Question 166674
# 1



Let's graph {{{y=2x}}}



Looking at {{{y=2x}}} 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=0}}}  note: {{{y=2x}}} really looks like {{{y=2x+0}}} 



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


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


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,0,.1)),
  blue(circle(0,0,.12)),
  blue(circle(0,0,.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 up 2  and over 1




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

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



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


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




<hr>



# 2


Let's graph {{{y=5-x}}}



First rearrange the terms to get {{{y=-x+5}}}



Looking at {{{y=-x+5}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-1}}} and the y-intercept is {{{b=5}}} 



Since {{{b=5}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,5\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,5\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,5,.1)),
  blue(circle(0,5,.12)),
  blue(circle(0,5,.15))
)}}}



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

{{{slope=rise/run}}}


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


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



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




So starting at *[Tex \LARGE \left(0,5\right)], go down 1 unit 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,5,.1)),
  blue(circle(0,5,.12)),
  blue(circle(0,5,.15)),
  blue(arc(0,5+(-1/2),2,-1,90,270))
)}}}


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(0,5,.1)),
  blue(circle(0,5,.12)),
  blue(circle(0,5,.15)),
  blue(circle(1,4,.15,1.5)),
  blue(circle(1,4,.1,1.5)),
  blue(arc(0,5+(-1/2),2,-1,90,270)),
  blue(arc((1/2),4,1,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-x+5}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-x+5),
  blue(circle(0,5,.1)),
  blue(circle(0,5,.12)),
  blue(circle(0,5,.15)),
  blue(circle(1,4,.15,1.5)),
  blue(circle(1,4,.1,1.5)),
  blue(arc(0,5+(-1/2),2,-1,90,270)),
  blue(arc((1/2),4,1,2, 0,180))
)}}} So this is the graph of {{{y=-x+5}}} through the points *[Tex \LARGE \left(0,5\right)] and *[Tex \LARGE \left(1,4\right)]




<hr>



# 3



In order to graph ANY linear equation, you MUST solve for "y" first.



{{{4x+6y=8}}} Start with the given equation.



{{{6y=8-4x}}} Subtract {{{4x}}} from both sides.



{{{6y=-4x+8}}} Rearrange the terms.



{{{y=(-4x+8)/(6)}}} Divide both sides by {{{6}}} to isolate y.



{{{y=((-4)/(6))x+(8)/(6)}}} Break up the fraction.



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



Now let's graph {{{y=-(2/3)x+4/3}}} 





In order to graph this equation, we only need two points to create a straight line





----------------Let's find the first point----------------------



{{{y=-(2/3)x+4/3}}} Start with the given equation



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



{{{y=-4/3+4/3}}} Multiply {{{-2/3}}} and {{{2}}} to get {{{-4/3}}}



{{{y=0/3}}} Add the fractions 



{{{y=0}}} Reduce




So when {{{x=2}}}, we have the value {{{y=0}}}. This means we have the first point *[Tex \LARGE \left(2,0\right)]





----------------Let's find the first point----------------------



{{{y=-(2/3)x+4/3}}} Start with the given equation



{{{y=-(2/3)(5)+4/3}}} Plug in {{{x=5}}} 



{{{y=-10/3+4/3}}} Multiply {{{-2/3}}} and {{{5}}} to get {{{-10/3}}}



{{{y=-6/3}}} Add  



{{{y=-2}}} Reduce 




So when {{{x=5}}}, we have the value {{{y=-2}}}. This means we have the second point *[Tex \LARGE \left(5,-2\right)]





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



So we have the two points: *[Tex \LARGE \left(2,0\right)] and *[Tex \LARGE \left(5,-2\right)]



Now plot these two points on a coordinate system


{{{drawing(500,500,-6.5,13.5,-6.5,13.5,
graph(500,500,-6.5,13.5,-6.5,13.5,0),
grid(1),
circle(2,0/3,0.1),
circle(2,0/3,0.12),
circle(2,0/3,0.15),
circle(5,-6/3,0.1),
circle(5,-6/3,0.12),
circle(5,-6/3,0.15)
) }}}




Now draw a straight line through the two points. This line is the graph of {{{y=-(2/3)x+4/3}}}


{{{drawing(500,500,-6.5,13.5,-6.5,13.5,
graph(500,500,-6.5,13.5,-6.5,13.5,-(2/3)x+4/3),
grid(1),
circle(2,0/3,0.1),
circle(2,0/3,0.12),
circle(2,0/3,0.15),
circle(5,-6/3,0.1),
circle(5,-6/3,0.12),
circle(5,-6/3,0.15)
) }}} Graph of {{{y=-(2/3)x+4/3}}} through the two points *[Tex \LARGE \left(2,0\right)] and *[Tex \LARGE \left(5,-2\right)]