Phorphyrion wrote: Solve the following system (over the real numbers): \[\begin{cases}5x+4y+5xy-2xy^2-2x^2y=20 &\\ 3x+3y+3xy+xy^2+x^2y=23&\end{cases}\] Adding first line plus twice second line, we get $11x+10y+11xy=66$ So $y\ne -1$ and $x=\frac 2{11}\frac{33-5y}{y+1}$ Plugging this in second line, we get $110y^4-749y^3+1133y^2-3179y+605=0$ This quartic can easily be solved with any classical method. It has two real roots whose closed form are (in my opinion) quite ugly. So I let you use these closed forms and I just give numerical result (including $x=\frac 2{11}\frac{33-5y}{y+1}$ application) : $\boxed{\text{S1 : }(x,y)\sim(4.83366115349,0.20309754518)}$ $\boxed{\text{S2 : }(x,y)\sim(0.09720998049,5.86583007195)}$

Sorry, typo. Fixed now.

Phorphyrion wrote: Solve the following system (over the real numbers): \[\begin{cases}5x+5y+5xy-2xy^2-2x^2y=20 &\\ 3x+3y+3xy+xy^2+x^2y=23&\end{cases}\] $11x+11y+11xy=66\Rightarrow x+y+xy=6\Rightarrow xy^2+x^y=5\Rightarrow xy(x+y)=5\Rightarrow (x+y-xy)^2=(x+y+xy)^2-4xy(x+y)=36-4.5=16\Rightarrow x+y-xy=4$ or $x+y-xy=-4\Rightarrow x+y=5,xy=1 $ or $x+y=1,xy=5$(impossible) $ \Rightarrow (x,y)=(\frac{5+\sqrt{21}}2,\frac{5-\sqrt{21}}2),(\frac{5-\sqrt{21}}2,\frac{5+\sqrt{21}}2)$

Just assume x+y = a and xy=b

5x+5y+5xy-2xy²-2x²y= 20 3x+3y+3xy+xy²+x²y=23 x+y= a, xy= b So, a+b= 6 a= 6-b 3a+3b+ab=23 b²-6b+5= 0 b=1 or b= 5 if b= 5, x²-x+5= 0 x is not real. So b= 1 a=5 x= 5+√21/2, 5-√21/2 y= 5-√21/2, 5+√21/2