Question 1199034
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There are several ways you could set this up directly using three variables for the amounts in each investment.  But that would require three equations, and some method for solving a system of three equations.<br>
I always prefer to take a little time to process the given information so that I can set up the problem using a single variable.  Because of the way the information is given in this problem, that takes some effort; but I feel it is worth it.<br>
I hope you get responses from other tutors showing different ways of setting up and solving the problem.<br>
Given the information that the interest from the investment at 8% is 6 times the interest from the investment at 6%, do some mental arithmetic (or use formal algebra) to determine that the amount invested at 8% is 4.5 times the amount invested at 6%.  (8% of 4.5x is 6 times 6% of x...)<br>
Then<br>
let 2x = amount invested at 6%
then 9x = amount invested at 8%
then 37000-11x = amount invested at 9%<br>
Note I chose 2x and 9x, instead of x and 4.5x, to represent the amounts invested at 6% and 8%.  That's just a personal preference for avoiding fractions or decimals in my equations....<br>
The total interest was $3030:<br>
{{{.06(2x)+.08(9x)+.09(37000-11x)=3030}}}<br>
{{{.12x+.72x+3330-.99x=3030}}}<br>
{{{300=.15x}}}
{{{x=300/.15=2000}}}<br>
The amounts invested at the three rates were<br>
ANSWERS:
2x = $4000 at 6%
9x = $18000 at 8%
$37000-11x = $15000 at 9%<br>
CHECK: .06(4000)+.08(18000)+.09(15000) = 240+1440+1350 = 3030<br>