Intersections {x^2/y+y^2/x=25 and 5-x^2=y}

Question

I did run in to this problem last week in school and it was supposed to be solved with algebra.

This isn't just two quadratic equations

Solve all intersections

x^2/y+y^2/x=25
5-x^2=y


Spoiler

I did get this two
(25-5x^2+x^3-5x^4-x^6)/(5x-5x^4)=25
100-75x^2+x^3+15x^4-x^6=0
I did get a difrent sulution on my CAS calkylator
(x⁶ - 15x⁴ - x³ + 75x² - 125) / (x³ - 5x) = 25

How do you solve the intersections and check if you missed anyone

The X values the solution should be at

{x = -2.19267344038, x = 0.7772329487424, x = 2.19261654432, x = 3.893011723522}

Answer 1

Substituting the second equation into the first one: $$\dfrac{x^2}{5-x^2}+\dfrac{(5-x^2)}{x}=25$$ $$\dfrac{x^3+(5-x^2)^2}{5x-x^3}=25$$ Gives: $$x=-2.192...\vee x=0.198...\vee x=2.192...$$ So the three solutions are: $$x=-2.192...,y=0.195...$$ $$x=0.198...,y=4.960...$$ $$x=2.192...,y=0.195...$$