SOLUTION: Helene invested a total of $1000 in 2 simple interest bank accounts. One paid %5 annual interest the other paid %6 annual interest. The total amount of interest she earned after 1

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Question 1109000: Helene invested a total of $1000 in 2 simple interest bank accounts. One paid %5 annual interest the other paid %6 annual interest. The total amount of interest she earned after 1 year was $58. Find the amount she invested in each account
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52887)   (Show Source): You can put this solution on YOUR website!
.
    x +     y = 1000,     (1)
0.05x + 0.06y =   58.     (2)


From (1), express y = 1000-x  and then substitute it into (2)

0.05x + 0.06*(1000-x) = 58

0.05x + 60 - 0.06x = 58

-0.01x = 58 - 60 = -2  ====>  x =  =  = 200.


Answer.  $200 was invested at 5%.   The rest  $1000 - $200 = $800 was invested at 6%.


Check.   200*0.05 + 800*0.06 = 58.   ! Correct !


Solved.   //   I used the Substitution method to solve the system of 2 equations in 2 unknowns.

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Also,  you have this free of charge online textbook in ALGEBRA-I in this site
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The referred lesson is the part of this online textbook under the topic "Systems of two linear equations in two unknowns".


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Answer by greenestamps(13209)   (Show Source): You can put this solution on YOUR website!


You can also solve "mixture" problems like this by comparing the actual amount of interest to the amounts of interest that would be obtained at each of the two rates.

$1000 at 5% interest would make $50 interest; at 6% it would make $60 interest.

$58 is 4/5 of the way from $50 to $60; that means 4/5 of the money should be invested at 6%.

So 4/5 of $1000 = $800 at 6%; the remaining $200 at 5%.

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