Question 805602
{{{2x=y/3 -4 }}}.........here you have both {{{x}}} and {{{y}}} first degree variable; so, you have a linear equation

you can solve it for {{{y}}} and get {{{slope-intercept}}} form {{{y=mx+b}}} where {{{m=slope}}} and {{{b=y-intercept}}}

then you one point that lie on a line and you will need only one more point to graph this line

{{{2x+4=y/3 }}}

{{{2*3x+4*3=y*3/3 }}}

{{{6x+12=y }}}

or

{{{y=6x+12 }}}

so, the slope {{{m=6}}} and y-intercept is {{{b=12}}} which means a line crosses {{{y-axis}}} at point ({{{ 0 }}},{{{12}}})

now find one more point; easiest is to find  x-intercept, so set {{{y=0}}} and solve equation for {{{x}}}


{{{0=6x+12 }}}


{{{-12=6x  }}}

{{{-12/6=x  }}}

{{{x=-2  }}}

so,x-intercept is at point ({{{ -2 }}},{{{ 0 }}})

plot both points and draw a line through:

{{{ drawing( 600, 600, -10, 10, -10, 15,circle(-2,0,0.2),circle(0,12,0.2), locate(-2.8,-0.5,p1(-2,0)),locate(0.3,12,p2(0,12)),graph( 600, 600, -10, 10, -10, 15, 6x+12)) }}}