System of equations $x + xy + y = 11$ and $yx^2 + xy^2 = 30$ I have problem with solving this one.
Total number of solutions from system of equations?
\begin{cases} 
x + xy + y = 11 \\ 
x^2y + y^2x = 30 
\end{cases}
There is a system of equation and I have tried to get some normal solutions, but I always get the fourth degree polynomial from which I do not know how to get simple 'x's and 'y's. I know that task asks me to just find total number, but I would like to know which solutions are those. This is adjusted for high school mathematics level. 
 A: make a change of variables $$x+y = u, xy = v. $$ then the equation in the new variables are $$\begin{align}u+v = 11\\uv = 30  \end{align}$$
this has solutions $$u = \frac{{11}^2\pm\sqrt{{11}^2-4 \times30}}{2}=6,5\quad v = 5,6$$
and $$x, y = \frac{u^2\pm\sqrt{u^2-4v}}{2}. $$
A: try substituting $p=x+y$ and $q=xy$
this will give you a simple quadratic for $p$ and/or $q$ and then some more quadratics for $x$ and $y$
A: Hint: 
This can be written as $xy(x+y)=30, xy+(x+y)=11$. 
First solve for $x+y$ and $xy$, then for each of those solutions, solve for $x,y$.
A: Hint: Use the substitution a=x+y
and b=xy.
Thus your two equations become a+b=11,   ab=30.
Can you solve the above equation, and from that solve for x and y?
A: There are exactly $4$ solutions (over characteristic zero), namely $(x,y)=(1,5),(2,3),(3,2),(5,1)$. To see this, note first that $y+1\neq 0$. In fact, for $y=-1$ the first equations implies $11=0$, a contradiction. It follows that $x=-\frac{y-11}{y+1}$. Substituting this into the second equation gives
$$
(y - 1)(y - 2)(y - 3)(y - 5)=0.
$$
A: x+xy+y=11             (i)
x^2y+xy^2=30          (ii)
from (ii)
xy(x+y)=30            (iii)
from (i)
xy= 11-(x+y)          (iv)
put value of xy in    (iii)
11-(x+y)=30
let x+y= z
therefore, (11-z)z=30
therefore, 11z-z^2=30
z^2-11x+30=0
solve for z
you will get z=5
      or  z=6

therefore, x+y can be equal to 5 or 6
as 11-(x+y)=xy
therefore, we will take cases
case 1
when x+y=5
then, xy=6
the only two cases thus formed are {2,3} and {3,2}
here we cannot take x or y = 6 because in (ii) it is clearly given that x^2y+xy^2=30. if x is 6 then {x^2y+xy^2} is greater than 30 which is not possible.
case 2
when x+y=6
then, xy=5
the only cases formed are {5,1} and {1,5}
so the final answer would be {1,5}; {5,1}; {2,3} and {3,2}.
hope you are clear with the solution..
