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
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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):
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 !
You will find there different approaches (using one equation or a system of two equations in two unknowns), as well as
different methods of solution to the equations (Substitution, Elimination).
