3425.sagews

#    Find x^2 + y^3
#    given that:
#        x + y = 1
#        x*y = 2

#    Substitute y = 1-x into x*y=2 and solve:
show(solve(x*(1-x)==2,x))

\newcommand{\Bold}[1]{\mathbf{#1}}\left[x = -\frac{1}{2} i \, \sqrt{7} + \frac{1}{2}, x = \frac{1}{2} i \, \sqrt{7} + \frac{1}{2}\right]

#    Let the first solution be x1 and the second solution be x2.
x1 = -1/2*i*sqrt(7)+1/2
x2 = 1/2*i*sqrt(7)+1/2

#    Use y = 1-x to find y1 and y2.
y1 = 1-x1
y2 = 1-x2

#    Display the pairs of points (x1,y1) and (x2,y2).
show((x1,y1))
show((x2,y2))

\newcommand{\Bold}[1]{\mathbf{#1}}\left(-\frac{1}{2} i \, \sqrt{7} + \frac{1}{2}, \frac{1}{2} i \, \sqrt{7} + \frac{1}{2}\right)

\newcommand{\Bold}[1]{\mathbf{#1}}\left(\frac{1}{2} i \, \sqrt{7} + \frac{1}{2}, -\frac{1}{2} i \, \sqrt{7} + \frac{1}{2}\right)

#    Compute x^2 + y^3 for both pairs of solutions
show(expand(x1^2+y1^3))
show(expand(x2^2+y2^3))

\newcommand{\Bold}[1]{\mathbf{#1}}-i \, \sqrt{7} - 4

\newcommand{\Bold}[1]{\mathbf{#1}}i \, \sqrt{7} - 4

#    Both of these answers are valid solutions to the original problem.
#    Also, if it was possible to graph y+x=1, x*y=2, and x^2+y^3=i*sqrt(7)-4
#    or x^2+y^3=-i*sqrt(7)-4, all the graphs would intersect at two points,
#    [x1,y1] and [x2,y2].
#
#    It is not possible to graph these in two or three dimensions though.
#    To show imaginary planes for x and y, four dimensions are needed.
#    I can try to explain this better if you would like.