Graph one inequality. First graph the boundary line using a table of values, intercepts, or any other method you prefer. The boundary line for [latex]x+y\geq1[/latex] is [latex]x+y=1[/latex], or [latex]y=−x+1[/latex]. Since the equal sign is included with the greater than sign, the boundary line is solid.

Find an ordered pair on either side of the boundary line. Insert the x and y-values into the inequality [latex]x+y\geq1[/latex] and see which ordered pair results in a true statement.

[latex]\begin{array}{r}\text{Test }1:\left(−3,0\right)\\x+y\geq1\\−3+0\geq1\\−3\geq1\\\text{FALSE}\end{array}[/latex]

[latex]\begin{array}{r}\text{Test }2:\left(4,1\right)\\x+y\geq1\\4+1\geq1\\5\geq1\\\text{TRUE}\end{array}[/latex]

Since [latex](4, 1)[/latex] results in a true statement, the region that includes [latex](4, 1)[/latex] should be shaded.

Do the same with the second inequality. Graph the boundary line, then test points to find which region is the solution to the inequality. In this case, the boundary line is [latex]y–x=5\left(\text{or }y=x+5\right)[/latex] and is solid. Test point [latex](−3, 0)[/latex] is not a solution of [latex]y–x\geq5[/latex] and test point [latex](0, 6)[/latex] is a solution.

The purple region in this graph shows the set of all solutions of the system.

Graph one inequality. First graph the boundary line, then test points.

Remember, because the inequality [latex]3x + 2y < 12[/latex] does not include the equal sign, draw a dashed border line. Testing a point like [latex](0, 0)[/latex] will show that the area below the line is the solution to this inequality.

The inequality [latex]-1 ≤ y ≤ 5[/latex] is actually two inequalities: [latex]−1 ≤ y[/latex], and [latex]y ≤ 5[/latex]. Another way to think of this is y must be between [latex]−1[/latex] and [latex]5[/latex]. The border lines for both are horizontal. The region between those two lines contains the solutions of [latex]-1 ≤ y ≤ 5[/latex]. We make the lines solid because we also want to include [latex]y = −1[/latex] and [latex]y = 5[/latex].

Graph this region on the same axes as the other inequality.

The purple region shows the set of all solutions of the system.

The boundary lines for this system are parallel to each other. Note how they have the same slopes.

[latex]\begin{array}{c}y=2x+1\\y=2x-3\end{array}[/latex]

Plotting the boundary lines will give the graph below. Note that the inequality [latex]y\lt2x-3[/latex] requires that we draw a dashed line, while the inequality [latex]y\ge2x+1[/latex] requires a solid line.

Now we need to shade the regions that represent the inequalities.  For the inequality [latex]y\ge2x+1[/latex], we can test a point on either side of the line to see which region to shade. Test [latex]\left(0,0\right)[/latex] to make it easy.

Substitute [latex]\left(0,0\right)[/latex] into [latex]y\ge2x+1[/latex]

[latex]\begin{array}{c}y\ge2x+1\\0\ge2\left(0\right)+1\\0\ge{1}\end{array}[/latex]

This is not true, so we know that we need to shade the other side of the boundary line for the inequality  [latex]y\ge2x+1[/latex]. The graph will now look like this:

Now shade the region that shows the solutions to the inequality [latex]y\lt2x-3[/latex].  Again, we can pick [latex]\left(0,0\right)[/latex] to test, because it makes easy algebra.

Substitute [latex]\left(0,0\right)[/latex] into [latex]y\lt2x-3[/latex]

[latex]\begin{array}{c}y\lt2x-3\\0\lt2\left(0,\right)x-3\\0\lt{-3}\end{array}[/latex]

This is not true, so we know that we need to shade the other side of the boundary line for the inequality [latex]y\lt2x-3[/latex]. The graph will now look like this:

This system of inequalities has no points in common so has no solution.

What would the graph look like if the system had looked like this?

[latex]\begin{array}{c}y\ge2x+1\\y\gt2x-3\end{array}[/latex].

Testing the point [latex]\left(0,0\right)[/latex] would return a positive result for the inequality [latex]y\gt2x-3[/latex], and the graph would then look like this:

The pink region is the region of overlap for both inequalities.