An actor invests some money at 7%, and $42000 more than four times the amount at 11 %. the total annual interest earned from the investment is $59190. how much did he invest at each amount? use the six-step method.

he invested $ at 7% and $
at 11%.

The answer is that the actor invested $ 107,000 at 7% and $ 470,000 at 11%.

Step-by-step explanation:

1. Let's check the information provided to us to answer the problem correctly:

Two different interest rates from the investments:

7% from some money

11% from US$ 42,000 + four times the amount invested at 7%

Total annual interest earnings from both investments = US$ 59,190

2. How much did the actor invest at each amount? Use the six-step method.

x = Money invested at 7%

4x + 42,000 = Money invested at 11%

7% = 0.07, 11% = 0.11

3. Now, we can elaborate our equation to solve for x, this way:

0.07x + 0.11 (4x + 42,000) = 59,190

0.07x + 0.44x + 4,620 = 59,190

0.51x = 59,190 - 4,620

0.51x = 54,570

x = 54,570/0.51 = 107,000

4. Now we can find the amount the actor invested at 11%, this way:

4x + 42,000 = Money invested at 11%

4 * 107,000 + 42,000 = Money invested at 11%

428,000 + 42,000 = Money invested at 11%

470,000 = Money invested at 11%

5. Let's prove that x = 107,000 is correct, this way:

0.07x + 0.11 (4x + 42,000) = 59,190

0.07 * 107,000 + 0.11 * (4 * 107,000 + 42,000) = 59,190

7,490 + 0.11 * 470,000 = 59,190

7,490 + 51,700 = 59,190

59,190 = 59,190

We proved that x = 107,000 is correct.

6. The answer is that the actor invested $ 107,000 at 7% and $ 470,000 at 11%.

fwms@lil_ben06

step-by-step explanation:

x-3y=4

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