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Question 170001: This is a systems of equations question.
If the equation is:
2x+6=y
3x+4y=24
__________
Please solve by substitution, elimination, AND graphing. ( the same equation)
Thank you so very much.
: This is a systems of equations question.
If the equation is:
2x+6=y
3x+4y=24
__________
Please solve by substitution, elimination, AND graphing. ( the same equation)
Thank you so very much.

Answer by jim_thompson5910(9911) About Me  (Show Source):
You can put this solution on YOUR website!


Table of Contents:
Substitution
Elimination
Graphing




Jump to Top
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.


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

Answer:

So the solutions are x=0 and y=6


Which forms the ordered pair (0,6)







Jump to Top
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.


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


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 . Note: this is the same answer as before.


This means that the system is consistent and independent.









Jump to Top

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 .Remember the y-intercept is the point where the graph intersects with the y-axis

So we have one point

drawing(500,500,-10,10,-10,10,<BR>
  grid(1),<BR>
  blue(circle(0,6,.1)),<BR>
  blue(circle(0,6,.12)),<BR>
  blue(circle(0,6,.15))<BR>
)


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 , go up 2 units
drawing(500,500,-10,10,-10,10,<BR>
  grid(1),<BR>
  blue(circle(0,6,.1)),<BR>
  blue(circle(0,6,.12)),<BR>
  blue(circle(0,6,.15)),<BR>
  blue(arc(0,6+(2/2),2,2,90,270))<BR>
)

and to the right 1 unit to get to the next point
drawing(500,500,-10,10,-10,10,<BR>
  grid(1),<BR>
  blue(circle(0,6,.1)),<BR>
  blue(circle(0,6,.12)),<BR>
  blue(circle(0,6,.15)),<BR>
  blue(circle(1,8,.15,1.5)),<BR>
  blue(circle(1,8,.1,1.5)),<BR>
  blue(arc(0,6+(2/2),2,2,90,270)),<BR>
  blue(arc((1/2),8,1,2, 180,360))<BR>
)


Now draw a line through these points to graph y=2x+6

drawing(500,500,-10,10,-10,10,<BR>
  grid(1),<BR>
  graph(500,500,-10,10,-10,10,2x+6),<BR>
  blue(circle(0,6,.1)),<BR>
  blue(circle(0,6,.12)),<BR>
  blue(circle(0,6,.15)),<BR>
  blue(circle(1,8,.15,1.5)),<BR>
  blue(circle(1,8,.1,1.5)),<BR>
  blue(arc(0,6+(2/2),2,2,90,270)),<BR>
  blue(arc((1/2),8,1,2, 180,360))<BR>
) So this is the graph of y=2x+6 through the points and


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


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 .Remember the y-intercept is the point where the graph intersects with the y-axis

So we have one point

drawing(500,500,-10,10,-10,10,<BR>
  grid(1),<BR>
  blue(circle(0,6,.1)),<BR>
  blue(circle(0,6,.12)),<BR>
  blue(circle(0,6,.15))<BR>
)


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 , go down 3 units
drawing(500,500,-10,10,-10,10,<BR>
  grid(1),<BR>
  blue(circle(0,6,.1)),<BR>
  blue(circle(0,6,.12)),<BR>
  blue(circle(0,6,.15)),<BR>
  blue(arc(0,6+(-3/2),2,-3,90,270))<BR>
)

and to the right 4 units to get to the next point
drawing(500,500,-10,10,-10,10,<BR>
  grid(1),<BR>
  blue(circle(0,6,.1)),<BR>
  blue(circle(0,6,.12)),<BR>
  blue(circle(0,6,.15)),<BR>
  blue(circle(4,3,.15,1.5)),<BR>
  blue(circle(4,3,.1,1.5)),<BR>
  blue(arc(0,6+(-3/2),2,-3,90,270)),<BR>
  blue(arc((4/2),3,4,2, 0,180))<BR>
)


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

drawing(500,500,-10,10,-10,10,<BR>
  grid(1),<BR>
  graph(500,500,-10,10,-10,10,-(3/4)x+6),<BR>
  blue(circle(0,6,.1)),<BR>
  blue(circle(0,6,.12)),<BR>
  blue(circle(0,6,.15)),<BR>
  blue(circle(4,3,.15,1.5)),<BR>
  blue(circle(4,3,.1,1.5)),<BR>
  blue(arc(0,6+(-3/2),2,-3,90,270)),<BR>
  blue(arc((4/2),3,4,2, 0,180))<BR>
) So this is the graph of y=-(3/4)x+6 through the points and


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


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



drawing(500,500,-10,10,-10,10,<BR>
  grid(1),<BR>
  graph(500,500,-10,10,-10,10,2x+6,-(3/4)x+6),<BR>
  circle(0,6,.05),<BR>
  circle(0,6,.08),<BR>
  circle(0,6,.10)  <BR>
) 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)