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$9.192 is invested, part at 10% and the rest at 9 %. If the interest earned from the amount invested at 10 % exceeds the interest earned from the amount invested at 9 % by $716.85, how much is invested at each rate? (Round to two decimal places if necessary.)

Respuesta :

x = amount invested at 10%

y = amount invested at 9%

x + y = 9192

0.10x = 0.09y + 716.85

we have a equation system

x= 9192-y

we substitute x in the second equation

0.10(9192-y)=0.09y+716.85

919.2-0.10y =0.09y+716.85

we clear y

919.2-716.85=0.09y+0.10y

202.35=0.19y

y=202.35/0.19=1065

then

x=9192-1065=8127

x = amount invested at 10%= 8127

y = amount invested at 9%=

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