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The answer: $1,000 was invested at %5. $2,500 was invested @ 8% It is done with a system of equations when "x" is a portion of $3,500 and "y" is a portion of $3,500: x + y = 3500AND .05x +.08y = 250 Use algebra and solve #2 for "x" as follows: .05x + .08y = 250 .05x = 250 - .08y x = 5000 - 1.6y plug the value of "x" into the first equation and solve for "y": 5000 - 1.6y + y = 3500 5000 - 3500 - 1.6y + y = 0 1500 - 1.6y + y = 0 1500 = 1.6y -y 1500 = .6y 2500 = y plug the value of "y" into the original equation and solve for "x" x + 2500 = 3500 x = 3500 - 2500 x = 1000 You can plug them both into the 2nd equation in the system to check it.

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Q: A sum of 3500 dollars is invested in two parts One part brings a return of 5 percent and the other a return of 8 percent The total annual return is 250 dollars Find the amount invested at each rate?
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