Question 1143754
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Here is a non-algebraic method for solving this kind of problem, treating it as a mixture problem.<br>
$29000 all invested at 6% would yield $1740 interest; all invested at 8% would yield $2320 interest.<br>
The way the money is split between the two investments is exactly determined by where the actual amount of interest lies between  those two amounts.<br>
So consider the three interest amounts on a number line: 1740, 2040, and 2320.<br>
(1) The difference between $1740 and $2320 is $580; the difference between $1740 and $2040 is $300.
(2) So we can say that the actual amount of interest is 300/580 = 30/58 = 15/29 of the way from $1740 to $2320.
(3) That means 15/29 of the total was invested at the higher rate.<br>
ANSWER: (15/29) of $29,000, or $15,000 at 8%; the rest, $14,000, at 6%.<br>
CHECK: .08(15000)+.06(14000) = 1200+840 = 2040