+44 Travis has a savings account that his parents opened for him. It pays 6% annual interest. His uncle also opened an account for him, but it pays 8% annual interest. If there is $500 more in the account that pays 6%, and the total interest from both accounts is $114, how much money is in each of the accounts? PLEASE HELP
+118696 Travis has a savings account that his parents opened for him. It pays 6% annual interest. His uncle also opened an account for him, but it pays 8% annual interest. If there is $500 more in the account that pays 6%, and the total interest from both accounts is $114, how much money is in each of the accounts?
It depends on how often the interest is paid but lets assume it is paid yearly, bank interest is compound interest.
$interestfrom$8%$account$=I1=P(1.08)n−P$interestfrom$%$account$=I2=(P+500)(1.06)n−(P+500)I1+I2=P(1.08)n−P+(P+500)(1.06)n−(P+500)=114P(1.08)n−P+(P+500)(1.06)n−(P+500)=114P(1.08)n−P+(P+500)(1.06)n−P−500=114P(1.08)n+(P+500)(1.06)n−2P=614P(1.08)n+P(1.06)n+500(1.06)n−2P=614P[(1.08)n+(1.06)n−2]+500(1.06)n=614P=614−500(1.06)n(1.08)n+(1.06)n−2
THE 8% ACCOUNT HAS $P AND THE 6% ACCOUNT HAS $(P+500)
This is saying that the original principals will depend on how long the money is invested for. 
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I'm just going to try it with a couple of values
1 YEAR
(614−500×1.061)(1.081+1.061−2)=600
Time 1 year, amounts $600 and $1100 Interest = $48+$66=$114
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2 YEARS
(614−500×1.062)(1.082+1.062−2)=180
Time 2 years, amounts $180 and $680 Interest = $29.952+$84.048=$114
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3 YEARS
(614−500×1.063)(1.083+1.063−2)=41.0269608278163327
Time 3 years, amounts $41.0269 and $541.0269 etc
YES, THAT WORKS!
THAT QUESTION WAS A BIT DIFFERENT FROM THE NORM. 
+118696 Travis has a savings account that his parents opened for him. It pays 6% annual interest. His uncle also opened an account for him, but it pays 8% annual interest. If there is $500 more in the account that pays 6%, and the total interest from both accounts is $114, how much money is in each of the accounts?
It depends on how often the interest is paid but lets assume it is paid yearly, bank interest is compound interest.
$interestfrom$8%$account$=I1=P(1.08)n−P$interestfrom$%$account$=I2=(P+500)(1.06)n−(P+500)I1+I2=P(1.08)n−P+(P+500)(1.06)n−(P+500)=114P(1.08)n−P+(P+500)(1.06)n−(P+500)=114P(1.08)n−P+(P+500)(1.06)n−P−500=114P(1.08)n+(P+500)(1.06)n−2P=614P(1.08)n+P(1.06)n+500(1.06)n−2P=614P[(1.08)n+(1.06)n−2]+500(1.06)n=614P=614−500(1.06)n(1.08)n+(1.06)n−2
THE 8% ACCOUNT HAS $P AND THE 6% ACCOUNT HAS $(P+500)
This is saying that the original principals will depend on how long the money is invested for.
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I'm just going to try it with a couple of values
1 YEAR
(614−500×1.061)(1.081+1.061−2)=600
Time 1 year, amounts $600 and $1100 Interest = $48+$66=$114
-------------------------------------------------------
2 YEARS
(614−500×1.062)(1.082+1.062−2)=180
Time 2 years, amounts $180 and $680 Interest = $29.952+$84.048=$114
-------------------------------------------
3 YEARS
(614−500×1.063)(1.083+1.063−2)=41.0269608278163327
Time 3 years, amounts $41.0269 and $541.0269 etc
YES, THAT WORKS!
THAT QUESTION WAS A BIT DIFFERENT FROM THE NORM.
