Question 531829
The point (46,0) has x=46 and y=0.
It indeed lies on the line {{{2x+10y=92}}}
because for those values of x and y
{{{2x+10y=2*42+10*0=92+0=92}}}
However, when we substitute in the other 2 equations, those values of x and y satisfy neither
{{{2xy+10y=2*46*0+10*0=0+0=0}}} is far from 92, and
{{{-6x - 2y =-6*46-2*0=-6*46-0=-6*46}}} is far from {{{-6*4=-24}}}