Question

Melody invested $10,500 into two accounts. One account paid 3.5% interest, and the other paid 6% interest. She earned $515 in simple interest at the end of one year. How much money did she invest in each account? Set up a system of equations, state both variables, and solve. Write your answer in a complete sentence and make sure to write the units. (6 points)

A. x = $3,000, \, y = $7,500
B. x = $7,500, \, y = $3,000
C. x = $5,000, \, y = $5,500
D. x = $5,500, \, y = $5,000

Answer 1

Melody invested $7,500 at 3.5% interest and $3,000 at 6% interest. The system of equations was solved using the substitution method.

Step-by-step explanation:

To solve the problem, let's set up a system of equations. We'll define x as the amount invested at 3.5% interest and y as the amount invested at 6% interest. We can create two equations based on the information given:

1. The total investment is $10,500: x + y = 10,500
2. The total simple interest earned in one year is $515: 0.035x + 0.06y = 515

Now, we can solve this system using substitution or elimination. Let's use the substitution method:

1. From the first equation, y = 10,500 - x.
2. Substitute y into the second equation: 0.035x + 0.06(10,500 - x) = 515.
3. Simplify and solve for x: x = 7,500.
4. Substitute x back into the first equation to find y: y = 3,000.

Therefore, Melody invested $7,500 at 3.5% interest and $3,000 at 6% interest.