 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Solve a generic set of equations (Posted on 2017-09-03) 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

 No Solution Yet Submitted by Ady TZIDON Rating: 3.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution (spoiler) Comment 1 of 1
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
----------------------
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

 Posted by Steve Herman on 2017-09-03 11:56:51 Please log in:
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