The annual interest on a $15,000 $ ⁢ 15,000 investment exceeds the interest earned on an $8000 $ ⁢ 8000 investment by $685 $ ⁢ 685 . The $15,000 $ ⁢ 15,000 is invested at a 0.6% 0.6 % higher rate of interest than the $8000 $ ⁢ 8000 . What is the interest rate of each investment?

The interest rate on the $ 8, 000 investment is 9. 657 % = 9.66%

While the interest rate on the $ 15, 000 investment is 10.257% = 10.26%

Step-by-step explanation:

Generally, the formular for calculating simple interest, I is given by

I = (principal x rate x time) / 100

Let the the annual interest on the $8,000 investment be represented as I. While the rate is represented as R

Since the time is 1year, interest on the $8, 000 can be expressed as

I = ( 8000 x R × 1) ÷ 100

I = 8000R/100

I = 80R. ( eqn 1)

Also, let the annual interest on $15,000 investment be K and the rate be V

Since interest on $15, 000 is greater than that of the $8,000 by $685

K = I + 685

Also rate V is 0.6% higher than R

V = (R + 0.6)%

This the interest on the $15, 000 is expressed as

K = ( 15000 × V × 1) ÷ 100 (eqn 2)

Substitute for K and V in eqn 2

That is

I + 685 = ( 15000 x (R + 0.6) x 1) ÷ 100

I + 685 = 15000 (R + 0.6) ÷ 100

I + 685 = 150 (R + 0.6)

Open the bracket

I + 685 = 150R + 15 × 0.6

I + 685 = 150R + 9.

This is the same as

150R + 9 = I + 685. ( eqn 3)

Substitute eqn 1 into eqn 3

That is put

I = 80R in eqn 3

150R + 9 = 80R + 685

Collect like terms with R to one side

150R - 80 R = 685 - 9

70R = 676

R = 676 ÷ 70

R = 9.657 %

Recall that V = (R + 0.6)%

V = 10.257 %

Thus the interest rate on the $ 8, 000 investment is 9. 657 % = 9.66%

While the interest rate on the $ 15, 000 investment is 10.257% = 10.26%

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,

the correct option is b

step-by-step explanation: