SOLUTION: 17,818 is invested, part at 11% and the rest at 6%. if the interest earned from the amount invested at 11% exceeds the interest earned from the amount invested at 6% by 490.33, how

Question 905901: 17,818 is invested, part at 11% and the rest at 6%. if the interest earned from the amount invested at 11% exceeds the interest earned from the amount invested at 6% by 490.33, how much is invested at each rate?
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
We know the total amount of money invested. $17818
x+y=17818,
We know that the difference in interest earned by the two accounts is $490.33
0.11*x-0.06*y=490.33
x=17818-y
We substitute for x
0.11*(17818-y)-0.06*y=490.33
We multiply out
1959.98-0.11y-0.06*y=490.33
We combine like terms.
1469.65=0.17*y
Isolate y
y=1469.65/0.17
y=8645 at 6%
Calculate x
x=17818-8645
x=9173 at 11%
Check
0.11*9173-0.06*8645=490.33
interest earned at 11%=1009.03
interest earned at 6%=518.70
1009.03-518.7=490.33
490.33=490.33
Since this statement is TRUE and neither amount is negative then all is well.

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