Ms. moran has started an investment club at bss. $8000 is invested, some at 10%
annual interest and the other at 12%. if the annual income from the investments is $900.00, then how much was invested at each rate?

solve using elimination or comparison

The answer can be readily calculated using a single variable, x:

Let x = the amount being invested at an annual rate of 10%
Let (8000 - x) = the amount being invested at an annual rate of 12%

The problem is then stated as:

(x * 0.10) + ((8000 - x) * 0.12) = 900
0.10(x) + ((8000 * 0.12) - 0.12(x)) = 900
0.10(x) + 960 - 0.12(x) = 900
0.10(x) - 0.12(x) = 900 - 960
-0.02(x) = -60
-0.02(x) * -100/2 = -60 * -100/2
x = 6000 / 2
x = 3000

Thus, $3,000 is invested at 10% = $300 annually; and $8,000 - $3,000 = $5,000 invested at 12% = $600 annually, which sum to $900 annual investment.