High School

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You invested $29,000 in two accounts paying 6% and 7% annual interest, respectively. If the total interest earned for the year was $1,940, how much was invested at each rate?

Answer :

Final answer:

The student invested $9,000 at 6% interest and $20,000 at 7% interest. We found these amounts by setting up a system of equations based on the total amount invested and the total interest earned, and then solving the equations.

Explanation:

To solve the problem of determining how much was invested at each interest rate, we can set up a system of equations based on the given information. Let's denote x as the amount invested at 6% and y as the amount invested at 7%. The total amount invested is $29,000, so we have the equation x + y = 29,000. We also know that the total interest earned for the year is $1,940. Using the formula for interest, which is principal times rate, we can set up the second equation: 0.06x + 0.07y = 1,940.

  • Equation 1: x + y = 29,000
  • Equation 2: 0.06x + 0.07y = 1,940

We can solve this system of equations using either substitution or elimination method. Let's use the elimination method: Multiply equation 1 by -0.06 to cancel out x when we add it to equation 2.

  • -0.06x - 0.06y = -1,740 (after multiplying equation 1 by -0.06)
  • 0.06x + 0.07y = 1,940 (equation 2)

Adding the two equations, we eliminate x and get: 0.01y = 200. Thus, y = 20,000. Now that we have y, we can substitute it back into equation 1 to find x: x + 20,000 = 29,000, hence x = 9,000.

So the student invested $9,000 at 6% and $20,000 at 7%.

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