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

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