Question 201739
Step 1: Graph {{{4x+2y=8}}}





{{{4x+2y=8}}} Start with the given equation.



{{{2y=8-4x}}} Subtract {{{4x}}} from both sides.



{{{2y=-4x+8}}} Rearrange the terms.



{{{y=(-4x+8)/(2)}}} Divide both sides by {{{2}}} to isolate y.



{{{y=((-4)/(2))x+(8)/(2)}}} Break up the fraction.



{{{y=-2x+4}}} Reduce.





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}}} (which is also {{{4x+2y=8}}}) through the points *[Tex \LARGE \left(0,4\right)] and *[Tex \LARGE \left(1,2\right)]



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



Step 2: Graph {{{y=-2x+6}}}



Looking at {{{y=-2x+6}}} 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=6}}} 



Since {{{b=6}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,6\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,6\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,6,.1)),
  blue(circle(0,6,.12)),
  blue(circle(0,6,.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(0,6\right)], go down 2 units 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,6,.1)),
  blue(circle(0,6,.12)),
  blue(circle(0,6,.15)),
  blue(arc(0,6+(-2/2),2,-2,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,4\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,6,.1)),
  blue(circle(0,6,.12)),
  blue(circle(0,6,.15)),
  blue(circle(1,4,.15,1.5)),
  blue(circle(1,4,.1,1.5)),
  blue(arc(0,6+(-2/2),2,-2,90,270)),
  blue(arc((1/2),4,1,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-2x+6}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-2x+6),
  blue(circle(0,6,.1)),
  blue(circle(0,6,.12)),
  blue(circle(0,6,.15)),
  blue(circle(1,4,.15,1.5)),
  blue(circle(1,4,.1,1.5)),
  blue(arc(0,6+(-2/2),2,-2,90,270)),
  blue(arc((1/2),4,1,2, 0,180))
)}}} So this is the graph of {{{y=-2x+6}}} through the points *[Tex \LARGE \left(0,6\right)] and *[Tex \LARGE \left(1,4\right)]



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



Step 3: Graph the two equations together on the same coordinate plane



{{{ drawing(500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,-2x+4,-2x+6)

)}}}


Graph of {{{4x+2y=8}}} (red) and {{{y=-2x+6}}} (green). 



From the graph, we can see that the two lines are parallel and will NEVER cross. 



So this means that there are no solutions.



So the system is inconsistent.