Question 1101023
{{{matrix(3,11,
2y,"-",3x,"=",13xy,"-->",4y,"-",6x,"=",26xy,
5y,"+",2x,"=",4xy,"-->",15y,"-",6x,"=",12xy,
" "," "," "," "," ","-->",19y," "," ","=",38xy)}}}
There are two ways that
{{{19y=38xy}}} could be true: {{{system(y=0,"or",x=1/2)}}} .
So, we can see two solutions:
{{{system(y=0,5y+2x=4xy)}}} --> {{{system(y=0,0+2x=0y)}}} --> {{{highlight(system(y=0,x=0))}}}
{{{system(x=1/2,5y+2x=4xy)}}} --> {{{system(x=1/2,5y+1=2y)}}} --> {{{system(x=1/2,3y+1=0)}}} --> {{{highlight(system(x=1/2,y=-1/3))}}}
Those are solutions.
Could there be another solution?
Does this make sense?
{{{2y-3x=13xy}}} --> {{{2y-13xy=3x}}} --> {{{(2-13x)y=3x}}} --> {{{y=3x/(2-13x)}}}
{{{5y+2x=4xy}}} --> {{{5y-4xy=-2x}}} --> {{{(5-4x)y=-2x}}} --> {{{y=-2x/(5-4x)}}}
We could graph those two rational functions
and their horizontal and vertical asymptotes:
{{{drawing(300,300,-2,4,-2,4,
graph(300,300,-2,4,-2,4,-3,3x/(2-13x),-2x/(5-4x)),
line(1.25,-2,1.25,4),line(2/13,-2,2/13,4),
line(-3,-3/13,4,-3/13),line(-3,0.5,4,0.5)
)}}}
It looks like we've got all the solutions.