SOLUTION: a machinist invested $15,000, part at 6% simple interest and the remainder at 4% simple interest for one year. how much was invested at each rate if each investment earned the same

Question 1111872: a machinist invested $15,000, part at 6% simple interest and the remainder at 4% simple interest for one year. how much was invested at each rate if each investment earned the same interest?
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.

Let  x  be the amount invested at 6%  (in dollars).


Then the remainder investment is (15000-x) dollars.


Next, write this equation

interest = interest    (as the condition states !)


0.06x    = 0.04*(15000-x)


0.06x = 0.04*15000 - 0.04x

0.06x + 0.04x = 0.04*15000

0.1x = 0.04*15000  ====>  x =  = 6000.


Answer.  $6000 was invested at 6%.  The rest,  15000-6000 = 9000 dollars was invested at 4%.


Check.   0.06*6000 = 360 dollars.

         0.04*9000 = 360 dollars.   ! The same value !  Solved correctly !

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To see many other similar solved problems on investment, look into the lesson
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in this site.

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different methods of solution to the equations (Substitution, Elimination).

Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic "Systems of two linear equations in two unknowns".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
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to your archive and use it when it is needed.

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