Expert: Bobby Soltani - 6/29/2009

Question
trying to start a new career, and was humming along through algebra until I
hit this problem...
An actor invests some money at 9% and $35000 more than twice that amount
at 10%. The total annual interest that he earned from the investment is
$22930. How much did he invest at each rate?
I've been staring at this for 4 hours with no success, so any help would be
greatly appreciated - thanks!

Answer
Hi,

Let the amount invested at 9% be x, and the amount invested at 10% be y.

An actor invests x at 9% and $35000 more than twice x
at 10%.

y is 35000 more than 2x

y = 2x + 35000     (equation 1)

The interest earned on an amount is that amount times the rate.  So, we can write the following equation:

22930 = .09x + .10y   (equation 2)

since we know that the total interest 22930 is equal to the amount invested at 9% times .09 plus the amount invested at 10% times 0.10.

Now we have two equations and two unknowns, and we can solve for x and y using any number of methods.  For example, you can substitute (2x+35000) in for y in equation 2, and then solve for x.  Once you have x, you can plug that value into either equation to solve for y.

Let me know if you need help solving for x and y.  The answers I get are 67,000 for x, and 169,000 for y.

Good luck in your new career!

Regards,

Bobby