Question 1167264
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Here is a non-algebraic method for solving a "mixture" problem like this.<br>
(1) Find the amounts of interest if the whole $13,000 had been invested at each rate.<br>
$13,000 at 7% would yield $910 interest; $13,000 at 3% would yield $390 interest.<br>
(2) Where the actual interest lies between those two amounts exactly determines how much was invested at each rate.<br>
From 390 to 910 on a number line is a difference of 520; from 390 to 870 is a difference of 480.  480/520 = 48/52 = 12/13.<br>
$870 is 12/13 of the way from $390 to $910; that means 12/13 of the total was invested at the higher rate.<br>
ANSWERS: 12/13 of $13,000, or $12,000, was invested at 7%; the other $1000 was invested at 3%.<br>
CHECK: .07(12000)+.03(1000) = 840+30 = 870<br>