An actor invests some money at 7​%, and ​$25000 more than twice the amount at 11%. The total annual interest earned from the investment is ​$45670. How much did he invest at each​ amount? Use the​ six-step method.

He invested $ ___ at 7% and $___ at 11%

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Respuesta :

Let [tex]x_1[/tex] be the amount of money invested at 7% and  [tex]x_2[/tex] the amount of money invested at 11%. The actor invests [tex]x_1[/tex] at 7% and 25000$ more than [tex]2x_1[/tex] at 11%, it means:
[tex]2x_1+25000=x_2[/tex] (Equation 1)
The total annual interest earned is 45670$:
[tex]\frac{7}{100}x_1+\frac{11}{100}x_2=45670[/tex] (Equation 2).
We remove [tex]x_2[/tex] from equation 2 by using equation 1:
[tex]\frac{7}{100}x_1+\frac{11}{100}(2x_1+25000)=45670\\ 7x_1+22x_1+275000 = 4567000\\ 29x_1 = 429200\\ x_1 = 4292000/29 = 148000\\ x_2 = 2x_1+25000 = 321000\\[/tex]

He invests 148000$ at 7% and 321000$ at 11%.