An investment of 109000 was made by a business club The inve

An investment of $109,000 was made by a business club. The investment was split 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 $8130. The interest from the first investment was 3 times the interest from the second. Find the amounts of the three parts of the investment. Write a linear system of equations. Let x be the first investment, y be the second investment, and z be the third investment. Choose the correct answer below. x + y + z = 3(0.06y) 0.08x + 0.06y + 0.09z = 8130 0.08x = 109,000 x + y + z = 109,000 0.08x + 0.06y + 0.09z = 8130 0.08x = 3(0.08y) x + y + z = 8130 0.08x + 0.06y + 0.09z = 109,000 0.07x = 0.06y x + y + z = 109,000 0.08x + 0.06y + 0.09z = 8130 0.08x = 3(0.06y) The club invested $ at 8%, $ at 6%, and $ at 9%

Solution

Let three parts be x , y and z

=> x + y + z = 109000

Interest from first investment => 0.08x

Interest from second investment => 0.06y

Interest from third investment => 0.09x

=> Total Interest , 0.08x + 0.06y + 0.09z = 8130

0.08x = 3( 0.06y ) = 0.18y

=> 0.24y + 0.09z = 8130

=> 2,25y + y + z = 109000

=> 3.25y + z = 109000

Solving we get , y = 32000 , z = 5000 => x = 72000