Algebra Help, Thank you :)

Question

You invested $7000 in two accounts paying 6% and 8% annual interest. If the total interest earned for the year was $730, how much was invested at each rate?

Answer 1

730/7000 = about 10.4% return. There is no combination of 6 and 8% that would average out to 10.4%

but odds are you mis-copied the problem or the problem itself was printed incorrectly

maybe you meant 18%? not 8%? IF so, then:

.18X + .06(7000-X) = 730 where X is the amount invested at 18% and 7000-X is that invested at 6%

12X = 730 - 420 = 310

X = 310/.12 = 31000/12 = 7750/3= $2,583.33 invested at 18%

7000-X = $4,416.67 invested at 6%

6% is almost twice as close to 10.4% as 18% is, so you need almost twice as much invested at 6 as 18%

try this method with 6% and 8%, you'll get a negative amount invested at 8%, which doesn't seem to have a real life counterpart. Maybe selling off an investment, using the proceeds plus another $7,000 invested at 6%. But it really doesn't make sense.