A small business borrows $50,000 to get started. Some of the money is borrowed at 9%, some at 7%, and some at 10% simple interest. The total of the annual interest they are charged is $4340, and the amount borrowed at 7% is two times the amount borrowed at 9%. Set up a system of equations to represent the situation, then find the amounts of money (x, y, and z) borrowed at each interest rate.

Amount borrowed at 9% = 9,428.57

Amount borrowed at 7% = $18,857.14

Amount borrowed at 10% = $ 21,714.29

Step-by-step explanation:

Formula for simple interest :

I = PRT,

P= Principal, R =rate , T= time

Let x = amount borrowed at 9%.

y = amount borrowed at 7%.

z = amount borrowed at 10%.

i.e. x+y+z= 50000    (i) [Given : amount invested = 50000]

Interest on x = x (0.09)(1)   [as P=x,  T=1 year, r= 0.09]

= 0.09x

Similarly

Interest on y = y(0.07)(1) = 0.07y

Interest on z = z(0.1)(1)=0.1z

Total interest = 0.09x+0.07y+0.1z = 4340   (ii)

Also, y=2x   [given]  (iii)

Put y=2x in (i) and (ii), we get

x+2x+z= 50000 ⇒ 3x+z=50000    (iv)

0.09x+0.07(2x)+0.1z=4340  ⇒ 0.09x+0.14x+0.1z=4340

⇒ 0.23x+0.1z=4340    (v)

Multiply 10 on both sides of (v)

2.3x+z=43400   (vi)

Eliminate (vi) from (v)

0.7x=6600

⇒ x= 9,428.57

From (iii), y= 2(9428.57) =18,857.14

From (vi),

2.3(9428.57)+z=43400

⇒ 21685.71+z=43400

⇒ z= 43400-21685.71

⇒ z=21714.29

Hence, Amount borrowed at 9% = 9,428.57

Amount borrowed at 7% = $18,857.14

Amount borrowed at 10% = $ 21,714.29