Question

cannot figure a way to do this problem:

An investment of $98,000 was made by a business club. The investment was split into three parts and lasted for one year. The first part of the investment earned 8% interest, the second 6%, and third 9%. The total interest from the investments was $7560. The interest from the first investment was 3 times the interest from the second. Find the amounts of the three parts of the investment.

so I let the three investment amounts be E, S, and N

E+S+N=98000

7500=E*.08+S*.06+N*.09
E*.08=3*S*.06

I don't know how to figure out answer past that

Answer 1

let the first Investment be x

let the second Investment be y

let the third Investment be z

given that total investment is x+y+z = 98000 ---- eq 1

let interest earned for first investment is 'a' = 0.08x

let interest earned for second investment is 'b' = 0.06y

let interest earned for third investment is 'c' = 0.09z

total interest is a + b + c = 7560

0.08x + 0.06y + 0.09z = 7560 ---- eq 2

also given that

a = 3b

=> 0.08x = 3 * 0.06y

=> 4x = 9y ---- eq 3

substituting eq 3 in eq 1 and eq 2 we get

(13/4)y + z = 98000

(0.24)y + 0.09z = 7560

solving the above two equations we get

y = 24000

z = 20000

=> x = (9/4)y = 54000

therefore

first investment = 54000 dollors

first investment = 24000 dollors

first investment = 20000 dollors

 

Answer 2

Since E = 2.25 S (from the last equation), you can eliminate E from the other two equations. Then solve two equations in two unknows (S and N).

Follow the advice that BobPursley gave to Samantha.