Question 170807
# 2





Start with the given system of equations:



{{{system(4x-6y=2,2x-3y=1)}}}



In order to graph these equations, we <font size="4"><b>must</b></font> solve for y first.



Let's graph the first equation:



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



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



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



{{{y=(2/3)x-1/3}}} Rearrange the terms and simplify.



Now let's graph the equation:



{{{drawing(500,500,-10,10,-10,10,
grid(0),
graph(500,500,-10,10,-10,10,(2/3)x-1/3)
)}}} Graph of {{{y=(2/3)x-1/3}}}.



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



Now let's graph the second equation:



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



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



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



{{{y=(2/3)x-1/3}}} Rearrange the terms and simplify.



Now let's graph the equation:



{{{drawing(500,500,-10,10,-10,10,
grid(0),
graph(500,500,-10,10,-10,10,(2/3)x-1/3)
)}}} Graph of {{{y=(2/3)x-1/3}}}.



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



Now let's graph the two equations together:



{{{drawing(500,500,-10,10,-10,10,
grid(1),
graph(500,500,-10,10,-10,10,(2/3)x-1/3,(2/3)x-1/3)
)}}} Graph of {{{y=(2/3)x-1/3}}} (red). Graph of {{{y=(2/3)x-1/3}}} (green)



From the graph, we can see that one line is right on top of the other one, which means that they intersect an infinite number of times. So there are an infinite number of solutions. This means that the system of equations is consistent and dependent.



<hr>



# 3





Start with the given system of equations:



{{{system(10x-4y=3,5x-2y=6)}}}



In order to graph these equations, we <font size="4"><b>must</b></font> solve for y first.



Let's graph the first equation:



{{{10x-4y=3}}} Start with the first equation.



{{{-4y=3-10x}}} Subtract {{{10x}}} from both sides.



{{{y=(3-10x)/(-4)}}} Divide both sides by {{{-4}}} to isolate {{{y}}}.



{{{y=(5/2)x-3/4}}} Rearrange the terms and simplify.



Now let's graph the equation:



{{{drawing(500,500,-10,10,-10,10,
grid(0),
graph(500,500,-10,10,-10,10,(5/2)x-3/4)
)}}} Graph of {{{y=(5/2)x-3/4}}}.



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



Now let's graph the second equation:



{{{5x-2y=6}}} Start with the second equation.



{{{-2y=6-5x}}} Subtract {{{5x}}} from both sides.



{{{y=(6-5x)/(-2)}}} Divide both sides by {{{-2}}} to isolate {{{y}}}.



{{{y=(5/2)x-3}}} Rearrange the terms and simplify.



Now let's graph the equation:



{{{drawing(500,500,-10,10,-10,10,
grid(0),
graph(500,500,-10,10,-10,10,(5/2)x-3)
)}}} Graph of {{{y=(5/2)x-3}}}.



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



Now let's graph the two equations together:



{{{drawing(500,500,-10,10,-10,10,
grid(1),
graph(500,500,-10,10,-10,10,(5/2)x-3/4,(5/2)x-3)
)}}} Graph of {{{y=(5/2)x-3/4}}} (red). Graph of {{{y=(5/2)x-3}}} (green)



From the graph, we can see that the two lines are parallel, which means that they will <font size="4"><b>NEVER</b></font> intersect. So there are no solutions. This means that the system of equations is inconsistent.