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There are no integer solutions, but there are complex (imaginary number) solutions, where y2 - 6y +10 = 0 for y and x = 6-y

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You have two equations:

x + y = 6

x * y = 10

x= 6-y and sub in

y*(6-y) = 10

y^2 - 6y + 10 = 0 (quadratic equation)

y = -b +/- sqrt(b^2-4ac)/2a

= 6/2 +/- sqrt(36-4*1*10) / 2

= 6/2 +/- imag * sqrt(4)/2

= 3 +/- imag * 1

number 1= 3 + i

number 2 = 3 - i

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Wiki User

13y ago
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Anonymous

4y ago
needed it for homework now i dont have answer

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Q: What two numbers add up to 6 but multiply to 10?
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