A women has a total of $9000 to invest. She invests part of money in an account that pays 8% per year and rest in account that pays9% per year, if the interest earned in the first year is$750, how much did she invest in each account

To solve this problem, we can use a system of equations. Let's assume that the amount of money the woman invested at 8% per year is 'x' dollars, and the amount she invested at 9% per year is 'y' dollars.

Based on the given information, we can establish the following equations:

Equation 1: x + y = $9000 (since the total amount invested is $9000)

Equation 2: (0.08x) + (0.09y) = $750 (since the interest earned in the first year is $750)

Now, we have a system of equations:

Equation 1: x + y = $9000
Equation 2: 0.08x + 0.09y = $750

To solve this system, we can use the substitution or elimination method.

For the sake of explanation, let's solve it using the elimination method. We will multiply Equation 1 by 0.08 to make the coefficients of x equal:

0.08x + 0.08y = $720 (0.08 * Equation 1)

0.08x + 0.09y = $750 (Equation 2)

Now, we can subtract the two equations:

(0.08x + 0.08y) - (0.08x + 0.09y) = $720 - $750

0.08y - 0.09y = -$30

-0.01y = -$30

To isolate 'y', we divide both sides of the equation by -0.01:

y = -$30 / -0.01

y = $3000

Now, substitute the value of 'y' back into Equation 1 to solve for 'x':

x + $3000 = $9000

x = $9000 - $3000

x = $6000

Therefore, the woman invested $6000 at 8% per year and $3000 at 9% per year to earn a total interest of $750 in the first year.