Answer:
$3100 is invested at 9%
$4900 is invested at 11%
Step-by-step explanation:
Let's take "x" be the amount invested at 9%.
(x + 1800) is invested in another account at 11%.
The interest amount earned by the two accounts is $818.
Here we can use the simple interest formula and find the amount invested in each account.
Simple interest (I) = [tex]\frac{PNR}{100}[/tex], where P- is the principal , N is the number of years and R is the interest rate.
Simple interest = [tex]\frac{x*1*9}{100} + \frac{(x+1800)*1*11}{100}[/tex]
0.09x + 0.11(x+1800) = 818
Now we have to simplify and find the value of x .
Use the distributive property and simplify the second term.
0.09x + 0.11x + 198 = 818
0.2x + 198 = 818
0.2x =818 - 198
0.2x = 620
x = 620/0.2
x = 3100.
So $3100 is invested at 9%
x + 1800 = 3100 + 1800
= $4900
$4900 is invested at 11%
