Question 170001
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Table of Contents:

<a href="#substitution">Substitution</a>
<a href="#elimination">Elimination</a>
<a href="#graphing">Graphing</a>




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Substitution:


Note: the first equation {{{2x+6=y}}} is the same as {{{y=2x+6}}}



{{{3x+4y=24}}} Start with the second equation




{{{3x+4(2x+6)=24}}} Plug in {{{y=2x+6}}}



{{{3x+8x+24=24}}} Distribute.



{{{11x+24=24}}} Combine like terms on the left side.



{{{11x=24-24}}} Subtract {{{24}}} from both sides.



{{{11x=0}}} Combine like terms on the right side.



{{{x=(0)/(11)}}} Divide both sides by {{{11}}} to isolate {{{x}}}.



{{{x=0}}} Reduce. So this is the first part of the answer.



{{{y=2x+6}}} Go back to the first equation



{{{y=2(0)+6}}} Plug in {{{x=0}}}.



{{{y=0+6}}} Multiply {{{2}}} and {{{0}}} to get {{{0}}}.



{{{y=6}}} Combine like terms. This is the second part of the answer.



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Answer:


So the solutions are {{{x=0}}} and {{{y=6}}}



Which forms the ordered pair (0,6)




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Elimination:



{{{2x+6=y}}} Start with the first equation



{{{2x=y-6}}} Subtract 6 from both sides



{{{2x-y=-6}}} Subtract "y" from both sides





Start with the given system of equations:

{{{system(2x-y=-6,3x+4y=24)}}}



{{{4(2x-y)=4(-6)}}} Multiply the both sides of the first equation by 4.



{{{8x-4y=-24}}} Distribute and multiply.



So we have the new system of equations:

{{{system(8x-4y=-24,3x+4y=24)}}}



Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:



{{{(8x-4y)+(3x+4y)=(-24)+(24)}}}



{{{(8x+3x)+(-4y+4y)=-24+24}}} Group like terms.



{{{11x+0y=0}}} Combine like terms. Notice how the y terms cancel out.



{{{11x=0}}} Simplify.



{{{x=(0)/(11)}}} Divide both sides by {{{11}}} to isolate {{{x}}}.



{{{x=0}}} Reduce.



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{{{8x-4y=-24}}} Now go back to the first equation.



{{{8(0)-4y=-24}}} Plug in {{{x=0}}}.



{{{0-4y=-24}}} Multiply.



{{{-4y=-24}}} Remove any zero terms.



{{{y=(-24)/(-4)}}} Divide both sides by {{{-4}}} to isolate {{{y}}}.



{{{y=6}}} Reduce.



So our answers are {{{x=0}}} and {{{y=6}}}.



Which form the ordered pair *[Tex \LARGE \left(0,6\right)]. Note: this is the same answer as before.



This means that the system is consistent and independent.



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Graphing:



{{{2x+6=y}}} Start with the first equation



{{{y=2x+6}}} Rearrange the equation





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 up 2  and over 1




So starting at *[Tex \LARGE \left(0,6\right)], go up 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(1,8\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(1,8,.15,1.5)),
  blue(circle(1,8,.1,1.5)),
  blue(arc(0,6+(2/2),2,2,90,270)),
  blue(arc((1/2),8,1,2, 180,360))
)}}}



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(1,8,.15,1.5)),
  blue(circle(1,8,.1,1.5)),
  blue(arc(0,6+(2/2),2,2,90,270)),
  blue(arc((1/2),8,1,2, 180,360))
)}}} So this is the graph of {{{y=2x+6}}} through the points *[Tex \LARGE \left(0,6\right)] and *[Tex \LARGE \left(1,8\right)]



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{{{3x+4y=24}}} Now move onto the second equation



{{{4y=24-3x}}} Subtract {{{3x}}} from both sides.



{{{4y=-3x+24}}} Rearrange the terms.



{{{y=(-3x+24)/(4)}}} Divide both sides by {{{4}}} to isolate y.



{{{y=((-3)/(4))x+(24)/(4)}}} Break up the fraction.



{{{y=-(3/4)x+6}}} Reduce.





Looking at {{{y=-(3/4)x+6}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-3/4}}} 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 {{{-3/4}}}, this means:


{{{rise/run=-3/4}}}



which shows us that the rise is -3 and the run is 4. This means that to go from point to point, we can go down 3  and over 4




So starting at *[Tex \LARGE \left(0,6\right)], go down 3 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+(-3/2),2,-3,90,270))
)}}}


and to the right 4 units to get to the next point *[Tex \LARGE \left(4,3\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(4,3,.15,1.5)),
  blue(circle(4,3,.1,1.5)),
  blue(arc(0,6+(-3/2),2,-3,90,270)),
  blue(arc((4/2),3,4,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-(3/4)x+6}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-(3/4)x+6),
  blue(circle(0,6,.1)),
  blue(circle(0,6,.12)),
  blue(circle(0,6,.15)),
  blue(circle(4,3,.15,1.5)),
  blue(circle(4,3,.1,1.5)),
  blue(arc(0,6+(-3/2),2,-3,90,270)),
  blue(arc((4/2),3,4,2, 0,180))
)}}} So this is the graph of {{{y=-(3/4)x+6}}} through the points *[Tex \LARGE \left(0,6\right)] and *[Tex \LARGE \left(4,3\right)]



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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+6,-(3/4)x+6),
  circle(0,6,.05),
  circle(0,6,.08),
  circle(0,6,.10)  
)}}} Graph of {{{y=2x+6}}} (red) and graph of {{{y=-(3/4)x+6}}} (green)



Notice how the two lines intersect at the point (0,6). So this means that the solution is {{{x=0}}} and {{{y=6}}} (which confirms our previous answers)