Let’s start this topic with a refresher from the past. When you first started learning Algebra you most likely started with simple equations like x + 1 = 3.  To solve the equation what you’re doing is finding the value of x (or the unknown) that makes the equation true.  Here, for example, x = 2 because 2 + 1 = 3. 

The next topic was likely solving inequalities like x + 1 > 3.  Here, there are multiple answers, infinitely many actually.  Still, when you’re solving what you are doing is finding the value(s) of x that make the statement true.  Here, x is any number larger than 2.  Since 2 + 1 = 3, x is greater than 2.  This is also when you most likely first graphed solutions. 

This is substantially different than x + 1 ≥ 3.  Here x could be two, and also anything larger than 2.  With x + 1 > 3, x could not be two because 2 + 1 > 3 is NOT a true statement.  When graphing each of these we had to use different symbols.  Remember, in this context, a graph can be thought of as a picture of all of the solutions.  The style varies regionally, but in many places an open circle is used for the strict inequality¸ which is > or <.  A solid dot is used for a ≥ or ≤.

These graphs have meaning and can be read.  Everywhere that is shaded (or covered by the line we drew in the direction of the solutions) is a solution.  Any number to the right of our mark is a solution.  Any number to the left of the mark is not a solution.  Graphing linear inequalities has the same concept at play.  Let’s see an example.

To the left is a graph of the inequality
y > ½x + 3.  Notice that it is shaded, above the dotted line.  The dotted line is similar to the open circle, it has the same meaning.  Both the dotted line and the open circle indicate that the value they contain is NOT included, but is instead a boundary.  You can get infinitely close to that value, but the value itself is not a solution. 

The shading above the line indicates all of the solutions.  With these equations and inequalities have two variables, an input and an output, unlike our original examples.   Here the solutions aren’t just one number, like x = 4, but a pair of numbers like x = 4 and y = 6, which we write as an ordered pair (4, 6). 

The coordinate (4, 6) is covered by the shaded region, so it is a solution to the inequality.  Just like in our original inequalities, x = 5 would be a solution because it was covered by the shading.  In the graph above, the ordered pair (0, 0) is NOT a solution because it is not in the shaded region.

No point on the dotted line is a solution.  The dotted line has the same meaning as an open circle.  It will be used with strict inequalities.  We use a solid line when graphing these inequalities just like we would use a solid circle, with a non-strict inequality.

How to Graph a Linear Inequalities

A good way of thinking of graphing inequalities is that we find the boundaries.  In this case, that is either the dotted or solid line.  Then, we figure out which side of the boundary contains the solutions. Those solutions get shaded. 

The boundary is just the line of the inequality, should it be an equation.  So, if we need to graph y  < 2x + 1, we first figure out where the graph of y = 2x + 1 lies.  Then, we determine if the line is solid or dotted.  Then, we figure out which side of the line contains the solutions.  There are short-cuts and tricks for this last part, but those are easily confused over time.  It is probably best to rely on your ability to read and think.  Let’s see an example.

Graph y ≤ 2x + 1

Step 1:  Graph the line. Since this is in slope-intercept form, plot the y – intercept (0, 1), and then count the slope (up 2, and to the right one). 

Step 2:  Determine if you will use a solid or dotted line.  Since this is a non-strict inequality (≤), we will use a solid line.  Connect the two points we found with a solid line.

Step 3:  Determine the direction of the shading. The inequality says, “y is less than or equal to 2x + 1.”  Less, for y is down.  So the shading will be below the line.  Another way to determine the direction is to pick one coordinate on each side of the line, and see which is a solution.

Note:  The point (0, 1) is where you’d start graphing this line, and the point (1, 3) is where you’d arrive if you counted the slope of 2/1.

The points (-2, 0) and (2, 0) are the points we will check.  While they look left and right of the line, think of them as above and below.  Let’s check each to see which is a solution and which is not.  The shading will be in the direction of the solutions.

Once you have determined where the solutions exist, shade that direction, as is done in the graph below.

All of the ways you know to graph a line are still valid.  Sometimes using slope-intercept form is best, sometimes a t-chart is easiest, sometimes you’ll want to find the x and y intercepts.

Let’s see one more example.

Graph 3x – 2y < 6

For the sake of clarity numbers that are easy to calculate have been chosen.  However, even if the numbers are a little tricky, the process is no different.  This is Standard Form of a linear equation (inequality in this case).  The easiest way to graph this equation is by finding the intercepts, or one intercept and another solution. 

To find the x – intercept, you replace y with zero and solve for x.

3x – 2(0) < 6

The only thing that’s different here is that we solve this inequality as though it was an equation because we are looking for a specific point.  Whether the line that passes through the points is solid or dotted will address the inequality, when we get to that point.

3x – 2(0) = 6

x = 2

x – intercept: (2, 0)

The y – intercept is the same way, except replacing x with zero.

3(0) – 2y = 6

y = 3

y – intercept: (0, 3)

The line passing through these intercepts will be dotted because we have a strict inequality.  Once you draw that line, pick a coordinate on each side of the line, but pick one that is easy to calculate.  A great idea is to use the origin if possible because arithmetic with zeros is easy.

Here we chose the origin and (5, 0).  Plug the coordinates in, separately, and see which is the solution.  If (5, 0) is a solution, the shading will be in that direction from the line.  If the origin is a solution, the shading will be that direction. 

3(0) + 2(0) < 6 is a solution, so the shading will be under the line, or to the left of the line.