The image shows a math problem involving simple interest. Here's a summary of the problem:

Problem Statement:

• Melissa has invested $8,200 in two accounts. One account pays 7% annual interest, and the other pays 8% annual interest.
• After one year, the total interest earned from both accounts is $609.

You need to determine how much money Melissa invested in each account. Let $x$ represent the amount invested at 7%, and $y$ represent the amount invested at 8%.

Equations:

1. $x + y = 8200$ (since the total amount invested is $8,200)
2. $0.07x + 0.08y = 609$ (since the total interest earned is $609)

To Solve:

You can use the substitution or elimination method to solve the system of equations.

Let's solve this step-by-step using substitution:

1. From the first equation:
   $y = 8200 - x$

2. Substitute $y$ into the second equation:
   $0.07x + 0.08(8200 - x) = 609$

3. Distribute and combine like terms:
   $0.07x + 656 - 0.08x = 609$

   Simplifying:
   $-0.01x + 656 = 609$

4. Isolate $x$:
   $-0.01x = 609 - 656$
   $-0.01x = -47$
   $x = \frac{-47}{-0.01} = 4700$

So, Melissa invested $4,700 at 7%.

5. Substitute $x = 4700$ into $y = 8200 - x$:
   $y = 8200 - 4700 = 3500$

Melissa invested $3,500 at 8%.

Final Answer:

• $4,700 was invested at 7%.
• $3,500 was invested at 8%.

Would you like a more detailed explanation, or do you have any questions?

Related Questions:

1. How can you use the elimination method to solve the same system of equations?
2. What would the interest be if Melissa had invested the entire $8,200 at 7%?
3. What is the significance of the different interest rates in this problem?
4. How would the investment amounts change if the total interest earned was $700 instead of $609?
5. Can this type of problem be solved graphically? If so, how?

Tip:

When dealing with percentage problems, always ensure that you convert percentages to their decimal forms before performing calculations.