Question 150559
I'll do the first one to get you started



First let's graph the equation {{{2x + y = 4}}}



{{{2x + y = 4}}} Start with the first equation.



{{{y=4-2x}}} Subtract {{{2x}}} from both sides.



{{{y=-2x+4}}} Rearrange the terms.





Looking at {{{y=-2x+4}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-2}}} and the y-intercept is {{{b=4}}} 



Since {{{b=4}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,4\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,4\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,4,.1)),
  blue(circle(0,4,.12)),
  blue(circle(0,4,.15))
)}}}



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

{{{slope=rise/run}}}


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


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



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




So starting at *[Tex \LARGE \left(0,4\right)], go down 2 units 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,4,.1)),
  blue(circle(0,4,.12)),
  blue(circle(0,4,.15)),
  blue(arc(0,4+(-2/2),2,-2,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,2\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,4,.1)),
  blue(circle(0,4,.12)),
  blue(circle(0,4,.15)),
  blue(circle(1,2,.15,1.5)),
  blue(circle(1,2,.1,1.5)),
  blue(arc(0,4+(-2/2),2,-2,90,270)),
  blue(arc((1/2),2,1,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-2x+4}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-2x+4),
  blue(circle(0,4,.1)),
  blue(circle(0,4,.12)),
  blue(circle(0,4,.15)),
  blue(circle(1,2,.15,1.5)),
  blue(circle(1,2,.1,1.5)),
  blue(arc(0,4+(-2/2),2,-2,90,270)),
  blue(arc((1/2),2,1,2, 0,180))
)}}} So this is the graph of {{{y=-2x+4}}} through the points *[Tex \LARGE \left(0,4\right)] and *[Tex \LARGE \left(1,2\right)]



------------------------------------------------



Now let's graph the equation {{{3x - 4y = 6}}}



{{{3x - 4y = 6}}} Start with the second equation.



{{{-4y=6-3x}}} Subtract {{{3x}}} from both sides.



{{{y=(6-3x)/(-4)}}} Divide both sides by {{{-4}}} to isolate {{{y}}}.



{{{y=(6)/(-4)-(3x)/(-4)}}} Break up the fraction.



{{{y=-3/2+(3x)/(4)}}} Reduce 



{{{y=(3/4)x-3/2}}} Rearrange the terms.






Looking at {{{y=(3/4)x-3/2}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=3/4}}} and the y-intercept is {{{b=-3/2}}} 



Since {{{b=-3/2}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-\frac{3}{2}\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,-\frac{3}{2}\right)]


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



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

{{{slope=rise/run}}}


Also, because the slope is {{{3/4}}}, this means:


{{{rise/run=3/4}}}



which shows us that the rise is 3 and the run is 4. This means that to go from point to point, we can go up 3  and over 4




So starting at *[Tex \LARGE \left(0,-\frac{3}{2}\right)], go up 3 units 

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


and to the right 4 units to get to the next point *[Tex \LARGE \left(4,\frac{3}{2}\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-3/2,.1)),
  blue(circle(0,-3/2,.12)),
  blue(circle(0,-3/2,.15)),
  blue(circle(4,3/2,.15,1.5)),
  blue(circle(4,3/2,.1,1.5)),
  blue(arc(0,-3/2+(3/2),2,3,90,270)),
  blue(arc((4/2),3/2,4,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(3/4)x-3/2}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,(3/4)x-3/2),
  blue(circle(0,-3/2,.1)),
  blue(circle(0,-3/2,.12)),
  blue(circle(0,-3/2,.15)),
  blue(circle(4,3/2,.15,1.5)),
  blue(circle(4,3/2,.1,1.5)),
  blue(arc(0,-3/2+(3/2),2,3,90,270)),
  blue(arc((4/2),3/2,4,2, 180,360))
)}}} So this is the graph of {{{y=(3/4)x-3/2}}} through the points *[Tex \LARGE \left(0,-\frac{3}{2}\right)] and *[Tex \LARGE \left(4,\frac{3}{2}\right)]



======================================



So together the two graphs look like 


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-2x+4,(3/4)x-3/2),
  blue(circle(2,0,.1)),
  blue(circle(2,0,.12)),
  blue(circle(2,0,.15))
)}}} Graph of {{{y=-2x+4}}} (red) and {{{y=(3/4)x-3/2}}} (green)



From the graph, we can see that the two lines intersect at the point (2,0). So the solution of the system is {{{x=2}}} and {{{y=0}}}