Question 120588
Use the same procedure to graph any linear equation.  I'll do one of them.


Select two values for x.  It doesn't really matter which values you select, but I like small, easy to manipulate numbers like 0 and 1.  Since these are linear equations, i.e. you get a straight line when you graph them, you only need to determine two points.


Once you have chosen your x values, evaluate the equation for each by substituting each of your chosen values for x and solving for y.


In your first problem, let x = 0.  {{{y=(1/3)(0)+1=1}}}.  So if x is 0, y is 1, and one of the points on our line is (0,1).  Now let x = 3.  (I chose three because 3 is the denominator of the coefficient fraction)  {{{y=(1/3)(3)+1=1+1=2}}}.  So if x is 3, y is 2, and another point on our line is (3,2).


Plot these two points on your graph.  I've labeled them A and B


{{{drawing(600,600,-6,6,-6,6,
grid(1),
blue(
circle(0,1,.05),
locate(0.15,.85,A(0,1))),
green(
circle(3,2,.05),
locate(3.15,1.85,B(3,2)))
)}}} 


Now construct a straight line through the two points with a straight edge and you are done.


{{{drawing(600,600,-6,6,-6,6,
grid(1),
blue(
circle(0,1,.05),
locate(0.15,.85,A(0,1))),
green(
circle(3,2,.05),
locate(3.15,1.85,B(3,2))),
red(
line(-6,-1,6,3))
)}}}


All you have to do is follow the same procedure for each of the other three problems.


John