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Question 908900: $10,637 is invested, part at 13% and the other at 6%. If the interest earned from the amount invested at 13% exceeds the interest earned from the amount invested at 6% by $880.07, How much is invested at each rate?
Found 2 solutions by mananth, richwmiller:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! $10,637 is invested, part at 13% and the other at 6%. If the interest earned from the amount invested at 13% exceeds the interest earned from the amount invested at 6% by $880.07, How much is invested at each rate?
$10,637 is invested,
part at 13% . let it be x
and the other at 6%. let it be y
x+y= 10,637
interest earned from the amount invested at 13%= 0.13x
interest earned from the amount invested at 6% = 0.06y
difference =$880.07,
0.13x-0.06y=880.07
multiply by 100
13x+6y=88007
x+y = 10,637
1 x + 1 y = 10637 .............1
13 x + 6 y = 88007 .............2
Eliminate y
multiply (1)by -6
Multiply (2) by 1
-6 x -6 y = -63822
13 x + 6 y = 88007
Add the two equations
7 x = 24185
/ 7
x = 3455
plug value of x in (1)
1 x + 1 y = 10637
3455 + y = 10637
y = 10637 -3455
y = 7182
y = 7182
part at 13% ----3455
and the other at 6%. 7182
1
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! We know the total amount of money invested. $10637
x+y=10637,
We know that the difference in interest earned by the two accounts is $880.07
0.13*x-0.06*y=880.07
x=10637-y
We substitute for x
0.13*(10637-y)-0.06*y=880.07
We multiply out
1382.81-0.13y-0.06*y=880.07
We combine like terms.
502.74=0.19*y
Isolate y
y=502.74/0.19
y=2646.0 at 6%
Calculate x
x=10637-2646.0
x=7991 at 13%
Interest earned at 13% is 1038.83
Interest earned at 6% is 158.76
We check
0.13*7991-0.06*2646.0=880.07
1038.83-158.76=880.07
880.07=880.07
Since this statement is TRUE and neither x nor y is negative then all is well
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