Question 1061088
{{{y=mx+b}}}


Expect m to be a rational number, {{{m=n/d}}}.


{{{y=(n/d)x+b}}}
{{{y-b=(n/d)x}}}
{{{(n/d)x=y-b}}}
{{{(n/d)x-y=-b}}}
{{{d((n/d)x-y)=d(-b)}}}
{{{highlight(nx-dy=-db)}}}, comparable to Ax+By=C



( Generally,  {{{B<>b}}}  ).