know that the cost of raisins plus the cost of the nuts should be the total cost of the trail mix. Our raisins are charged at $3.50 per pound. So, I take $3.50 per pound times the number of pounds of raisins, I'll get my cost of the raisins. The same is true for the nuts. We have n pounds of nuts times $2 per pound. So, this is my cost. The trail mix is going to cost $2.60 per pound. But we need to figure out how much pounds in total we have of trail mix. If we used r pounds of raisins and n pounds of nuts, then our total pounds of trail mix would be, r plus n. This last part is probably one of the hardest steps in this problem. But if you remember that you we're making the trail mix, from pounds of raisins and pounds of nuts, it should make sense. I'm going to solve this system using substitution since I can easily solve for one of these variables. If I subtract r from both sides I'll get n is equal to 15 minus r. Now, I can perform substitution. I'll take 15 minus r and plug it in for n in my second equation. I could also take 15 minus r and plug it in for n right here in the equation. But I already know the value of r plus n. I know r plus n is equal to 15, so I can replace this entire parentheses with 15. Replacing this parentheses with 15 would be much simpler than replacing this n with 15 minus r because here, we'll get a number instead of a bunch of variables. Next, I group my like terms together. Then, I subtract coefficient of r to get r is equal to 6. R stood for the number of pounds of raisins we used, so really this is 6 pounds. The pounds of raisins and the pounds of nuts should add up to 15. So, using some mental math or solving an equation we can get n is equal to 9 pounds, the pounds of nuts. As always, let's go back and make sure these answers make sense. Our pounds of raisins was 6 and our pounds of nuts was 9. So, the grocer had 6 pounds and 9 pounds which did, indeed total 15 pounds. Next, you can replace 6 for r and 9 for n to determine the cost and make sure that our second equation is true. So, replace each of the variables and then I multiply each of these numbers together, making sure to add yes, this equation is true as well. So, I have $21 plus $18 is equal to $39. So, yes, both of these equations are true.
For the third practice problem, we were wondering how many tickets were sold. How many adult tickets and how many kid tickets. So, these must be the variables. We can let a equal the number of adult tickets, and we can let k equal the number of kid tickets. It doesn't matter what variables you use. You could use a smiley face or even a diamond. It just matters that you use the same variable for the number you're representing throughout the problem. These are the two equations we need to set up. The number of adult tickets and the number of kid tickets should equal our total number of tickets. Using our variables, I have a plus k is equal to 500. 500 was the total number of tickets sold for the fundraiser. The second equation involves the amount of money earned from adult tickets and the amount of money earned from kid tickets. If we add those two amounts together, we should get the total amount of money, which was $3312.50. I've written our two equations down here and this time, I'm going to solve it using elimination. We don't always have to use substitution. I'm going to multiply this equation through by a negative 4 so I can end up eliminating the ks. Here are the twp equations, and I can solve this equation for k. This allows me to do substitution that can plug in 500 minus a in for k. Now, we don't always need to do substitution, I could have multiplied this first equation by a negative 4 and that would have let me eliminate the ks. I could of solved it using elimination instead. Either way you do it, you should wind up with the same answer in the end. First, I distribute the positive 4 to the 500 and the negative a. Then, I combine like terms and subtract 2000. Our last step is to divide by the coefficient of a, 3 and a half. And we get a is equal to 375. This was the number of adult tickets that were sold for the fundraiser. We know the total number of tickets must add up to 500. So, I can replace a with 375 to find the number of kid tickets. Using some mental math here or an equation, we can get k is equal to 125. This is the number of kid tickets we sold. And to be sure that your answers are right, plug them in into your second equation to make sure that the amounts add up to $3,312.50.