SOLUTION: You invest $20000 in two accounts paying 7% and 9% annual interest, resectively. If the total annual interest earned is $1550, then how much was invested at each rate?

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Question 1009777: You invest $20000 in two accounts paying 7% and 9% annual interest, resectively. If the total annual interest earned is $1550, then how much was invested at each rate?
Found 2 solutions by macston, josmiceli:
Answer by macston(5194) (Show Source):
You can put this solution on YOUR website!
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S=amount invested at 7%; N=amount invested at 9%
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S+N=$20000
S=$20000-N
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0.07S+0.09N=$1550
0.07($20000-N)+0.09N=$1550
$1400-0.07N+0.09N=$1550
0.02N=$150
N=$7500
ANSWER 1: $7500 was invested at 9%.
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S=$20000-N=$20000-$7500=$12500
ANSWER 2: $12500 was invested at 7%.
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CHECK:
0.07S+0.09N=$1550
0.07($12500)+0.09($7500)=$1550
$875+$675=$1550
$1550=$1550
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Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!
Let = amounmt invested at 7%
= amount invested at 9%
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and

$12,500 was invested at 7%
$7,500 was invested at 9%
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check:

OK