You invested $19,000 in two accounts paying 4% and 8% annual interest, respectively. If the total interest earned for the year was $1160, how much was invested at each rate?
The amount invested at 4% is $.

$9000 at 4$

and

$10000 at 8%

Step-by-step explanation:

Let's assume that "x" is the amount deposited in the 4% account and "y" is the amount deposited in the 8% account.

Recall the formula for interest as :

where I is the interest, R is the annual rate of interest and t is the number of years.

Since there are two investments, we need to add both interests at the end of the one year: I1 = x (0.04) (1) = 0.04 x   and   I2 = y (0.08) (1) = 0.08 y

Total Interest = Interest (from the 4% account) + Interest (from the 8% account)

Total Interest = $1160 = 0.04 x +  0.08 y

we also know that the total invested (x + y) adds to $19,000, that is:

$19,000 = x + y

Then we can solve these system of two equations by substitution, for example solving for y in the second equation and using the y substitution in the first equation;

y = 19000 - x

1160 = 0.04 x + 0.08 (19000 - x)

1160 = 0.04 x + 1520 - 0.08 x

0.08 x - 0.04 x = 1520 - 1160

0.04 x = 360

x = 360/0.04 = $9000

Then the other investment was : y = $19000 - $9000 = $10000

the correct answer is the second option