The annual interest on a $14,000 investment exceeds the interest earned on a $7000 investment by $595. The $14,000 is invested at a 0.5% higher rate of interest than the $7000. What is the interest rate of each investment?

x = 8.00 Interest rate on $14000

y = 7.50 Interest rate on $7000

Step-by-step explanation:

Let interest rate of $14000 be x%

and Interest rate for $7000 be y %

According to the first condition

14000 * x% - 7000 * y% = 595

multiply by 100

14000x-7000y = 59500

/700

20x-10y=85 ... (1)

II condition

x%=y%+0.5%

x=y+0.5

x-y=0.5 ... (2)

solve (1) & (2)

20 x - 10 y = 85 ... 1

Total value

1 x - 1 y = 0.50 ... 2

Eliminate y

multiply (1) by 1

Multiply (2) by - 10

20.00 x - 10.00 y = 85.00

-10.00 x + 10.00 y = - 5.00

Add the two equations

10.00 x = 80.00

/ 10.00

x = 8.00

plug value of x in (1)

20.00 x - 10.00 y = 85.00

160.00 - 10.00 y = 85.00

-10.00 y = 85.00 - 160.00

-10.00 y = - 75.00

y = 7.50

x = 8.00 Interest rate on $14000

y = 7.50 Interest rate on $7000