Questions on Algebra: Linear Equations, Graphs, Slope answered by real tutors!

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Question 170422: Plot the graph of the equations 4x - 6y = 2 and 2x - 3y = 1 and interpret the result.
I am having such a difficult time solving this equation. Can someone please help me understand how to do this problem, any help will be greatly appreciated.
: Plot the graph of the equations 4x - 6y = 2 and 2x - 3y = 1 and interpret the result.
I am having such a difficult time solving this equation. Can someone please help me understand how to do this problem, any help will be greatly appreciated.

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

Graphing the first equation




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




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




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


So we have the two points: and


Now plot these two points on a coordinate system

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



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,<BR>
graph(500,500,-8,12,-8,12,(2/3)x-1/3),<BR>
grid(1),<BR>
circle(5,9/3,0.1),<BR>
circle(5,9/3,0.12),<BR>
circle(5,9/3,0.15),<BR>
circle(-1,-3/3,0.1),<BR>
circle(-1,-3/3,0.12),<BR>
circle(-1,-3/3,0.15)<BR>
) Graph of y=(2/3)x-1/3 through the two points and






Graphing the second equation




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,<BR>
graph(500,500,-8,12,-8,12,(2/3)x-1/3),<BR>
grid(1)<BR>
)


Now graphing the two equations together gives you

drawing(500,500,-8,12,-8,12,<BR>
graph(500,500,-8,12,-8,12,(2/3)x-1/3),<BR>
grid(1)<BR>
)


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.