Question

Raul invests $1500 into two savings accounts, one earning 4% annual interest and the other earning 6% annual interest. At the end of one year, Raul has earned $72 in interest. How much did he invest at each rate?

Answer 1

To solve the problem, we need to set up and solve a system of equations using the given information. After solving the system of equations, we find that Raul invested $600 at 4% and $900 at 6%.

Step-by-step explanation:

To solve this problem, we can set up two equations using the given information. Let's say Raul invested x dollars at 4% and y dollars at 6%. The interest earned from the 4% investment would be 0.04x and the interest earned from the 6% investment would be 0.06y. According to the problem, the total interest earned is $72, so we can write the equation:

0.04x + 0.06y = 72

Since Raul invested a total of $1500, we can write another equation:

x + y = 1500

We now have a system of equations that we can solve to find the values of x and y. Using substitution or elimination, we can find that x = $600 and y = $900. Therefore, Raul invested $600 at 4% and $900 at 6%.