Question 127399
I'll do the first one to get you on track



# 1


a)




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



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



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



{{{y=(-5x+5)/(-2)}}} Divide both sides by {{{-2}}}



{{{y=(-5/-2)x+(5)/(-2)}}} Break up the fraction



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




So the equation is now in slope-intercept form ({{{y=mx+b}}}) where the slope is {{{m=5/2}}} and the y-intercept is {{{b=-5/2}}}



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b)




{{{5*x-2*y=5}}} Start with the equation in standard form


Let's find the x-intercept


To find the x-intercept, let y=0 and solve for x:

{{{5*x-2*(0)=5}}} Plug in {{{y=0}}}


{{{5*x=5}}} Simplify


{{{x=5/5}}} Divide both sides by 5



{{{x=1}}} Reduce




So the x-intercept is *[Tex \Large \left(1,0\right)] (note: the x-intercept will always have a y-coordinate equal to zero)




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{{{5*x-2*y=5}}} Start with the equation  in standard form


Now let's find the y-intercept


To find the y-intercept, let x=0 and solve for y:

{{{5*(0)-2*y=5}}} Plug in {{{x=0}}}


{{{2*y=5}}} Simplify


{{{x=5/-2}}} Divide both sides by -2




{{{y=-5/2}}} Reduce




So the y-intercept is *[Tex \Large \left(0,-\frac{5}{2}\right)] (note: the y-intercept will always have a x-coordinate equal to zero)


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So we have these intercepts:

x-intercept: *[Tex \Large \left(1,0\right)]


y-intercept: *[Tex \Large \left(0,-\frac{5}{2}\right)]




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c)


Now let's use the equation {{{y=(5/2)x-5/2}}} to graph





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



Since {{{b=-5/2}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-\frac{5}{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{5}{2}\right)] note: the point also looks like (0,-2.5)


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



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

{{{slope=rise/run}}}


Also, because the slope is {{{5/2}}}, this means:


{{{rise/run=5/2}}}



which shows us that the rise is 5 and the run is 2. This means that to go from point to point, we can go up 5  and over 2




So starting at *[Tex \LARGE \left(0,-\frac{5}{2}\right)], go up 5 units 

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


and to the right 2 units to get to the next point *[Tex \LARGE \left(2,\frac{5}{2}\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-5/2,.1)),
  blue(circle(0,-5/2,.12)),
  blue(circle(0,-5/2,.15)),
  blue(circle(2,5/2,.15,1.5)),
  blue(circle(2,5/2,.1,1.5)),
  blue(arc(0,-5/2+(5/2),2,5,90,270)),
  blue(arc((2/2),5/2,2,2, 180,360))
)}}}



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


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,(5/2)x-5/2),
  blue(circle(0,-5/2,.1)),
  blue(circle(0,-5/2,.12)),
  blue(circle(0,-5/2,.15)),
  blue(circle(2,5/2,.15,1.5)),
  blue(circle(2,5/2,.1,1.5)),
  blue(arc(0,-5/2+(5/2),2,5,90,270)),
  blue(arc((2/2),5/2,2,2, 180,360))
)}}} So this is the graph of {{{y=(5/2)x-5/2}}} through the points *[Tex \LARGE \left(0,-\frac{5}{2}\right)] and *[Tex \LARGE \left(2,\frac{5}{2}\right)]