Question 130092


Start with the given system of equations:


{{{2x+y=5}}}

{{{3x-y=10}}}





In order to graph these equations, we need to solve for y for each equation.




So let's solve for y on the first equation


{{{2x+y=5}}} Start with the given equation



{{{y=5-2x}}}  Subtract {{{2 x}}} from both sides



{{{y=-2x+5}}} Rearrange the equation




Now lets graph {{{y=-2x+5}}} 





Looking at {{{y=-2x+5}}} 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=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 {{{-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(0,5\right)], go down 2 units 

{{{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+(-2/2),2,-2,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,3\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(1,3,.15,1.5)),
  blue(circle(1,3,.1,1.5)),
  blue(arc(0,5+(-2/2),2,-2,90,270)),
  blue(arc((1/2),3,1,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-2x+5}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-2x+5),
  blue(circle(0,5,.1)),
  blue(circle(0,5,.12)),
  blue(circle(0,5,.15)),
  blue(circle(1,3,.15,1.5)),
  blue(circle(1,3,.1,1.5)),
  blue(arc(0,5+(-2/2),2,-2,90,270)),
  blue(arc((1/2),3,1,2, 0,180))
)}}} So this is the graph of {{{y=-2x+5}}} through the points *[Tex \LARGE \left(0,5\right)] and *[Tex \LARGE \left(1,3\right)]



<hr>



So let's solve for y on the second equation


{{{3x-y=10}}} Start with the given equation



{{{-y=10-3x}}}  Subtract {{{3 x}}} from both sides



{{{-y=-3x+10}}} Rearrange the equation



{{{y=(-3x+10)/(-1)}}} Divide both sides by {{{-1}}}



{{{y=(-3/-1)x+(10)/(-1)}}} Break up the fraction



{{{y=3x-10}}} Reduce




Now lets  graph {{{y=3x-10}}} 






Looking at {{{y=3x-10}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=3}}} and the y-intercept is {{{b=-10}}} 



Since {{{b=-10}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-10\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,-10\right)]


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



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

{{{slope=rise/run}}}


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


{{{rise/run=3/1}}}



which shows us that the rise is 3 and the run is 1. This means that to go from point to point, we can go up 3  and over 1




So starting at *[Tex \LARGE \left(0,-10\right)], go up 3 units 

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


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,-7\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-10,.1)),
  blue(circle(0,-10,.12)),
  blue(circle(0,-10,.15)),
  blue(circle(1,-7,.15,1.5)),
  blue(circle(1,-7,.1,1.5)),
  blue(arc(0,-10+(3/2),2,3,90,270)),
  blue(arc((1/2),-7,1,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=3x-10}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,3x-10),
  blue(circle(0,-10,.1)),
  blue(circle(0,-10,.12)),
  blue(circle(0,-10,.15)),
  blue(circle(1,-7,.15,1.5)),
  blue(circle(1,-7,.1,1.5)),
  blue(arc(0,-10+(3/2),2,3,90,270)),
  blue(arc((1/2),-7,1,2, 180,360))
)}}} So this is the graph of {{{y=3x-10}}} through the points *[Tex \LARGE \left(0,-10\right)] and *[Tex \LARGE \left(1,-7\right)]


<hr>



Now let's graph the two equations together on the same coordinate system




{{{ drawing(500,500,-10,10,-10,10,
grid(1),
graph( 500,500,-10,10,-10,10, -2x+5,3x-10),
  blue(circle(3,-1,.1)),
  blue(circle(3,-1,.12)),
  blue(circle(3,-1,.15))
)
}}} Graph of {{{y=-2x+5}}}(red) and {{{y=3x-10}}}(green)



From the graph, we can see that the two lines intersect at the point (3,-1)