How many points are common to the graphs of the two equations $(x-y+2)(3x+y-4)=0$ and $(x+y-2)(2x-5y+7)=0$? How many points are common to the graphs of the two equations $(x-y+2)(3x+y-4)=0$ and $(x+y-2)(2x-5y+7)=0$?
\begin{align*}
  (x-y+2)(3x+y-4) &= 0\tag{1}\\ 
 (x+y-2)(2x-5y+7) &= 0\tag{2}
\end{align*}
In equation $1$, it is not possible for both $x-y+2$ and $3x+y-4$ to be nonzero. Like-wise in equation $2$, it is not possible for both $x+y-2$ and $2x-5y+7$ to be nonzero.
Therefore the LHS of either equation must involve a product of $0$ and some number.
This means there are infinitely many solutions to both equations.
However this doesn't shead any light on which points are common between the two equations. I'm confused. It seems I've made an error in my judgement. What should I reconsider?

If $x-y+2=0,y=?$
Replace the value of$y$ in terms of $x,$ in the second equation to find
$$0=(x+x+2-2)(3x+x+2-4)=2x(4x-2)$$
$$x=?,?$$
Had there been $n$ factors, we would have $n$ degree equation in $x$
Similarly for $3x+y-4=0\implies y=4-3x$
Can you take it from here?
