###### Carl Horowitz

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!

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# Applied Linear Equations: Investment Problem - Problem 1

Carl Horowitz
###### Carl Horowitz

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!

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For this example, we're going to be dealing with a linear equation problem, that is disguised as a interest problem. So, my uncle deposited some money into an account that pays 5% interest. He made a second deposit of \$2000, than the first, into an account that yielded 9%. In total he made \$310, how much did he invest at 5%?

So let's take a look at this. I will always take a word problem, and turn it into a diagram, to sort out the information. So, my uncle puts in a certain amount of money into an account that pays 5%. We don't know how much it is, so let's call it x. He then puts in some more money, into an account that pays 9%. Again, we don't know how much he inputs into this account, but we do know that he puts in \$2000 more than he did before. So if he put \$1000, he'd put \$3000 here. So if he put x over here, it has to be x plus 2000, and in total, we know that he makes \$310, interest from one, plus interest from the other, is equal to 310.

So he has \$x invested at 5%, if he invested \$100, he'd get 5% of that, he would get \$5 back, so to find the amount of interest he makes, you take the amount you invest, times your interest rate. So in this case, it's .05x. The same idea for this account, 9% x plus 2000 and that 310 still stays the same.

So, we now have turned our word problem into a diagram, into an equation. We found this as we would any other equation, distribute through, again, if you wanted to multiply by 100, you could give it a value of decimals, it doesn't make a difference. So you have .5x stays the same, so you have .9x stays the same. 2000 times .09 just to make sure I don't make a decimal mistake, let's multiply it out equals 180, so this is plus 180 equals 310.

Combine our variables, so we end up with .05 plus .09, 1.4x, subtract 180 is equal to 130 and to solve for x, divide by .14, 130 divided by .14 it is not the nicest number, but x is equal to 928.57, not the nicest number in the world, but it's our answer.

So the question is asking, for how much did he invest at 5%? So we found x, which is actually the amount that he invested at 5%, and so, our variable actually matches up, that would be our answer. If it asked for the amount of 9%, we would just add \$2000 to that. So, our answer is that x value \$928.57 cents. So, took a interest problem, turned it from a problem to a diagram, and then solved it out.