SOLUTION: the annual interest on a $13000 investment exceeds the interest earnes on a $6000 investment by $301. the $13000 is invested at a .7% higher rate of interest than the $6000. what i
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Question 906132: the annual interest on a $13000 investment exceeds the interest earnes on a $6000 investment by $301. the $13000 is invested at a .7% higher rate of interest than the $6000. what is the interest rate of each investment? Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! We have both amts invested.13000 and 6000
we know the total interest earned 301
We know the difference in rates is 0.007
x*13000-y*6000=301
x=y-0.007
substitute for x
(y-0.007)*13000-y*6000=301
y*13000-0.007*13000-y*6000=301
y*7000-91.0=301
y*7000=91.0+301
y*7000=392.0
y=0.056
y=5.6%
x=0.056-0.007
x=0.049
x=4.9%
check
x*13000-y*6000=301
0.049*13000-0.056*6000=301
637.0-336.0=301
301.0=301
ok
