Before to going to college, Carlos had saved up $3000. Part of it was invested at 7% and the remainder at 9%. At the end of one year, he had earned $242 in interest. How much did he invest at each rate?

Answer 1

Let x be the amount Carlos invested at 7% interest.
He invested $3000 - x at 9% interest.
The interest from his 7% investment is x * 7/100 = 7x/100.
The interest from his 9% investment is ($3000 - x) * 9/100 = (27000 - 9x)/100.
The total interest earned is 7x/100 + (27000 - 9x)/100 = ($242 * 100)/100.
Combining like terms, we get 7x + 27000 - 9x = 242.
Subtracting 27000 from both sides, we get -2x = -26758.
Dividing both sides by -2, we get x = 1337.
Carlos invested $<<1337=1337>>1337 at 7% interest and $3000 - $1337 = $1663 at 9% interest. Answer: \boxed{1337, 1663}.