a woman ivests in two bank accounts, one with a 3% percent interest and the other paying 9% simple interest per year.She puts twice as much in the lower account. Her annual interest is 7110. How much did she invest in each account?

higher account --- x

lower account ---- 2x

.09x + .03(2x) = 7110
.15x = 7110
x = 47400

$47400 at 9% and $94800 at 3%

check:
.09(47400) = 94800(.03) = indeed 7110

To solve this problem, let's call the amount invested in the lower interest account "x". Since the amount invested in the higher interest account is twice as much, we can say that the amount invested in the higher interest account is "2x".

Now, let's calculate the interest earned from each account. The interest earned from the lower interest account is 3% of the amount invested, which is 0.03*x. The interest earned from the higher interest account is 9% of the amount invested, which is 0.09*(2x).

We are given that the total annual interest is $7110. So, we can set up the equation:
0.03*x + 0.09*(2x) = 7110

Now, let's simplify and solve for x:
0.03x + 0.18x = 7110
0.21x = 7110
x = 7110 / 0.21
x = 33857.14

Now that we have the value of x, we can find the amount invested in the higher interest account:
2x = 2 * 33857.14
2x = 67714.29

Therefore, the woman invested $33,857.14 in the lower interest account and $67,714.29 in the higher interest account.