You invested $4000 between two accounts paying 4% and 8% annual interest respectively if the total interest earned for the year was $200 how much was invested at each rate

We can set up a system of two equations in two variables.
First, we define two variables.

Let 
x = the amount invested at 4%
y = the amount invested at 8%
The total amount invested was $4000, so we can write the first equation.

x + y = 4000

Now we deal with the interest earned by the two accounts.

x amount invested at 4% earns 0.04x interest in one year.
y amount invested at 8% earns 0.08y interest in one year.
The total interest earned in one year was $200, so now we can write the second equation.

0.04x + 0.08y = 200

Here is our system, of equations.

x + y = 4000
0.04x + 0.08y = 200

I will solve the system of equations using the substitution method.
Solve the first equation for x.
x + y = 4000
x = 4000 - y

Now we substitute 4000 - y for x in the second equation.
0.04(4000 - y) + 0.08y = 200
160 - 0.04y + 0.08y = 200
0.04y = 40
y = 1000

Now substitute 1000 for y in the first equation, and solve for x.
x + y = 4000
x + 1000 = 4000
x = 3000

$3000 was invested at 4%, and $1000 was invested at 8%