Question 134288

{{{5y=10-x}}}  Subtract {{{ x}}} from both sides

{{{5y=-x+10}}} Rearrange the equation

{{{y=(-x+10)/(5)}}} Divide both sides by {{{5}}}

{{{y=(-1/5)x+(10)/(5)}}} Break up the fraction

{{{y=(-1/5)x+2}}} Reduce

Looking at {{{y=-(1/5)x+2}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-1/5}}} and the y-intercept is {{{b=2}}}

Since {{{b=2}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,2\right)].Remember the y-intercept is the point where the graph intersects with the y-axis

So we have one point *[Tex \LARGE \left(0,2\right)]

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

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

{{{slope=rise/run}}}

Also, because the slope is {{{-1/5}}}, this means:

{{{rise/run=-1/5}}}

which shows us that the rise is -1 and the run is 5. This means that to go from point to point, we can go down 1  and over 5

So starting at *[Tex \LARGE \left(0,2\right)], go down 1 unit

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

and to the right 5 units to get to the next point *[Tex \LARGE \left(5,1\right)]

{{{drawing(500,500,-10,10,-10,10,
grid(1),
blue(circle(0,2,.1)),
blue(circle(0,2,.12)),
blue(circle(0,2,.15)),
blue(circle(5,1,.15,1.5)),
blue(circle(5,1,.1,1.5)),
blue(arc(0,2+(-1/2),2,-1,90,270)),
blue(arc((5/2),1,5,2, 0,180))
)}}}

Now draw a line through these points to graph {{{y=-(1/5)x+2}}}

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