interest in systems

Question

You invest $7200 in two accounts paying 6% and 10% annual interest. At the end of the year the accounts earn the same interest. Set up a system and solve for how much was invested at each rate 

Answer 1

Let x = the amount in the 6% interest account and let y = the amount in the 10% interest account.

 

So first you know that x + y = 7200

 

Now how much interest is earned in the 6% interest account?  That's just 6% of x, or .06x.

 

Similarly the amount of interest earned in the 10% interest account is 10% of y, or .1y.

 

Since the amount of interest earned was the same in both accounts, you know that .06x = .1y

 

So that gives you a system of two equations in x and y:

 

x + y = 7200

.06x = .1y

 

A simple way to solve would be to multiply both sides of the second equation by 10.  That gives you:

.6x = y

 

That's a simple expression for y that you can substitute back into the first equation:

x + .6x = 7200

1.6 x = 7200

x = 4500

 

Since x + y =7200, y must equal 2700.

 

So there was $4500 invested in the 6% interest account and $2700 invested in the 10% interest account.