jack invested $4000 in two accounts. one pays simple interest at the rate of 5% and the other pays simple interest at the rate of 6%. if jack earned a total of $224 in interest in the first year, how much did he invest in each account?

He invested:

$X @ 5%.
$(4000-X) @ 6%.

.05x + 0.06(4000-x) = 224.
0. 05x - 0.06x + 240 = 224.
-0.01x = 224 - 240 = -16.
X = $1600.
4000-X = 4000-1600 = $2400.

To find out how much Jack invested in each account, let's assign variables to the unknowns.

Let's say Jack invested x dollars in the account that pays 5% interest, and y dollars in the account that pays 6% interest.

We know that the total amount Jack invested is $4000. So we can write the equation: x + y = 4000.

Now, let's calculate the interest earned on each account and set up another equation using the given information.

The interest earned on the account with 5% interest is (x * 0.05) = 0.05x, and the interest earned on the account with 6% interest is (y * 0.06) = 0.06y.

According to the problem, Jack earned a total of $224 in interest. So, we can write another equation: 0.05x + 0.06y = 224.

Now, we have a system of two equations:

Equation 1: x + y = 4000

Equation 2: 0.05x + 0.06y = 224

We can solve this system using substitution or elimination method.

Let's use the elimination method to solve this system:

Multiply Equation 1 by 0.05 to align the coefficients of x or y in both equations:

0.05x + 0.05y = 0.05 * 4000
0.05x + 0.06y = 224

These equations become:

0.05x + 0.05y = 200
0.05x + 0.06y = 224

Now subtract Equation 1 from Equation 2 to eliminate x:

(0.05x + 0.06y) - (0.05x + 0.05y) = 224 - 200
0.06y - 0.05y = 24
0.01y = 24
y = 24 / 0.01
y = 2400

Now substitute the value of y into Equation 1 to find x:

x + y = 4000
x + 2400 = 4000
x = 4000 - 2400
x = 1600

Therefore, Jack invested $1600 in the account that pays 5% interest, and $2400 in the account that pays 6% interest.