Jessica earned a total of $647 in simple interest from two seperate accounts. In an account earning 9% interest, Jessica invested $2,300 more than twice the amount she invested in an account earning 2%. How much did she invest in each account?
Invested:
Po @ 2%.
(2Po+2300) @ 9%.
t = 1 yr.?
I = Po*r*t + (2Po+2300)*r*t = 647.
Po*0.02*1 + (2Po+2300)*0.09*1 = 647,
0.02Po + 0.18Po + 207 = 647,
0.20Po = 440, Po = $2200.
2Po+2300 = 2*2200 + 2300 = $6700.
To solve this problem, let's assume that Jessica invested x dollars in the account earning 2% interest.
According to the problem, she invested $2,300 more than twice the amount invested in the 2% interest account in the 9% interest account. So, the amount invested in the 9% interest account would be (2x + $2,300).
Now, let's calculate the interest earned from each account:
Interest from the 2% account = (0.02x)
Interest from the 9% account = (0.09)(2x + $2,300) = 0.18x + $207
According to the problem, the total interest earned by Jessica from both accounts is $647. Therefore, we can set up the following equation:
0.02x + 0.18x + $207 = $647
We can simplify the equation:
0.20x + $207 = $647
Now, let's isolate x by subtracting $207 from both sides:
0.20x = $647 - $207
0.20x = $440
Divide both sides of the equation by 0.20 to solve for x:
x = $440 / 0.20
x = $2200
Hence, Jessica invested $2,200 in the account earning 2% interest.
To find the amount invested in the 9% interest account, we substitute the value of x back into the expression (2x + $2,300) to calculate:
Amount invested in the 9% interest account = 2($2,200) + $2,300 = $4,400 + $2,300 = $6,700
Therefore, Jessica invested $2,200 in the 2% interest account and $6,700 in the 9% interest account.