A person invested $8400 for 1​ year, part at 6​%, part at 9​%, and the remainder at 1%

Question

A person invested $8400 for 1​ year, part at 6​%, part at 9​%, and the remainder at 1%.The total annual income from these investments was ​$939. The amount of money invested at 14​% was ​$600 more than the amounts invested at 6​% and 9​% combined. Find the amount invested at each rate.

Answer 1

$1400 invested at 6%

$2500 invested at 9%

$4500 invested at 14%

.06S + .09N + .14F = 939

S + N + F = 8400

F = 600 + S+N

3 equations 3 unknowns

solve by substitution and elimination or with matrix algebra

S+N + 600 +S+N = 8400

2S +2N = 8400-600 = 7800

6S + 9N + 14F = 93900

6S +9N +14(600) +14S +14N = 93900

20S +23N = 939-84 = 85500

20S +20N = 78000

3N = 7500

N = 7500/3 = $2,500 invested at 9%

S+F = 84-25 = 59

F-S = 31

2F = 59+31 = 90

F =90/2 = $4,500 invested at 14%

84-25-45 = $1,400 invested at 6%

Answer 2

$1400 invested at 6%

$2500 invested at 9%

$4500 invested at 14%

.06S + .09N + .14F = 939

S + N + F = 8400

F = 600 + S+N

3 equations 3 unknowns

solve by substitution and elimination or with matrix algebra

S+N + 600 +S+N = 8400

2S +2N = 8400-600 = 7800

6S + 9N + 14F = 93900

6S +9N +14(600) +14S +14N = 93900

20S +23N = 939-84 = 85500

20S +20N = 78000

3N = 7500

N = 7500/3 = $2,500 invested at 9%

S+F = 84-25 = 59

F-S = 31

2F = 59+31 = 90

F =90/2 = $4,500 invested at 14%

84-25-45 = $1,400 invested at 6%