Larry Mitchell invested part of his advance at annual simple interest and the rest at annual simple interest. If his total yearly interest from both accounts was , find the amount invested at each rate. The amount invested at is $ nothing. The amount invested at is $ nothing.

Complete question :

Larry Mitchell invested part of his $32000 advance at 4% annual simple interest and the rest at 9% annual simple interest. If his total yearly interest from both accounts was 2430, find the amount invested at each rate. The amount invested at is $ nothing. The amount invested at is $ nothing.

Investment at 4% = $9000

Investment at 9% = $23,000

Step-by-step explanation:

Given that :

Total investment = 32000

Let;

amount invested at 4% = a

amount invested at 9% = 32000 - a

Total interest earned = 2430

Simple interest = principal * rate * time

Investment at 4%:

(a * 0.04) + [(32000 - a) * 0.09] = 2430

0.04a + 2880 - 0.09a= 2430

-0.05a + 2880 = 2430

-0.05a = 2430 - 2880

-0.05a = - 450

a = 9000

Hence,

Investment at 4% = $9000

Investment at 9% = $32,000 - $9000 = $23,000