Questions: Colan invests 7300 in two different accounts. The first account paid 6%, the second account paid 4% in interest. At the end of the first year he had earned 396 in interest. How much was in each account? at 6% at 4%

Colan invests 7300 in two different accounts. The first account paid 6%, the second account paid 4% in interest. At the end of the first year he had earned 396 in interest. How much was in each account?
at 6%
at 4%

Transcript text: Colan invests $\$ 7300$ in two different accounts. The first account paid $6 \%$, the second account paid $4 \%$ in interest. At the end of the first year he had earned $\$ 396$ in interest. How much was in each account? \$ $\square$ at $6 \%$ \$ $\square$ at 4\%

Solution

Solution Steps

To solve this problem, we need to set up a system of equations based on the information given. Let $ x $ be the amount invested at 6% and $ y $ be the amount invested at 4%. We know that the total investment is $7300, so $ x + y = 7300 $. The total interest earned is $396, so $ 0.06x + 0.04y = 396 $. We can solve this system of equations to find the values of $ x $ and $ y $.

Step 1: Set Up the Equations

Let $ x $ be the amount invested at 6% and $ y $ be the amount invested at 4%. We have the following equations based on the problem statement: