$ 2000 is invested at 3 % interest and $ 4000 is invested at 8 % interest
Solution:
Given that, total of $6000 was invested
Let "x" be the amount invested at 3 % interest
Then, (6000 - x) is the amount invested at 8 % interest
Given that,
The total yearly interest amounted to $380
Then, we can frame a equation as:
![x * 3 \% + (6000 - x) * 8 \% = 380](https://img.qammunity.org/2021/formulas/mathematics/middle-school/xxa9wj6ermcm7cny7prhonj2u93354era0.png)
Solve the above expression for "x"
![x * (3)/(100) + (6000-x) * (8)/(100) = 380\\\\0.03x + (6000-x) * 0.08 = 380\\\\0.03x + 480 - 0.08x = 380\\\\-0.05x = -100\\\\\text{Divide both sides by -0.05 }\\\\x = 2000](https://img.qammunity.org/2021/formulas/mathematics/middle-school/q6wy2b2gsg5xxfaaj0ejqw5dhi3lxmcb29.png)
Thus, $ 2000 is invested at 3 % interest
(6000 - x) = 6000 - 2000 = 4000
$ 4000 is invested at 8 % interest