Question

$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? (Round to two decimal places if necessary.)

Answer 1

Answer: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.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.