What couples of numbers satisfy the following set of equations:
x^2+xy=t
y^2+xy=t*k ?
List all the qualifying couples.
Please verify for t=20 & k=2
Answer
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For any given t and k
a) if t=0 then there are an infinite number of solutions of the form x = -y
b) if t <> 0, then the solutions (x,y) are
( sqrt(t/(1+k)), k*sqrt(t/(1+k)) ) and ( -sqrt(t/(1+k)), -k*sqrt(t/(1+k)) )
Method
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y(x+y) = tk
x(x+y) = t
If t <> 0 then divide the 1st equation by the 2nd, giving
y/x = k
substitute y = xk into the 2nd equation, giving
x(x+xk) = t
x^2* (1+k) = t
x = +/- sqrt(t/(1+k))
Requested verification
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If t = 20 and k = 2
then x = +/- sqrt(20/3)
If x is positive, then y = 2(sqrt(20/3))
x^2 + xy = 20/3 + 40/3 = 20 = t
y^2 + xy = 80/3 + 40/3 = 40 = tk
same if x and y are negative
Solution (spoiler)