How To Solve A Non-Linear System of Equations

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[@JosBergervoet]

If you allow complex solutions at 5:02 in the video, you should also have allowed them at 3:18, where there should be two real and two imaginary 4th roots!

[@ManjulaMathew-wb3zn]

You are absolutely correct. I noticed the same. Now you have to square root i and -i too.

[@user-xh6ez2ko7c]

🌹🌹🌹👍👍👍

[@SrisailamNavuluri]

Nice explanation.

[@ShortsOfSyber]

Glad you liked it

[@-basicmaths862]

Solving the above equations We get x=2y.
x{{-4,4},y={-2,2}

[@toveirenestrand3547]

xy + 24 = x^3/y and xy - 6 = y^3/x:
xy^2 + 24y = x^3, so x(x^2 - y^2) = 24y, thus x^2 - y^2 = 24y/x, and (x^2)y - 6x = y^3, so y(x^2 - y^2) = 6x, thus x^2 - y^2 = 6x/y. From this we have that 24y/x = 6x/y, hence 4y^2 = x^2 which yields that (2y)^2 = x^2. Therefore x = +/- 2y.
Then (+/- 2y)y + 24 = +/- 2y^2 + 24 = (+/- 2y)^3/y = [(+/- 2)^3](y^3)/y = +/- 8(y^2) which gives +/- 6y^2 = 24, so y^2 = +/- 4.
If y^2 = 4, y = +/- 2. If y = 2, x = 4, and if y = - 2, x = - 4.
If y^2 = - 4, y = +/- 2i. If y = 2i, x = - 4i, and if y = - 2i, x = 4i.
Answer: (x, y) = {(- 4, - 2), (4, 2), (- 4i, 2i), (4i, - 2i)}.

[@trojanleo123]

x,y = 4,2 or
x,y = -4,-2

[@petrileskinen2988]

At time 2:50 my solution was to multiply both equations with xy = 8:
x^3 / y *xy = 32 * 8 => x^4 = 256 => x = (4, -4, 4i, -4i)
and
y^3 / x * xy = 2 * 8 => y^4 = 16 => y = (2, -2, 2i, -2i)