Question 700783
ONE WAY TO FIND {{{y}}}:
{{{2x-y=1}}} --> {{{2x-y-1=1-1}}} --> {{{2x-y-1=0}}} --> {{{2x-y-1+y=0+y}}} --> {{{2x-1=y}}} --> {{{highlight(y=2x-1)}}}
ANOTHER WAY:
{{{2x-y=1}}} --> {{{2x-y-2x=1-2x}}} --> {{{-y=1-2x}}} --> {{{(-1)(-y)=(-1)(1-2x)}}} --> {{{y=-1+2x}}} --> {{{highlight(y=2x-1)}}}
Using {{{y=2x-1}}} you can calculate the coordinates of two points on the line,
and then you would just plot the points and connect them with a straight line.
 
However, you can graph {{{2x-y=1}}} without transforming the equation into {{{y=2x-1}}}.
To get the coordinates of a point on the line,
you can give a value to {{{x}}} (or to {{{y}}}),
and solve for the other variable.
For {{{x=0}}}, substituting into {{{2x-y=1}}} we get
{{{2*0-y=1}}} --> {{{0-y=1}}} --> {{{-y=1}}} --> {{{y=-1}}}.
That gives us the point (0,-1), with {{{x=0}}} and {{{y=-1}}}.
For {{{x=3}}}, we get
{{{2*3-y=1}}} --> {{{6-y=1}}} --> {{{-y=1-6}}} --> {{{-y=-5}}} --> {{{y=5}}}
That gives us the point (3,5).
Plotting (0,-1) and (3, 5} we get {{{drawing(300,300,-5,5,-4,6,
grid(1),
blue(circle(0,-1,0.2)),blue(circle(3,5,0.2))
)}}}
Adding the line, we get
{{{drawing(300,300,-5,5,-4,6,
grid(1),blue(line(-2,-5,4,7)),
blue(circle(0,-1,0.2)),blue(circle(3,5,0.2))
)}}}
 
On the line for {{{y=-3}}}, all the points have {{{y=-3}}}, for any and all values of {{{x}}}.
Adding that line (in green), we get:
{{{drawing(300,300,-5,5,-4,6,
grid(1),blue(line(-2,-5,4,7)),
blue(circle(0,-1,0.2)),blue(circle(3,5,0.2)),
green(line(-5,-3,5,-3))
)}}}