 This video will talk about solving systems of equations applications. Simple interest. So it's the amount invested multiplied by the interest rate. This is simple interest, as simple as it gets. And remember your rate is going to be as a decimal. So for example, we have a thousand dollars invested at six and a half percent. It would be a thousand dollars and if we take the six and a half percent remember you move the decimal two places. So it would be .065 and A would be when I multiply 65. So the amount of interest earned on a thousand dollars at six and a half percent for one year is going to be 65. There should probably be a T in here for time, how long it's going to be, but we're just going to use one year. So how much is being invested in this problem? Let's see. Ian is retiring and has 225,000 to invest in two accounts. So he's investing 225,000. Would like to earn $7,900 and he's going to do that, an account paying three percent and one and doing four percent and those are simple interest. So how much is he investing? Well, he's investing the 225,000. Assigned a variable so the amounts invested in each, we're going to say that he has x at three percent and we're going to say that he has y at four percent. Using the two variables that represents the total amount invested. Remember x is an amount and y is an amount. So this is really just going to be x plus y is equal to the 225,000. So let's go on. How much did he need to earn? If you go back and look at the problem, it said that he had $7,900 that he wanted to earn. So how would you represent the amount of interest Ian makes? On the first variables invested amount, he had x dollars at three percent. We can call that 0.03x. And for the second amount, he had y dollars that he was investing at four percent. So we're going to call that 1.04y. So to write the equation using the two amounts that he earns for each, we're going to have 0.03x plus 0.04y. This is an interest, this is an interest, $7,900 is an interest. So interest is equal to interest plus interest. All right. So we want to solve the system. And here's our two equations that we found. Where x is the amount at three percent, and y is the amount at four percent. When you look at the system, this one looks like it would be very nice to use substitution either solving for x or solving for y. I'm just going to solve for x. I'm going to say x is equal to 225,000 minus the y. And then I'm going to substitute that in for the x in the bottom equation. So I have 7,900 is equal to 0.03 times my x, which is 225,000 minus y, and then plus the 0.04y. 7,900 then is equal to when I distribute here, I get 6,750. When I distribute here, I get minus 0.03y, and then plus 0.04y. 7,900 is equal to the 6,750 plus 0.01y when I combine my light terms. Subtracting the 6,750, I'm going to have 1150, 1,150 is equal to 0.01y. Y is going to be equal to 115,000 dollars. That's the y. We have to find the x. But if I take this y value and plug it in here for this y, then I'm going to be able to find x. X is equal to 225,000 minus my 115,000. So x is equal to 110,000 dollars at 3%, and 115,000 is at 4%.