Question 278209
I'll do the first one to show you how to do these.



First, we must solve for 'y' (if not already solved for y).



{{{x+2y=6}}} Start with the given equation.



{{{2y=6-x}}} Subtract {{{x}}} from both sides.



{{{2y=-x+6}}} Rearrange the terms.



{{{y=(-x+6)/(2)}}} Divide both sides by {{{2}}} to isolate y.



{{{y=((-1)/(2))x+(6)/(2)}}} Break up the fraction.



{{{y=-(1/2)x+3}}} Reduce.



Looking at {{{y=-(1/2)x+3}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-1/2}}} and the y-intercept is {{{b=3}}} 



Since {{{b=3}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,3\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,3\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15))
)}}}



Now since the slope is comprised of the "rise" over the "run" this means

{{{slope=rise/run}}}


Also, because the slope is {{{-1/2}}}, this means:


{{{rise/run=-1/2}}}



which shows us that the rise is -1 and the run is 2. This means that to go from point to point, we can go down 1  and over 2




So starting at *[Tex \LARGE \left(0,3\right)], go down 1 unit 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15)),
  blue(arc(0,3+(-1/2),2,-1,90,270))
)}}}


and to the right 2 units to get to the next point *[Tex \LARGE \left(2,2\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15)),
  blue(circle(2,2,.15,1.5)),
  blue(circle(2,2,.1,1.5)),
  blue(arc(0,3+(-1/2),2,-1,90,270)),
  blue(arc((2/2),2,2,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-(1/2)x+3}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-(1/2)x+3),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15)),
  blue(circle(2,2,.15,1.5)),
  blue(circle(2,2,.1,1.5)),
  blue(arc(0,3+(-1/2),2,-1,90,270)),
  blue(arc((2/2),2,2,2, 0,180))
)}}} So this is the graph of {{{y=-(1/2)x+3}}} through the points *[Tex \LARGE \left(0,3\right)] and *[Tex \LARGE \left(2,2\right)]