SOLUTION: $9500 is invested,part of it at 11% and part of it at 9%. For a certain year, the total yield is $973. How much was invested at each rate?
Algebra ->
Expressions-with-variables
-> SOLUTION: $9500 is invested,part of it at 11% and part of it at 9%. For a certain year, the total yield is $973. How much was invested at each rate?
Log On
Question 1093598: $9500 is invested,part of it at 11% and part of it at 9%. For a certain year, the total yield is $973. How much was invested at each rate? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! x invested at 11%=0.11x interest
9500-x invested at 9%=855-.09x interest
that sum is $973
0.02x+855=973
0.02x=118
multiply by 50,
x=$5900 @ 11%=$649
9500-x=$3600@9%=$324
sum is $973.
