Linear inequalities can be graphed top top a A aircraft formed by the intersection that a horizontal number line referred to as the x-axis and also a upright number line dubbed the y-axis.

You are watching: Shade above or below the line inequalities

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. The options for a direct inequality room in a region of the coordinate plane. A boundary line, i beg your pardon is the related linear equation, serves together the border for the region. You have the right to use a visual representation to number out what values make the inequality true—and also which ones do it false. Let’s have actually a look at inequalities by return to the name: coordinates plane.

Linear inequalities are different than direct equations, although friend can apply what you know around equations to assist you know inequalities. Inequalities and also equations space both mathematics statements the compare two values. Equations use the symbol =; inequalities will be stood for by the symbols , and also ≥.

One method to visualize two-variable inequalities is come plot them on a coordinate plane. Below is what the inequality x > y watch like. The solution is a region, i beg your pardon is shaded. There are a few things to notification here. First, look in ~ the dashed red boundary line: this is the graph the the related straight equation x = y. Next, look in ~ the irradiate red an ar that is to the appropriate of the line. This an ar (excluding the heat x = y) to represent the entire collection of solutions for the inequality x > y. Remember just how all clues on a line are remedies to the straight equation the the line? Well, every points in a an ar are solutions to the A mathematical statement in 2 variables using the inequality symbols , >, ≤, or ≥ to show the relationship between two expressions. As soon as the inequality price is changed by an equal sign, the resulting related equation will graph as a right line.

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representing that region.

Let’s think about it for a moment—if x > y, climate a graph the x > y will present all ordered pairs (x, y) for which the x-coordinate is higher than the y-coordinate.

The graph below shows the region x > y and also some ordered pairs on the coordinate plane. Look at each ordered pair. Is the x-coordinate higher than the y-coordinate? walk the ordered pair sit inside or exterior of the shaded region? The ordered pairs (4, 0) and (0, −3) lie within the shaded region. In this ordered pairs, the x-coordinate is bigger than the y-coordinate. This ordered pairs are in the solution collection of the equation x > y.

The ordered pairs (−3, 3) and (2, 3) are exterior of the shaded area. In this ordered pairs, the x-coordinate is smaller sized than the y-coordinate, so they are not had in the collection of remedies for the inequality.

The notified pair (−2, −2) is on the boundary line. It is not a equipment as −2 is not greater than −2. However, had the inequality been x ≥ y (read as “x is greater than or equal to y"), then (−2, −2) would have been included (and the heat would have been stood for by a solid line, not a dashed line).

Let’s take it a look in ~ one much more example: the inequality 3x + 2y ≤ 6. The graph listed below shows the region of worths that makes this inequality true (shaded red), the boundary line 3x + 2y = 6, as well as a handful of bespeak pairs. The border line is hard this time, because points top top the border line 3x + 2y = 6 will make the inequality 3x + 2y ≤ 6 true. As friend did with the ahead example, you can substitute the x- and also y-values in every of the (x, y) bespeak pairs right into the inequality to discover solutions. While friend may have been able to carry out this in her head for the inequality x > y, occasionally making a table that values provides sense for more complex inequalities.

 Ordered Pair Makes the inequality 3 x + 2y ≤ 6 a true statement Makes the inequality 3 x + 2y ≤ 6 a false statement (−5, 5) 3(−5) + 2(5) ≤ 6 −15 +10 ≤ 6 −5 ≤ 6 (−2, −2) 3(−2) + 2(–2) ≤ 6 −6 + (−4) ≤ 6 –10 ≤ 6 (2, 3) 3(2) + 2(3) ≤ 6 6 + 6 ≤ 6 12 ≤ 6 (2, 0) 3(2) + 2(0) ≤ 6 6 + 0 ≤ 6 6 ≤ 6 (4, −1) 3(4) + 2(−1) ≤ 6 12 + (−2) ≤ 6 10 ≤ 6

If substituting (x, y) into the inequality returns a true statement, then the ordered pair is a systems to the inequality, and the allude will be plotted within the shaded an ar or the suggest will be component of a solid boundary line. A false statement method that the bespeak pair is not a solution, and also the point will graph external the shaded region, or the suggest will be component of a dotted boundary line.

 Example Problem Use the graph to determine which ordered pairs plotted listed below are services of the inequality x – y Solutions will be situated in the shaded region. Because this is a “less than” problem, ordered bag on the boundary line space not contained in the systems set. (−1, 1) (−2, −2) These worths are located in the shaded region, so space solutions. (When substituted into the inequality x – y (1, −2) (3, −2) (4, 0) These values space not located in the shaded region, so space not solutions. (When substituted into the inequality x – y Answer (−1, 1), (−2, −2)

 Example Problem Is (2, −3) a systems of the inequality y −3x + 1? y −3x + 1 If (2, −3) is a solution, climate it will certainly yield a true statement as soon as substituted into the inequality y −3x + 1. −3 −3(2) + 1 Substitute x = 2 and y = −3 into inequality. −3 −6 + 1 Evaluate. −3 −5 This explain is not true, for this reason the bespeak pair (2, −3) is not a solution. Answer (2, −3) is not a solution.

 Which ordered pair is a equipment of the inequality 2y - 5x A) (−5, 1) B) (−3, 3) C) (1, 5) D) (3, 3) Show/Hide Answer A) (−5, 1) Incorrect. Substituting (−5, 1) right into 2y – 5x B) (−3, 3) Incorrect. Substituting (−3, 3) right into 2y – 5x C) (1, 5) Incorrect. Substituting (1, 5) into 2y – 5x D) (3, 3) Correct. Substituting (3, 3) right into 2y – 5x

Graphing Inequalities

So just how do you get from the algebraic form of an inequality, prefer y > 3x + 1, come a graph of that inequality? plotting inequalities is fairly straightforward if you follow a pair steps.

 Graphing Inequalities To graph an inequality: o Graph the related boundary line. Change the , ≤ or ≥ sign in the inequality with = to discover the equation that the boundary line. o identify at the very least one notified pair on either next of the boundary line and substitute those (x, y) values right into the inequality. Shade the an ar that contains the bespeak pairs the make the inequality a true statement. o If clues on the border line are solutions, then use a solid heat for illustration the border line. This will occur for ≤ or ≥ inequalities. o If points on the border line no solutions, then use a dotted line for the border line. This will occur for inequalities.

Let’s graph the inequality x + 4y ≤ 4.

To graph the border line, uncover at the very least two values that lied on the line x + 4y = 4. You can use the x- and also y- intercepts because that this equation by substituting 0 in for x very first and finding the value of y; then substitute 0 in for y and find x.

 x y 0 1 4 0

Plot the point out (0, 1) and (4, 0), and also draw a line v these 2 points for the border line. The heat is solid because ≤ way “less 보다 or same to,” so every ordered pairs along the heat are included in the solution set. The following step is to discover the an ar that includes the solutions. Is it above or below the border line? To recognize the region where the inequality stop true, you have the right to test a pair of ordered pairs, one on each side of the border line.

If you substitute (−1, 3) into x + 4y ≤ 4:

 −1 + 4(3) ≤ 4 −1 + 12 ≤ 4 11 ≤ 4

This is a false statement, due to the fact that 11 is not much less than or same to 4.

On the other hand, if you instead of (2, 0) into x + 4y ≤ 4:

 2 + 4(0) ≤ 4 2 + 0 ≤ 4 2 ≤ 4

This is yes, really! The region that includes (2, 0) must be shaded, as this is the an ar of solutions. And there you have it—the graph that the set of remedies for x + 4y ≤ 4.

 Example Problem Graph the inequality 2y > 4x – 6.   Solve for y. x y 0 −3 2 1

Create a table of worths to uncover two clues on the line, or graph it based on the slope-intercept method, the b value of the y-intercept is -3 and also the steep is 2.

Plot the points, and graph the line. The line is dotted because the authorize in the inequality is >, no ≥ and therefore points on the line are not solutions to the inequality. 2y > 4x – 6

Test 1: (−3, 1)

2(1) > 4(−3) – 6

2 > –12 – 6

2 > −18 True!

Test 2: (4, 1)

2(1) > 4(4) – 6

2 > 16 – 6

2 > 10 False!

Find an bespeak pair top top either side of the border line. Insert the x- and also y-values into the inequality 2y > 4x – 6 and see which ordered pair results in a true statement.

Since (−3, 1) outcomes in a true statement, the region that includes (−3, 1) have to be shaded. 