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학습 가이드 > Intermediate Algebra

3B.2 Graph Linear Inequalities

3b.2 Learning Objectives

  • Identify and follow steps for graphing a linear inequality in two variables
  • Identify whether an ordered pair is in the solution set of a linear inequality
Did you know that you use linear inequalities when you shop online? When you use the option to view items within a specific price range, you are asking the search engine to use a linear inequality based on price. Essentially, you are saying "show me all the items for sale between $50 and $100," which can be written as 50≤x≤100{50}\le {x} \le {100}, where x is price. In this section, you will apply what you know about graphing linear equations to graphing linear inequalities. So how do you get from the algebraic form of an inequality, like y>3x+1y>3x+1, to a graph of that inequality? Plotting inequalities is fairly straightforward if you follow a couple steps.

Graphing Inequalities

To graph an inequality:
  • Graph the related boundary line. Replace the <, >, ≤ or ≥ sign in the inequality with = to find the equation of the boundary line.
  • Identify at least one ordered pair on either side of the boundary line and substitute those (x,y)(x,y) values into the inequality. Shade the region that contains the ordered pairs that make the inequality a true statement. 
  • If points on the boundary line are solutions, then use a solid line for drawing the boundary line. This will happen for ≤ or ≥ inequalities.
  • If points on the boundary line aren’t solutions, then use a dotted line for the boundary line. This will happen for < or > inequalities.
Let’s graph the inequality x+4y≤4x+4y\leq4. To graph the boundary line, find at least two values that lie on the line x+4y=4x+4y=4. You can use the x- and y-intercepts for this equation by substituting 0 in for x first and finding the value of y; then substitute 0 in for y and find x.
x y
0 1
4 0
Plot the points (0,1)(0,1) and (4,0)(4,0), and draw a line through these two points for the boundary line. The line is solid because ≤ means “less than or equal to,” so all ordered pairs along the line are included in the solution set. Solid downward-sloping line that crosses the points (0,1) and (4,0). The point (-1,3) and the point (2,0) are also plotted. The next step is to find the region that contains the solutions. Is it above or below the boundary line? To identify the region where the inequality holds true, you can test a couple of ordered pairs, one on each side of the boundary line. If you substitute (−1,3)(−1,3) into x+4y≤4x+4y\leq4:

−1+4(3)≤4−1+12≤411≤4\begin{array}{r}−1+4\left(3\right)\leq4\\−1+12\leq4\\11\leq4\end{array}

This is a false statement, since 11 is not less than or equal to 4. On the other hand, if you substitute (2,0)(2,0) into x+4y≤4x+4y\leq4:

2+4(0)≤42+0≤42≤4\begin{array}{r}2+4\left(0\right)\leq4\\2+0\leq4\\2\leq4\end{array}

This is true! The region that includes (2,0)(2,0) should be shaded, as this is the region of solutions. Solid downward-sloping line marked x+4y=4. The region below the line is shaded and is labeled x+4y is less than or equal to 4. And there you have it—the graph of the set of solutions for x+4y≤4x+4y\leq4.

3B.2.1 Graphing Linear Inequalities in Two Variables

https://youtu.be/2VgFg2ztspI

Example 3B.2.A

Graph the inequality 2y>4x–62y>4x–6.

Answer: Solve for y.

2y>4x−62y2>4x2−62y>2x−3 \displaystyle \begin{array}{r}2y>4x-6\\\\\frac{2y}{2}>\frac{4x}{2}-\frac{6}{2}\\\\y>2x-3\\\end{array}

Create a table of values to find two points on the line y=2x−3 \displaystyle y=2x-3, or graph it based on the slope-intercept method, the b value of the y-intercept is −3-3 and the slope is 2. Plot the points, and graph the line. The line is dotted because the sign in the inequality is >, not ≥ and therefore points on the line are not solutions to the inequality. Dotted upward-sloping line that crosses the points (2,1) and (0,-3). The points (-3,1) and (4,1) are also plotted.

y=2x−3 \displaystyle y=2x-3

x y
0 −3−3
2 1
Find an ordered pair on either side of the boundary line. Insert the x- and y-values into the inequality 2y>4x–62y>4x–6 and see which ordered pair results in a true statement. Since (−3,1)(−3,1) results in a true statement, the region that includes (−3,1)(−3,1) should be shaded.

2y>4x–6Test 1:(−3,1)2(1)>4(−3)–6       2>–12–6       2>−18TRUETest 2:(4,1)2(1)>4(4)–6      2>16–6      2>10FALSE\begin{array}{l}2y>4x–6\\\\\text{Test }1:\left(−3,1\right)\\2\left(1\right)>4\left(−3\right)–6\\\,\,\,\,\,\,\,2>–12–6\\\,\,\,\,\,\,\,2>−18\\\text{TRUE}\\\\\text{Test }2:\left(4,1\right)\\2(1)>4\left(4\right)– 6\\\,\,\,\,\,\,2>16–6\\\,\,\,\,\,\,2>10\\\text{FALSE}\end{array}

Answer

The graph of the inequality 2y>4x–62y>4x–6 is: The dotted upward-sloping line of 2y=4x-6, with the region above the line shaded.

A quick note about the problem above—notice that you can use the points (0,−3)(0,−3) and (2,1)(2,1) to graph the boundary line, but that these points are not included in the region of solutions, since the region does not include the boundary line! https://youtu.be/Hzxc4HASygU

3B.2.2 Solution Sets of Inequalities

The graph below shows the region of values that makes the inequality 3x+2y≤63x+2y\leq6 true (shaded red), the boundary line 3x+2y=63x+2y=6, as well as a handful of ordered pairs. The boundary line is solid because points on the boundary line 3x+2y=63x+2y=6 will make the inequality 3x+2y≤63x+2y\leq6 true. A solid downward-sloping line running. The region below the line is shaded and is labeled 3x+2y is less than or equal to 6. The region above the line is unshaded and is labeled 3x+2y=6. The points (-5,5) and (-2,-2) are in the shaded region. The points (2,3) and (4,-1) are in the unshaded region. The point (2,0) is on the line. You can substitute the x- and y-values in each of the (x,y)(x,y) ordered pairs into the inequality to find solutions. Sometimes making a table of values makes sense for more complicated inequalities.
Ordered Pair Makes the inequality 3x+2y≤63x+2y\leq6 a true statement Makes the inequality 3x+2y≤63x+2y\leq6 a false statement
(−5,5)(−5, 5) 3(−5)+2(5)≤6−15+10≤6−5≤6\begin{array}{r}3\left(−5\right)+2\left(5\right)\leq6\\−15+10\leq6\\−5\leq6\end{array}
(−2,−2)(−2,−2) 3(−2)+2(–2)≤6−6+(−4)≤6–10≤6\begin{array}{r}3\left(−2\right)+2\left(–2\right)\leq6\\−6+\left(−4\right)\leq6\\–10\leq6\end{array}
(2,3)(2,3) 3(2)+2(3)≤66+6≤612≤6\begin{array}{r}3\left(2\right)+2\left(3\right)\leq6\\6+6\leq6\\12\leq6\end{array}
(2,0)(2,0) 3(2)+2(0)≤66+0≤66≤6\begin{array}{r}3\left(2\right)+2\left(0\right)\leq6\\6+0\leq6\\6\leq6\end{array}
(4,−1)(4,−1) 3(4)+2(−1)≤612+(−2)≤610≤6\begin{array}{r}3\left(4\right)+2\left(−1\right)\leq6\\12+\left(−2\right)\leq6\\10\leq6\end{array}
If substituting (x,y)(x,y) into the inequality yields a true statement, then the ordered pair is a solution to the inequality, and the point will be plotted within the shaded region or the point will be part of a solid boundary line. A false statement means that the ordered pair is not a solution, and the point will graph outside the shaded region, or the point will be part of a dotted boundary line.

Example 3B.2.B

Use the graph to determine which ordered pairs plotted below are solutions of the inequality x–y<3x–y<3. Upward-sloping dotted line. The region above the line is shaded and labeled x-y<3. The points (4,0) and (3,-2) are in the unshaded region. The point (1,-2) is on the dotted line. The points (-1,1) and (-2,-2) are in the shaded region.

Answer: Solutions will be located in the shaded region. Since this is a “less than” problem, ordered pairs on the boundary line are not included in the solution set. These values are located in the shaded region, so are solutions. (When substituted into the inequality x–y<3x–y<3, they produce true statements.)

(−1,1)(−1,1)

(−2,−2)(−2,−2)

These values are not located in the shaded region, so are not solutions. (When substituted into the inequality x−y<3x-y<3, they produce false statements.)

(1,−2)(1,−2)

(3,−2)(3,−2)

(4,0)(4,0)

Answer

(−1,1)   (−2,−2)(−1,1)\,\,\,(−2,−2)

The following video show an example of determining whether an ordered pair is a solution to an inequality. https://youtu.be/GQVdDRVq5_o

Example 3B.2.C

Is (2,−3)(2,−3) a solution of the inequality y<−3x+1y<−3x+1?

Answer: If (2,−3)(2,−3) is a solution, then it will yield a true statement when substituted into the inequality y<−3x+1y<−3x+1.

y<−3x+1y<−3x+1

Substitute x=2x=2 and y=−3y=−3 into inequality.

−3<−3(2)+1−3<−3\left(2\right)+1

Evaluate.

−3<−6+1−3<−5\begin{array}{l}−3<−6+1\\−3<−5\end{array}

This statement is not true, so the ordered pair (2,−3)(2,−3) is not a solution.

Answer

(2,−3)(2,−3) is not a solution.

The following video shows another example of determining whether an ordered pair is a solution to an inequality. https://youtu.be/-x-zt_yM0RM

Summary

When inequalities are graphed on a coordinate plane, the solutions are located in a region of the coordinate plane, which is represented as a shaded area on the plane. The boundary line for the inequality is drawn as a solid line if the points on the line itself do satisfy the inequality, as in the cases of ≤ and ≥. It is drawn as a dashed line if the points on the line do not satisfy the inequality, as in the cases of < and >. You can tell which region to shade by testing some points in the inequality. Using a coordinate plane is especially helpful for visualizing the region of solutions for inequalities with two variables.

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