Simplfiying $x(5xy+2x-1)=y(5xy+2y-1)$ I want to simplify: $x(5xy+2x-1)=y(5xy+2y-1)$ to $(x-y)(5xy+something-1)=0$ but I can't figure out what to do with the $2x$ and $2y$ on both sides.
 A: $$x(5xy+2x-1)=y(5xy+2y-1)$$
$$(x-y)5xy+2x^2-x-2y^2+y=0$$
$$(x-y)5xy+2(x^2-y^2)-(x-y)=0$$
$$(x-y)5xy+2(x-y)(x+y)-(x-y)=0$$
$$(x-y)(5xy+2(x+y)-1)=0$$
A: You know there will be a factor $x-y$ because setting $x=y$ makes the two sides equal.
First multiply everything out to get $$5x^2y+2x^2-x=5xy^2+2y^2-y$$
Now you say you know what to do with the $5x^2y$ term and the $x$ term (and equivalents on the right-hand side) - so bring the terms you can handle over to one side, and leave the terms you are unsure about on the other:
$$5xy(x-y)-(x-y)=2y^2-2x^2$$
Now it is easier to spot $y^2-x^2=(y+x)(y-x)=-(x+y)(x-y)$ so that $$(5xy-1)(x-y)=(-2x-2y)(x-y)$$ and finally, bringing it all together $$(5xy+2x+2y-1)(x-y)=0$$
A: we have $x(5xy+2x-1)-y(5xy+2y-1)=5x^2y+2x^2-x-5xy^2-2y^2+y=5xy(x-y)+2(x-y)(x+y)-(x-y)=(x-y)(5xy+2(x+y)-1)$
A: \begin{eqnarray}
&&x(5xy+2x-1)=y(5xy+2y-1)\\
&\Rightarrow&(5x^2y-5xy^2)+(2x^2-2y^2)-(x-y)=0\\
&\Rightarrow&5xy(x-y)+2(x-y)(x+y)-(x-y)=0\\
&\Rightarrow&(x-y)(5xy+2x+2y-1)=0
\end{eqnarray}
Moreover, note that $5xy+2x+2y-1=5(x+\frac{2}{5})(y+\frac{2}{5})-\frac{9}{5}$, you can simplifying as following (if you want):
$(x-y)[(x+\frac{2}{5})(y+\frac{2}{5})-\frac{9}{25}]=0$.
