How do you graph #2x + y = 3# by plotting points?

1 Answer
Mar 25, 2017

Answer:

Find a few #(x,y)# pairs that satisfy the equation; plot these points; connect them.

Explanation:

Pick some values for #x#, and use the equation to find what #y# must be for each of those #x#'s.

For example, if #x# is 0, then we have:

#"         "2x+y=3#
#=>2(0)+y=3#
#=>"     "0+y=3#
#=>"            "y=3#

So when #x=0#, we have #y=3#. Meaning the point #(x,y)=(0,3)# is a solution to #2x+y=3#, and so our graph of the equation will pass through #(0,3)#.

We can get a few more points:

If #x=1#, then

#"         "2x+y=3#
#=>2(1)+y=3#
#=>"     "2+y=3#
#=>"            "y=1#

So #(x,y)=(1,1)# is also on our graph.

Similarly, if #x=2#, then we get #y="–"1#, giving us the point #(x,y)=(2,"–"1)#.

We then take the points we've computed, #(0,3), (1,1), (2,"–"1)#, and plot them on a graph:

graph{((x)^2+(y-3)^2-.02)((x-1)^2+(y-1)^2-.02)((x-2)^2+(y+1)^2-.02)=0 [-10, 10, -5, 5]}

Last, since we know this equation is linear, we just need to connect the dots with a line:

graph{((x)^2+(y-3)^2-.02)((x-1)^2+(y-1)^2-.02)((x-2)^2+(y+1)^2-.02)(2x+y-3)=0 [-10, 10, -5, 5]}

And we're done!