Question 740308: "A sum of $3300 is invested, part of it at 10% interest and the remainder at 12%. If the interest earned by the 12% investment is $143 more than the interest earned by the 10% investment, find the amount invested at each rate."
I would like help with setting up a proper equation and solving.
Thank you!
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! 12%-----------x
10%----------y
x+y=3000,,,,,,,,,,,,,,,,,,,,,,,,(1)
12%x=10%y+143
12%x-10%y=143
multiply by 100
12x-10y=14300.....................(2)
1.00 x + 1.00 y = 3300.00 .............1
Total value
12.00 x -10.00 y = 14300.00 .............2
Eliminate y
multiply (1)by 10.00
Multiply (2) by 1.00
10.00 x 10.00 y = 33000.00
12.00 x -10.00 y = 14300.00
Add the two equations
22.00 x = 47300.00
/ 22.00
x = 2150.00
plug value of x in (1)
1.00 x + 1.00 y = 3300.00
2150.00 + y = 3300.00
y = 3300.00 -2150.00
y = 1150.00
y = 1150.00
12%---------$2150
10%---------$1150
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