Question 636217

System of linear equations word problem. Help please!

An Investment of 54,000 was made by a club. It was split into three parts lasting only year. The first part of the investment earned 8% interest, the second 6%, and the third 9%. Total interest from investments was $4260. The interest from the first was 6 times from the second. Find the amounts of the three parts of the investments.


Let the amounts invested at 8%, 6%, and 9% be E, S, and N, respectively
Forming equations based on info provided, we have:
E + S + N = 54,000 ------ eq (i)


.08E + .06S + .09N = 4,260 
8E + 6S + 9N = 426,000 ---- Multiplying by 100 ------ eq (ii)


.08E = 6(.06S) ----- .08E = .36S 
8E = 36S ---- Multiplying by 100 ------ eq (iii)


36S + 6S + 9N = 426,000 ------ Substituting 36S for 8E in eq (ii)
42S + 9N = 426,000 ------ eq (iv) 


– 8E – 8S – 8N = - 432,000 ----- Multiplying eq (i) by – 8 ----- eq (v)
8E + 6S + 9N = 426,000 ----- eq (ii) 
– 2S + N = - 6,000 ----- Adding eqs (v) & (ii) ----- eq (vi)


42S + 9N = 426,000 ----- eq (iv)
18S – 9N = 54,000 ------- Multiplying eq (vi) by – 9 ----- eq (vii)
60S = 480,000 ------ Adding eqs (iv) & (vii)
S, or amount invested at 6% = {{{480000/60}}}, or ${{{highlight_green(8000)}}}


8E = 36(8,000) ----- Substituting 8,000 for S in eq (iii)
8E = 288,000
E, or amount invested at 8% = {{{288000/8}}}, or ${{{highlight_green(36000)}}}


36,000 + 8,000 + N = 54,000 ----- Substituting 8,000 for S, and 36,000 for E in eq (i)
44,000 + N = 54,000
N, or amount invested at 9% = 54,000 – 44,000, or ${{{highlight_green(10000)}}}


I’m sure you can do your own check!!


If you know and are more comfortable with matrices, you could also determine the three unknown amounts using this method. 


Send comments and “thank-yous” to “D” at  MathMadEzy@aol.com