SOLUTION: Anne invested $3000 more at 12% interest than she did at 25% interest. For the year she earned $35 more on her 12% vinvestment than she did from the 25% investment. How much did sh

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: Anne invested $3000 more at 12% interest than she did at 25% interest. For the year she earned $35 more on her 12% vinvestment than she did from the 25% investment. How much did sh      Log On

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Question 1088596: Anne invested $3000 more at 12% interest than she did at 25% interest. For the year she earned $35 more on her 12% vinvestment than she did from the 25% investment. How much did she invest at each rate?
Answer by ikleyn(52771) About Me  (Show Source):
You can put this solution on YOUR website!
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Let x be the amount (in dollars) invested at 12% and y be the amount invested at 25%.

From the condition,

    x -     y = 3000,       (1)
0.12x - 0.25y =   35.       (2)


This is the key moment in the solution called "the setup". You must look and carefully think on this system of equations.

If you understand its meaning (the meaning of each equation) then the rest is just a technique.


From (1), express x = y + 3000 and substitute it into (2). You will get a single equation for a single unknown y

0.12(y+3000) - 0.25y = 35,   or

0.12y + 360 - 0.25y = 35,

-0.13y = 35 - 360,

-0.13y = -325  ===>  y = %28-325%29%2F%28-0.13%29 = 2500.


Answer.  Anne invested $2500 at 25% and $2500 + $300 = $5500 at 12%.

Solved.

You check the solution by substituting the found values into equations (1) and (2).