# 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.

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 $$x, y = \frac{u^2\pm\sqrt{u^2-4v}}{2}.$$

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 $y$

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 $x,y$.

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?

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 $$(y - 1)(y - 2)(y - 3)(y - 5)=0.$$

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..