Investment Math Question

Question

An investment of $75,000 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 the third 9%. Total interest from the investments was $5670. The interest from the first investment was 2 times the interest from the second. Find the amounts of the three parts of the investment.

Answer 1

solve by substitution, elimination or matrix algebra

3 equations, 3 unknowns

1st equation

.08x +.06y +.09z =5670

multiply by 100

8x + 6y +9z = 567,000

2nd equation

x+ y + z = 75,000

3rd equation

.08x =2(.06y)

.08x -.12y = 0

8x - 12y = 0

4x - 6y = 0

add 1st equation to eliminate y, to get

12x + 9z = 567,000

multiply 2nd by 6

6x + 6y + 6z =6(75,000)= 450,000

8x + 6y + 9z = 567,000, subtract

6x+6y+6z= 450000

2x + 3z =567000-450000= 117,000

multiply by 3

6x + 9z =3(117000)=351,000

12x+9z = 567000 subtract

6x +9z =450000

6x = 567000-351000= 216,000

x =216000/6= $36,000 invested at 8%

y=2x/3=24000

= $24,000 invested at 6%

z=75000-x-y= $15,000 invested at 9%

check the answers

.08(36)+.06(24)+.09(15=2.88+1.44+1.35= 5.67 thousand =5670

it works