Question 170422
<h3>Graphing the first equation</h3>



In order to graph {{{4x-6y=2}}}, we need to solve for "y" first.



{{{4x-6y=2}}} Start with the first equation.



{{{-6y=2-4x}}} Subtract {{{4x}}} from both sides.



{{{-6y=-4x+2}}} Rearrange the terms.



{{{y=(-4x+2)/(-6)}}} Divide both sides by {{{-6}}} to isolate y.



{{{y=((-4)/(-6))x+(2)/(-6)}}} Break up the fraction.



{{{y=(2/3)x-1/3}}} Reduce.





In order to graph this equation, we only need two points to create a straight line





--------------------------------Let's find the first point--------------------------------


{{{y=(2/3)x-1/3}}} Start with the given equation





{{{y=(2/3)(5)-1/3}}} Plug in {{{x=5}}} 





{{{y=10/3-1/3}}} Multiply {{{2/3}}} and {{{5}}} to get {{{10/3}}}





{{{y=9/3}}} Subtract 




So when {{{x=5}}}, we have the value {{{y=3}}}. This means we have the first point *[Tex \LARGE \left(5,3\right)]





--------------------------------Let's find the second point--------------------------------


{{{y=(2/3)x-1/3}}} Start with the given equation





{{{y=(2/3)(-1)-1/3}}} Plug in {{{x=-1}}} 





{{{y=-2/3-1/3}}} Multiply {{{2/3}}} and {{{-1}}} to get {{{-2/3}}}





{{{y=-1}}} Subtract




So when {{{x=-1}}}, we have the value {{{y=-1}}}. This means we have the second point *[Tex \LARGE \left(-1,-1\right)]





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



So we have the two points: *[Tex \LARGE \left(5,\frac{9}{3}\right)] and *[Tex \LARGE \left(-1,-\frac{3}{3}\right)]



Now plot these two points on a coordinate system


{{{drawing(500,500,-8,12,-8,12,
graph(500,500,-8,12,-8,12,0),
grid(1),
circle(5,9/3,0.1),
circle(5,9/3,0.12),
circle(5,9/3,0.15),
circle(-1,-3/3,0.1),
circle(-1,-3/3,0.12),
circle(-1,-3/3,0.15)
) }}}




Now draw a straight line through the two points. This line is the graph of {{{y=(2/3)x-1/3}}}


{{{drawing(500,500,-8,12,-8,12,
graph(500,500,-8,12,-8,12,(2/3)x-1/3),
grid(1),
circle(5,9/3,0.1),
circle(5,9/3,0.12),
circle(5,9/3,0.15),
circle(-1,-3/3,0.1),
circle(-1,-3/3,0.12),
circle(-1,-3/3,0.15)
) }}} Graph of {{{y=(2/3)x-1/3}}} through the two points *[Tex \LARGE \left(5,3\right)] and *[Tex \LARGE \left(-1,-1\right)]



<hr>



<h3>Graphing the second equation</h3>



{{{2x-3y=1}}} Start with the second equation.



{{{-3y=1-2x}}} Subtract {{{2x}}} from both sides.



{{{-3y=-2x+1}}} Rearrange the terms.



{{{y=(-2x+1)/(-3)}}} Divide both sides by {{{-3}}} to isolate y.



{{{y=((-2)/(-3))x+(1)/(-3)}}} Break up the fraction.



{{{y=(2/3)x-1/3}}} Reduce.



Take note that this equation is EXACTLY identical to the equation we just plotted. So the graph of {{{y=(2/3)x-1/3}}} (and {{{2x-3y=1}}}) is 




{{{drawing(500,500,-8,12,-8,12,
graph(500,500,-8,12,-8,12,(2/3)x-1/3),
grid(1)
) }}} 



Now graphing the two equations together gives you


{{{drawing(500,500,-8,12,-8,12,
graph(500,500,-8,12,-8,12,(2/3)x-1/3),
grid(1)
) }}} 



Note: there are really two equations here. One is just right on top of the other (which hides it)



Since one equation is right on top of the other, this tells us that there are an infinite number of solutions (since there are an infinite number of intersections).



So the system is dependent.