SOLUTION: Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate?

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Question 120538: Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate?
Answer by tutor_paul(519) About Me  (Show Source):
You can put this solution on YOUR website!
Set it up as two equations and two unknowns. Let x be the amount invested at 9%, and let y be the amount invested at 11%. Then you can write:
Eqn #1: x%2By=6000
Eqn #2: x%280.09%29%2By%280.11%29=624
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Now you have 2 equations and 2 unknowns. Any time you have at least as many equations as you have unknowns, you can solve the problem!
Solve Eqn #1 for x:
x=6000-y
Substitute this value of x into Eqn #2:
%286000-y%29%2A0.09%2By%280.11%29=624
Solve for y:
540-0.09y%2B.11y=624
540%2B.02y=624
0.02y=84
highlight%28y=4200%29
Now that you have the answer for y, plug that into Eqn #1 and solve for x:
x%2B4200=6000
highlight%28x=1800%29
You can check these answer by making sure each equation holds true when you plug in these values for x and y.
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Good Luck,
tutor_paul@yahoo.com