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Larry Mitchell invested part of his $19,000 advance at 6% annual simple interest and the rest at 3% annual simple interest. If his total yearly interest from both accounts was $720, find the amount invested at each rate.

The amount invested at 6% is $?
The amount invested at 3% is $?

User Therii
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1 Answer

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Final answer:

Larry Mitchell invested $5,000 at 6% simple interest and $14,000 at 3% simple interest to achieve a total yearly interest of $720 from his $19,000 total investment.

Step-by-step explanation:

Larry Mitchell needs to determine the amount invested at both 6% and 3% annual simple interest rates when his total investment is $19,000, and his total yearly interest from both accounts is $720.

Let's assume Larry invested $x at 6% simple interest. Therefore, he invested $(19,000 - x) at 3% simple interest. The total interest from both investments is $720 per year. To find the amounts invested at each rate, we can set up the following equations:

Interest from 6% investment: 0.06x

Interest from 3% investment: 0.03(19,000 - x)

Total interest: 0.06x + 0.03(19,000 - x) = $720

Solving this equation:

  • 0.06x + 0.03(19,000) - 0.03x = 720
  • 0.06x - 0.03x + 570 = 720
  • 0.03x = 720 - 570
  • 0.03x = 150
  • x = 150 / 0.03
  • x = $5,000

Therefore, Larry invested $5,000 at 6% and $19,000 - $5,000 = $14,000 at 3% interest.

User Jinzu
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