Question

moe jackson has a total of $6,500 invested in two simple - interest accounts. the annual simple interest rate on one account is 9%. the annual simple interest rate on the second account is 6%. how much is invested in each account if both accounts earn the same amount of simple interest? at 9% $ at 6% $

Explanation:

Step1: Let the amount invested at 9% be $x$.

Then the amount invested at 6% is $(6500 - x)$.

Step2: Set up the interest - equal equation.

The simple - interest formula is $I=Prt$. Since $t = 1$ year, the interest from the 9% account is $0.09x$ and the interest from the 6% account is $0.06(6500 - x)$. Set them equal: $0.09x=0.06(6500 - x)$.

Step3: Expand the right - hand side.

$0.09x=0.06\times6500-0.06x$.
$0.09x = 390-0.06x$.

Step4: Add $0.06x$ to both sides.

$0.09x + 0.06x=390$.
$0.15x = 390$.

Step5: Solve for $x$.

$x=\frac{390}{0.15}=2600$.

Step6: Find the amount invested at 6%.

The amount invested at 6% is $6500 - 2600 = 3900$.

Answer:

at 9%: $\$2600$
at 6%: $\$3900$