SOLUTION: Melanie invested $5000 for 18 months, part at a rate of 12% annual interest and the rest at a rate of 9% annual interest. The interest from the investment at 12% was $348.75 more t

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: Melanie invested $5000 for 18 months, part at a rate of 12% annual interest and the rest at a rate of 9% annual interest. The interest from the investment at 12% was $348.75 more t      Log On

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Question 28285: Melanie invested $5000 for 18 months, part at a rate of 12% annual interest and the rest at a rate of 9% annual interest. The interest from the investment at 12% was $348.75 more than the interest from the investment at 9%. How much money did she invest at each rate? (Interest formula: Interest=principal x rate x time [expressed in years]) All steps in the solution must be shown and each step must be accompanied by a verbal/written explanation to justify your work and/or explain your reasoning.
No idea where to start let alone write an explanation.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
After you've done enough of these interest and other types of word problems
you get a sense of where to start.
As soon as I saw that they're doing 2 different things with $5000 what popped
into my head was x and 5000 - x
This is a great way to divide 5000 in two, the sum adds up to 5000 as it should
and your unknown is neatly tucked into each part. Very useful
What do you do with it?
Let x be the amount invested at 12%
So the rest of the $5000, namely 5000 - x, is invested at 9%
Following the problem,
.12+%2A+x+=+.09+%2A+%285000+-+x%29+%2B+348.75
Solve for x
.12+%2A+x+=+450+-+.09+%2A+x++%2B+348.75
.21+%2A+x+=+450+%2B+348.75
.21+%2A+x+=+798.75
I get
x = 3803.60
and
5000 - x = 1196.40
Those would be the amounts invested at 12% and 9%
checking
.12+%2A+3803.60+=+.09+%2A+1196.40++%2B+348.75
456.43+=+107.68+%2B+348.75
456.43+=+456.43
Unless I goofed that's it