Question 657547
Since there is an {{{x}}} and a {{{y}}} in your system, you need an x-axis and a y-axis in your graph.
Each inequality will be represented as the half of the x-y plane to one side of a boundary line.
 
The equation {{{x+y=2}}} represents one of those boundary lines, and is graphed as a solid line, because all the points on that line are part of the solution to 
{{{x+y>=2}}}. Any point (x,y) in that line satisfies {{{x+y=2}}} and also satisfies {{{x+y>=2}}}.
{{{graph(300,300,-2,8,-2,8,2-x)}}} is the graph (line) for {{{x+y=2}}}, and the inequality {{{x+y>=2}}} graphs as {{{graph(300,300,-2,8,-2,8,2-x,x+y>=2)}}}
 
The equation {{{y=2x-1}}} represents the boundary line for {{{y<2x-1}}}, and is graphed as a dashed line, because the points (x,y) with {{{y=2x-1}}} do not satisfy {{{y<2x-1}}}
 
To graph each line you need 2 points.
To get a point you chose a value for {{{x}}} and find the corresponding {{{y}}} or the other way around. A value of zero is often a good choice.
 
FOR {{{x+y=2}}}:
{{{x=0}}} gives you {{{0+y=2}}} --> {{{y=2}}} for the point (0,2).
{{{y=0}}} gives you {{{x+0=2}}} --> {{{x=2}}} for the point (2,0).
You plot the two points and connect them with a line to graph {{{x+y=2}}}
{{{drawing(300,300,-2,8,-2,8,
grid(1),
blue(circle(2,0,0.2)),blue(circle(0,2,0.2)),
blue(line(4,-2,-2,4))
)}}}
 
FOR {{{x+y>=2}}}:
You graph the line as shown above. It is a solid line because there is a "=" in the inequality, and that causes the solutions to {{{x+y=2}}} to be solutions of {{{x+y>=2}}}.
Next, you figure out which side of the line to shade or color.
An easy way to do it is to pick a point on one side of the line and check if it is part of the solution.
The easiest choice (if not on the line) is the point (0,0), with {{{x=y=0}}}
For that point {{{x+y=0+0<2}}}, so that point is not part of the solution.
 
FOR {{{y<2x-1}}}:
You plot the line for {{{y=2x-1}}} as a dashed line, because it is not part of the solution.
{{{x=0}}} gives you {{{y=2*0-1}}}--> {{{y=-1}}} for point (0,-1).
Choosing {{{y=0}}} would not make the calculations or the graphing easy, so I pick {{{x=3}}} for the second point.
{{{x=3}}} gives me {{{y=2*3-1}}} --> {{{y=5}}} for point (3,5).
I plot the points and connect them with a dashed line:
{{{drawing(300,300,-2,8,-2,8,
grid(1),
green(circle(0,-1,0.2)),green(circle(3,5,0.2)),
green(line(-0.5,-2,-0.25,-1.5)),green(line(0,-1,0.25,-0.5)),
green(line(0.5,0,0.75,0.5)),green(line(1,1,1.25,1.5)),
green(line(1.5,2,1.75,2.5)),green(line(2,3,2.25,3.5)),
green(line(2.5,4,2.75,4.5)),green(line(3,5,3.25,5.5)),
green(line(3.5,6,3.75,6.5)),green(line(4,7,4.25,7.5))
)}}} Finally, I decide which side of the line is the solution to {{{y<2x-1}}}.
Using (0,0), the origin, as a test point, I figure that the solution is the side that does not contain the origin,
because substituting {{{x=y=0}}} into {{{y<2x-1}}}, I get
{{{0<2*0-1=-1}}}, which is not true.
The colored graph of the inequality will look like this:
{{{graph(300,300,-2,8,-2,8,y<2x-1)}}}
 
The solution to the system is the part of the x-y plane that is a solution to both inequalities. It graphs as the colored or shaded pie wedge bounded by the lines, as shown below.
{{{drawing(300,300,-2,8,-2,8,
grid(1),
blue(line(1,1,4,-2)),green(line(1,1,1.25,1.5)),
green(line(1.5,2,1.75,2.5)),green(line(2,3,2.25,3.5)),
green(line(2.5,4,2.75,4.5)),green(line(3,5,3.25,5.5)),
green(line(3.5,6,3.75,6.5)),green(line(4,7,4.25,7.5)),
red(line(4.4,7.8,8,7.8)),red(line(4.3,7.6,8,7.6)),
red(line(4.2,7.4,8,7.4)),red(line(4.1,7.2,8,7.2)),
red(line(4,7,8,7)),red(line(3.9,6.8,8,6.8)),
red(line(3.8,6.6,8,6.6)),red(line(3.7,6.4,8,6.4)),
red(line(3.6,6.2,8,6.2)),red(line(3.5,6,8,6)),
red(line(3.4,5.8,8,5.8)),red(line(3.3,5.6,8,5.6)),
red(line(3.2,5.4,8,5.4)),red(line(3.1,5.2,8,5.2)),
red(line(3,5,8,5)),red(line(2.9,4.8,8,4.8)),
red(line(2.8,4.6,8,4.6)),red(line(2.7,4.4,8,4.4)),
red(line(2.6,4.2,8,4.2)),red(line(2.5,4,8,4)),
red(line(2.4,3.8,8,3.8)),red(line(2.3,3.6,8,3.6)),
red(line(2.2,3.4,8,3.4)),red(line(2.1,3.2,8,3.2)),
red(line(2,3,8,3)),red(line(1.9,2.8,8,2.8)),
red(line(1.8,2.6,8,2.6)),red(line(1.7,2.4,8,2.4)),
red(line(1.6,2.2,8,2.2)),red(line(1.5,2,8,2)),
red(line(1.4,1.8,8,1.8)),red(line(1.3,1.6,8,1.6)),
red(line(1.2,1.4,8,1.4)),red(line(1.1,1.2,8,1.2)),
red(line(1,1,8,1)),red(line(1.2,0.8,8,0.8)),
red(line(1.4,0.6,8,0.6)),red(line(1.6,0.4,8,0.4)),
red(line(1.8,0.2,8,0.2)),red(line(2,0,8,0)),
red(line(2.2,-0.2,8,-0.2)),red(line(2.4,-0.4,8,-0.4)),
red(line(2.6,-0.6,8,-0.6)),red(line(2.8,-0.8,8,-0.8)),
red(line(3,-1,8,-1)),red(line(3.2,-1.2,8,-1.2)),
red(line(3.4,-1.4,8,-1.4)),red(line(3.6,-1.6,8,-1.6)),
red(line(3.8,-1.8,8,-1.8))
)}}}