You can
put this solution on YOUR website!Sam invested $5000 for one year, part at 9% annual interest and the rest at 12% annual interest.
The interest from the investment at 9% was $198 more than the interest from the investment at 12%.
How much money did she invest at 9%?
:
Let x = amt invested at %9
Total invested given as 5000, therefore
(5000-x) = amt invested at 12%
:
.09x - 298 = .12(5000-x)
.09x - 298 = 600 - .12x
.09x + .12x = 600 + 298
.21x = 898
x =

x = $4,276.19 invested at 9%
then
5000 - 4276.19 = $723.08 invested at %12
:
:
Is this true? find the interest of each
.09*4276.19 = 384.86
.12 * 723.08 = 86.86
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difference = 298; confirms these solutions
You can
put this solution on YOUR website!
Let the amount invested at 9% be N
Then the amount invested at 12% is 5,000 - N
Now, the interest earned at 9%, less $198 equals the einterest earned at 12%
Therefore, we'll have: .09N - 198 = .12(5,000 - N)
.09N - 198 = 600 - .12N
.09N + .12N = 600 + 198
.21N = 798

= $3,800
We can conclude that N, or the amount invested at 9% is: $
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Check
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Amount invested at 9% = $3,800, and interest earned at 9% = (.09 * 3,800) = $342
Amount invested at 12% = $1,200 (5,000 - 3,800), and interest earned at 12% = (.12 * 1,200) = $144
Difference: $198 ($342 - $144), with 9% interest earned ($342) being $198 more than 12% interest earned ($144).