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Applied Linear Equations: Investment Problem - Concept
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
In Algebra II and in the real world, sometimes we need to solve investment math problems by using linear equations. When solving word problems using linear equations, we first need to pull out the relevant information and put it into equation form. When working with investment math problems, we are usually asked to calculate amounts earned by interest on original principal amounts.
This example we're going to be dealing with linear equations in a Investment problem. So we have here is I have $3,000 to invest I put some of it to an account that pays 6 percent and the rest into an account that pays 8 percent. These are both simple interest which basically means that you take interest once a year. So later in your units and your years you're going to talk about sort of more complicated interest. This is just we take it once you're done with it. So invest some at 6 some at 8 and after a year I hope to make $216. How much should I invest in each account?
Okay so let's take a look at this. As I do with a lot of my problems is I actually take the word problem and then turn it to some sort of diagram that I can sort all the information out. So we have one account, that's the first one we'll call this one being the 6 percent. I'm investing a certain amount in this account, we don't know what it is so let's call it x. Okay we then have another account, this one is at 8 percent and we don't know how much we actually have to invest in this one. But we don know that together they are $3,000 so if I invested 1,000 here I'll know that I would have to invest 2,000 here. If I invested 500 I'll have to invest 2,500 put together they have to add up to $3,000.
So this one is just going to be 3,000 minus whatever I invested over here, and together we end up with $216 so the interest from one account plus the interest from the other account is equal to 216. Okay so once we have our diagram we can go ahead and make our equation, so we get 6 percent from this amount we invested. We invested $100 we get 6 percent as interest, we'll get 6 percent of a 100, $6. So basically we just multiply 6 percent as the decimal times the amount we invested. So this turn to 0.06x, 6 percent remember to change your decimal over there, okay this one is just 8 percent times the amount we invested, same idea we 0.08 times 3000 minus x and that's going to equal the total amount of interest we make $216.
Okay we have a linear equation so solve it as we would any other equation, distribute this through and again if you want you can multiply by 100 to clear our decimals, for this example I'm going to leave it in there either way it's just fine. So 06x stays the same 0.8 times 3000 it's a little bit big so we can use our calculator. 0.08 times 3000 is 240 0.08 times x and then that's 216 is still there. So in your equation combine like terms find 0.06 minus 0.08 will end up with -0.02x is 216. Can bring our 240 around -0.02x equals 216 minus 240, 216 minus 240 I should be able to do that in my head but just to make sure you end up with -24 divide by -0.02 to isolate that variable and -24 divided by 0.02 is $1200 okay make sure we always answer our questions correctly.
So the questions is asking for how much that I invested in each account so x is here was the amount we invested at 6 percent, so we know that we need to get 1200 at 6 percent out total amount of money was $3000 that 3000 minus this 12000 we invested will be 1800, 1800 at 8 percent. So in order to get $216 interest you'll have to invest 1200 at 6 percent and 1800 at 8 percent.
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