Algebra 1/Unit 8: Graphing Inequalities

Welcome to Week 8 of the Algebra 1 course! This unit builds on our previous work with inequalities from Unit 7. For this unit, we will focus on:

Graphing solutions of linear inequalities in one variable on a number line (visualizing the range of solutions and using proper notation).
Graphing compound inequalities (“and” / “or” inequalities) on a number line and understanding their solution sets.
Graphing linear inequalities in two variables on the coordinate plane (drawing boundary lines, using solid/dashed lines, and shading solution regions).
Understanding different ways to denote solution sets, including inequality notation vs interval notation, and how graphs help visualize these solutions.

Introduction

In previous lessons, we solved inequalities algebraically to find their solution sets. Unlike linear equations (which might have a single solution like $x=3$ ), an inequality usually has infinitely many solutions, which forms a range or interval of values. Graphing inequalities is important because it gives us a visual picture for all of these solutions, making it easier to understand which values satisfy the inequality. By representing solutions on a number line or a coordinate plane, we visualize the solution set and are able to see at a glance the extent of values that are included and excluded from the solution. This is especially helpful in real-life contexts (like budget limits, temperature ranges, etc.), where a range of values satisfies a condition rather than one exact number.

Recall: There are multiple ways to represent the solution set of an inequality. For example, consider the inequality $x<-3$ :

Inequality notation: $x<-3$ (just writing the inequality as is)
Set-builder notation: { $x|x<-3$ } (the set of all real numbers $x$ such that $x$ is less than −3).
Interval notation: (−∞,−3) (all real numbers from −∞ up to −3, not including −3), remember if you have a closed bracket ([ or ]) next to a number, that means that is included in the solution set while open parentheses ( ) are not included/excluded from the solution set.
Graphical representation: a number line with an open point at −3 and shading to the left (for $x<-3$ )

Each notation conveys the same idea, just in a different style and format. In this unit, we will emphasize the graphical representation of inequalities and interval notation.

Graphing One Variable Linear Inequalities

When we have a linear inequality in one variable (such as $x>5$ or $y\leq 3$ ), we can represent its solutions on a number line. The number line graph gives a clear visual of all the real numbers that satisfy the inequality:

Endpoint: First identify the critical value (the number on the right side of the inequality). Mark this point on the number line.
Open vs. Closed Circle: Use an open circle if the inequality is strict ( $<$ or $>$ ), meaning the endpoint itself is not included in the solution. Use a closed (filled-in) circle if the inequality is inclusive ( $\leq$ or $\geq$ ), meaning the endpoint is included in the solution. An open circle indicates “not including this value,” while a closed circle indicates “including this value.” For example, $x>5$ requires an open circle at 5 (since 5 is not a solution), whereas $x\geq 5$ would use a closed dot at 5 (since 5 is included).
Shading/Arrow: Draw a line or arrow to indicate all the numbers that satisfy the inequality. For $x>5$ , we shade the line to the right of 5 (toward larger numbers); for $x<5$ , we shade to the left (toward smaller numbers). The arrow shows that the solution set extends indefinitely in that direction.

Inequality vs Interval Notation: The graph nicely complements algebraic notation. For

Number Line Graph of x > 5

instance, if we have $x>5$ , we graph an open circle at 5 with a ray extending to the right (open image to the right in another window if you want a better look at it). In inequality notation we write $x>5$ , and in interval notation we write (5,∞) to represent the same set of values. Similarly, $x\leq -2$ would be a closed dot at −2 with a ray to the left, corresponding to (−∞,−2] in interval notation. It’s important to be comfortable moving between these notations.

Example 1: Graph a Simple Inequality

Problem: Graph the solution of the inequality $x\geq -2$ on a number line, and write the solution set in interval notation.

Solution:

Identify the endpoint and type of inequality: The critical value here is 2. The inequality is $x\geq -2$ , which is “greater than or equal to 2.” Because of the “equal to” part ( $\geq$ ), 2 is included in the solution set.
Mark the endpoint: Draw a number line and locate 2 on it. Mark 2 with a closed circle (since x can equal 2, we include this point). thirdspacelearning.com explains that a closed circle shows the value is included in the solution set.
Determine which direction to shade: $x\geq 2$ means all numbers greater than or equal to 2. These are numbers to the right of 2 on the number line (including 2 itself). Draw a solid line or arrow extending to the right from the point at 2 to indicate all those larger values.
Interval notation: Because 2 is the smallest value in the solution (and it is included), and the solutions extend without bound to the right, we write the interval as [2,∞). The “[” bracket at 2 means 2 is included (matching the closed dot), and ∞ always gets a parenthesis since infinity isn’t a number we reach.

Number line graph of inequality x >= 2

So, graphically we have a filled dot at 2 and a ray going right. In inequality notation the solution is $x\geq 2$ , and in interval notation it’s [2,∞).

Example 2: Graphing a Negative Inequality

Problem: Graph the inequality $x<-4$ on a number line and express the solution in interval notation.

Solution:

The critical value/point of emphasis here is $-4$ . Since the inequality symbol is the less than sign: $<$ , that means we have to consider solutions that are strictly less than -4. In other words, $-4$ is not included in the solution.
Draw a number line and mark $-4$ with an open circle (since $x$ cannot equal $-4$ ). This signifies that $-4$ is a boundary and not part of the solution (always remember this if a value has an open circle on top of it)
Since $x<-4$ means all values less than $-4$ , we will have to shade the line to the left of $-4$ with an arrow point leftwards to show that the solution continues indefinitely.
In interval notation, this solution set is $(-\infty ,-4)$ . Notice that we use an open parenthesis to correspond to the open circle for $-4$ (to indicate it is not included in the solution set) and we also use the open parenthesis for $-\infty$ to show that it is not included.

Number line Graph of Inequality x < -4

We read this number line as "all $x$ such that $x$ is less than $-4$ ." The number line/graph on the right shows this by everything to the left of $-4$ being shaded.

Graphing Compound Inequalities on a Number Line

Now that you have gotten a taste of what its like to graph linear inequalities on a number line, let's move on to the next step: graphing compound inequalities. Graphing these types of inequalities is relatively the same as graphing linear inequalities, but typically within a specific bound of numbers. In its essence, a compound inequality combines two conditions on the the variable, connected by the keywords "and" or "or." More specifically:

An “and” compound inequality means both conditions must be true at the same time. The solution is the intersection of the two solution sets (the overlap of values that satisfy both). Often “and” inequalities can be written in a simpler, compact form. For example,

Inequality of 1 < x <= 5
$1<x\leq 5$ means “ $x>1$ and $x\leq 5$ ” simultaneously. The solution set is all $x$ values between 1 and 5, and we graph it as a segment between 1 and 5 on the number line, instead of having arrows pointing to one or both directions. See the image above to see how it is graphed. Notice how 1 has an open circle above it to signify that it is not included in the solution set while the 5 has a closed circled above it to show that it is included in the solution set for the inequality $1<x\leq 5$ . See? It's pretty simple once you get the hang of it.
An “or” compound inequality means either condition (or inequalities in this case) can be true. The solution is the union of the two solution sets (any value that satisfies at least one of the inequalities). For example, $x<1$ or $x>4$ means the solution is any number that is either less than 1 or greater than 4 (this will produce two separate rays/arrows going opposite directions on the number line). Take a look at the graphical representation of the inequality on the number line below to get an idea of what its like to graph "or" compound inequalities. Also notice the type of circle that is placed above the critical values.

Here are some practical steps to help you graph both "and" and "or" compound inequalities:

Graphing "AND" Inequalities:

Solve/understand each part: If the inequality is not already in a combined form, consider each inequality separately. For example, suppose we want to graph $-1<x\leq 4$ . This can be understood as two conditions: $-1<x$ and $x\leq 4$ .
Graph each condition: Graphing $-1<x$ is the same as saying $x>-1$ , so we would put an open circle at $-1$ (as it is not included with the strict less than/greater than symbol) and shade to the right. $x\leq 4$ means we would put a closed circle at $4$ (as it is included in the solution set with the less than or equal to sign) and shade to the left.
Find the overlap: The “and” requires both conditions at once, so the solution is the overlapping region. In this example, the overlap is all $x$ values between $-1$ and $4$ . On the number line, that’s the segment starting just above $-1$ and going up to $4$ . We would draw a heavy line from $-1$ to $4$ . Remember to use an open circle at $-1$ and a closed circle at $4$ based on the reasoning in the last step. This segment represents all numbers $x$ such that $-1<x\leq 4$ .
(Extra) Notation: In inequality form we might leave the solution as $-1<x\leq 4$ . If you wanted to express the solution in interval notation, however, it would be $(-1,4]$ . Notice the parenthesis at $-1$ (corresponding to $-1<x$ or $x>-1$ depending on how you want to view it), to show that $-1$ is not included and the bracket at $4$ (corresponding to $x\leq 4$ ), to show that $4$ is included.

Heres the complete number line graph representation of the inequality:

Compound Inequality of -1 < x <= 4

Graphing "OR" Inequalities:

Graph each inequality separately: For example, consider $x\leq -3$ or $x>2$ . Graph $x\leq -3$ on a number line (closed circle/dot at $-3$ , shading to the left) and $x>2$ (open circle/dot at $2$ , shading to the right).
Combine the graphs: Because it’s an “or,” compound inequality, any point that satisfies either condition is part of the solution. This means we take all the shaded parts from both graphs together. In our example, that results in two distinct regions: one ray/arrow extending left from $-3$ , and another ray/arrow extending to the right from $2$ . There will be a gap between $-3$ and $2$ that is not shaded (since numbers in between these two values don’t satisfy either inequality, and hence are not solutions).
Check endpoints: At $-3$ we have a closed circle (because of “ $\leq -3$ ”) and at $2$ we have an open circle (because of “ $>2$ ”).
(Extra) Notation: We cannot write “or” solutions as a single chained inequality. Instead we use the word “or” or use set notation/union. In interval notation, we write this solution as ${\bigl (}-\infty ,-3]\cup {\bigl (}2,\infty {\bigr )}$ . This denotes the union of two intervals: all numbers $\leq -3$ or $x>2$ . (Using ${\bigl (}-\infty ,-3]$ for the left part, and ${\bigl (}2,\infty {\bigr )}$ for the right part.)

Heres the visual representation of graphing this or compound inequality on a number line:

Compound Inequality of x <= -3 or x > 2

Now that you have gotten a hang of what the number line representation of these compound inequalities should look like, feel free to practice drawing/graphing the number lines of both linear inequalities and compound inequalities in one variable, based on the sample problems below.

Practice Problems (Graphing Inequalities on a Number Line); Linear + Compound

Graphing Linear Inequalities in Two Variables

Now we move to linear inequalities in two variables (like $y>2x+1$ or $x+y\leq 4$ ). These inequalities represent a region of the coordinate plane, rather than just a segment on a line. Graphing them helps us see all solution pairs (x,y) that satisfy the inequality.

A linear inequality in two variables can be written in forms like $y>mx+b$ , $y\geq mx+b$ , $Ax+By<C$ , etc. For example, $y>2x+1$ or $x+y\leq 4$ . To graph such inequalities, follow these steps:

Rewrite in a convenient form (optional): It often helps to solve the inequality for y (if it’s not already) to easily identify the boundary line’s slope and intercept. For instance, $x+y\leq 4$ . can be rewritten as $y\leq -x+4$ . This tells us the boundary line equation and makes shading direction clearer.
Graph the boundary line: Replace the inequality sign with an equals sign to get the equation of the boundary line. In our example, the boundary line is $y=-x+4$ . Plot this line on the coordinate plane. Use techniques from linear graphing (find the y-intercept and slope, or x- and y-intercepts, etc.).
- Solid or Dashed? If the original inequality is $\geq$ or $\leq$ (inclusive), draw the boundary as a solid line (because points on the line satisfy the inequality). If the inequality is $>$ or $<$ (strict), draw the boundary as a dashed line – this indicates the line itself is not part of the solution. Example: $y>2x+1$ would use a dashed line for $y=2x+1$ but on the other hand, $y\geq 2x+1$ would use a solid line. (Think of it like the open/closed circle idea extended to an entire line.)
Determine which side to shade (the solution region): The inequality will be true for one side of the boundary line. To figure out which side, pick a test point that is not exactly on the line – a convenient choice is often (0,0) unless the line passes through (0,0). Plug the test point’s coordinates into the inequality:
- If the inequality holds true with that point, then the region containing that point is the solution region.
- If the inequality is false for that point, then the solution region is the opposite side of the line. For example, suppose we are graphing $y<2x+1$ . The boundary is $y=2x+1$ (dashed). Take the test point (0,0). Plugging in the x and y values: 0<2(0)+1 gives 0<1, which is true. So (0,0) satisfies the inequality, meaning the side of the line that contains (0,0) (the “below” side, in this case) is the solution region. We would shade the half-plane that contains (0,0). If instead the test point had failed, we’d shade the opposite side. Another quick clue: If the inequality is solved for $y$ (like $y>$ .. or $y<$ ..), you can often tell the shading direction immediately: $y>$ means shade above the line, $y<$ means shade below the line (assuming “above” means larger y-values). But be careful if you have flipped the inequality or if it’s in standard form; the test-point method always works.
Shade the appropriate region: Once you know which side of the boundary contains the solutions, shade that entire half-plane. Every point in this shaded area (and on the boundary if it’s solid) is a solution to the inequality. Every point on the other side is not a solution.

Let’s work through a concrete example:

Example 3: Graphing a Two-Variable Inequality

Problem: Graph the inequality $y>2x+1$ on the coordinate plane.

Solution:

Boundary line: First, graph $y=2x+1$ . This is a line with slope 2 and y-intercept 1. So it passes through (0,1). Using the slope, another easy point is (1,3) (since from (0,1), rise 2 and run 1). Draw a line through these points.
Line type: The inequality is $>$ (greater than, not equal), so we draw the line dashed to show that points on $y=2x+1$ are not included in $y>2x+1$ . (Any point on the line would satisfy $y=2x+1$ , which is not strictly greater, so those are not solutions.)
Test a point: We need to find which side of this line to shade. Take a test point not on the line. A convenient choice is (0,0) (the origin). Plug (0,0) into the inequality $y>2x+1$ : substitute x=0,y=0 to get 0>2(0)+1, which yields 0 > 1. This statement is false (0 is not greater than 1). This tells us that (0,0) is not in the solution region. Therefore, the half-plane containing (0,0) is not the solution; we must shade the opposite side of the line. The line $y=2x+1$ divides the plane into two halves: one half has points above the line, and the other has points below the line. Since (0,0) (which is below the line) didn’t satisfy the inequality, we shade the other side, which is the region above the line.
Shade: Shade all points above (but not on) the line $y=2x+1$ . This represents $y>2x+1$ . Any point in this shaded region will satisfy the inequality. For example, a point like (0,2) is in the shaded region: check 2 > 2(0)+1, which yields 2 > 1. This is a true statement. So this point would be included in the shaded region of the graph. A point like (0,0) in the unshaded side we already saw is false.

The final graph has a dashed line through (0,1) and (1,3), and the region above that line is shaded. There is no shading below the line, indicating those points don’t satisfy $y>2x+1$ .

In summary, $y>2x+1$ means all points (x,y) lying above the line $y=2x+1$ . We used a dashed line to exclude the boundary, and shaded above. If the inequality had been $y\geq 2x+1$ , we would use a solid line and also shade above (including the line).

Interval/Set Notation: It’s trickier to express two-variable solutions in interval notation, because the solution is a two-dimensional region (infinitely many points in the plane). We usually just leave it in inequality form or describe it in words (like saying “the set of all (x,y) such that $y>2x+1$ ”). You might see set-builder notation: {(x,y)∣ $y>2x+1$ }. But graphing is the most intuitive way to understand it.

And heres the visual of the graph below. Go through the steps above to see how we arrived at this graph (also note the shaded region indicating the solution set to $y>2x+1$ is in red). Also, the dashed line is $y=2x+1$ :

Graph of linear inequality y > 2x + 1

Now, if we were to graph one inequality that goes in the opposite direction with a negative slope, let's say $x+y\leq 4$ , we would first change it by isolating y to make $y\leq -x+4$ and then we essentially do the same steps as the first example, except keeping in mind that the boundary line ( $y=-x+4$ ) in this case would be a solid line instead of a dashed one. Also, we would shade below the boundary line in this example as we are using the $\leq$ (less than or equal to) inequality sign. Plug in points like in the example up above to see if the inequality for this example ( $x+y\leq 4$ ) holds true. For instance, if you use the coordinate point (1,1) for (x, y) respectively, the inequality would state (after plugging in the coordinate values for x and y) $1+1\leq 4$ which is $2\leq 4$ , and that is a true statement. Hence, you know that the coordinate point (1,1) will be included in the shaded region as it is a solution set to the inequality. Do the same process for other points if you want to confirm whether certain points exist in the solution set to the inequality or not. Here's the visual for the inequality $x+y\leq 4$ or $y\leq -x+4$ below (keeping in mind the red shaded region is the solution set to the inequality):

Graph of inequality x + y <= 4

Practice Problems (Linear Inequalities in Two Variables)

Graphing Compound Inequalities in the Coordinate Plane

"And" Compound Inequalities

When two inequalities are joined by “and,” you only keep the points that satisfy both conditions simultaneously.

Steps:

Graph each boundary line
- Rewrite each inequality as an equation (replace “ $<$ ” or “ $>$ ” with “ $=$ ”).
- Draw that line on the plane.
  - Use a solid line if the inequality is “ $\leq$ ” or “ $\geq$ ” (points on the line count).
  - Use a dashed line if it’s “ $<$ ” or “ $>$ ” (points on the line do not count).
Shade each half-plane
- For an inequality like $y\geq x-2$ , shade the region above the line $y=x-2$ .
- For an inequality like $y<2x+3$ , shade the region below the line $y=2x+3$ .
Find the overlap
- The solution set is the area where the two shadings overlap. Everything in that overlapping region satisfies both inequalities.

Example:

Graph $y\geq x-2$ and $y<2x+3$

Boundary 1: $y=x-2$ , this is a solid line because of the $\geq$ sign in the first inequality $y\geq x-2$ . Hence, we shade above this boundary line.
Boundary 2: $y=2x+3$ , this is a dashed line because of the strict $<$ sign in the second inequality $y<2x+3$ . Hence, we shade below this boundary line.
Solution: the red wedge-shaped region between those two lines where both shadings coincide. (notice that the line for $y\geq x-2$ on the right side of the wedge is solid while the line on the left signifying $y<2x+3$ is not solid (sometimes it can be dashed or not solid like in this visual below):

Graph of "and" compound inequality 𝑦 ≥ 𝑥 − 2 and 𝑦 < 2 𝑥 + 3

"Or" Compound Inequalities

When joined by “or,” you include any point that satisfies at least one of the inequalities.

Steps:

Graph each boundary line (solid for “ $\leq$ or $\geq$ ", dashed for “ $<$ or $>$ ").
Shade each half-plane according to its own inequality.
Take the union of the shaded regions, meaning that every point that is shaded for either inequality is part of the overall solution.

Example:

Graph $x\leq 1$ or $y>-1$

Boundary 1: vertical line $x=1$ , which is solid because of the $\leq$ in the first inequality $x\leq 1$ . Also, we will shade to the left because of the less than or equal to symbol of this inequality.
Boundary 2: horizontal line $y=-1$ , which is dashed because of the $>$ in the second inequality $y>-1$ . Also, we will shade above it because of the greater than symbol of this inequality.
Solution: the full solution set will be all points to the left of $x=1$ plus all points above $y=-1$ . The two shaded regions may overlap, but you include both. (this is different than "and" because we are not trying to find the intersection of the two as it is an "or" inequality).

Here is a visual representation of this "or" compound inequality below. Note the dashed blue line at $y=-1$ to signify the $>$ sign in the second inequality $y>-1$ . And also note that the whole blue region above that is the solution set to the second inequality $y>-1$ . Also, there is a solid line at $x=1$ to signify the equal part of the $\leq$ sign in the first inequality $x\leq 1$ and that to the left of $x=1$ is the whole solution set for the inequality $x\leq 1$ . You can also notice that the two solution sets of both inequalities intersect in the dark green/teal colored region. Had this been an "and" compound inequality, that teal/dark green region would be the solution set/graph, but since this inequality is an "or" compound inequality, we include all regions/solution sets for each inequality, along with the intersection:

Graph of compound "or" inequality 𝑥 ≤ 1 (green) or 𝑦 > − 1 (blue)

And there you have it! You have figured out how to graph compound inequalities as well. With that, you now know how to graph all the relevant inequality types. Try out some problems of graphing compound inequalities below to gain a stronger understanding and practice.

Practice Problems (Compound Inequalities)

Graph each compound inequality. Remember to draw the appropriate boundary lines (solid/dashed), shade the half-planes, and clearly identify the solution region.

$y>-x+1$ and $y\leq {\frac {1}{2}}x+2$ (Hint: its the same process as any other compound inequality, first just graph the two linear inequalities separately, as you would regularly, and then find the intersection region (where they overlap) of that.)
$x>-2$ or $x+y\leq 4$
$y<3$ and $x\geq 0$
$2x-y\geq 1$ or $y+x<-1$
(BONUS CHALLENGE): $x\leq 3$ and ( $y>1$ or $y\leq -2$ )

TIP: When in doubt, pick a simple test point (like (0, 0)) it it is not on one of your boundary lines) and then decide which side of each boundary to shade depending on if plugging in (0,0) makes the inequality true or false. ALSO WE STRONGLY ENCOURAGE TO CHECK SOLUTIONS USING Desmos' Inequality Grapher. To look at the intersection of an "and" inequality specifically, you can input (ineq1) {ineq2} into the blank input line.

Now you have completed the section of how to graph different kinds of inequalities. With practice you will become an inequality expert! If you want a refresher of inequality rules and how to solve them, check out the Unit 7: Inequalities (also listed below), or if you feel content, feel free to move on to the next unit: Six rules of Exponents.

References

Khan Academy “Graphing Inequalities on a Number Line” & “Graphing Inequalities in Two Variables.” https://www.khanacademy.org/math/algebra/inequalities (Step-by-step methods for number-line and coordinate-plane graphing.)
Purplemath “Graphing Inequalities.” https://www.purplemath.com/modules/ineqgrph.htm (Visual examples and tips on dashed vs. solid boundaries and shading half-planes.)
OpenStax Stewart, J., et al. Intermediate Algebra (OpenStax, 2016). (Further illustrative examples of graphing linear inequalities in two variables.)

← Unit 7: Inequalities

Algebra 1

Unit 9: Six rules of Exponents →

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Algebra 1/Unit 8: Graphing Inequalities

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