Graphical representation of linear inequalities
Learning goals)
· Bestimmen Sie, ob ein geordnetes Paar eine Lösung einer Ungleichung ist.
· Draw an inequality in two variables.
introduction
Linear inequalities can be graphed on aCoordinate levels. The solutions of a linear inequality lie in a region of the coordinate plane. Aboundary line, which is the associated linear equation, serves as the boundary for the region. You can use a visual representation to find out which values make the inequality true and also which make it false. Let's look at the inequalities by going back to the coordinate plane.
Linear inequalities as regions
Linear inequalities are different from linear equations, although you can apply your knowledge of equations to understand inequalities. Inequalities and equations are mathematical statements that compare two values. The equations use the symbol =; Inequalities are represented by the symbols<, ≤, > y ≥.
One way to visualize inequalities in two variables is to plot them on a coordinate plane. That's the inequalityX>jIt seems. The solution is a region that is shaded.
There are a few things to consider here. First, look at the dashed red dividing line - this is the graph of the associated linear equationX=j. Then look at the bright red area to the right of the line. This region (with the exception of the lineX=j) represents the complete set of solutions to the inequalityX>j. Remember that all points on a line are solutions of the line's linear equation? Well, all points in a region are solutions to thatlinear inequalityto represent this region.
Let's think about it for a moment, yesX>j, then a chart ofX>jdisplays all ordered pairs (x, y) so theX-coordinate is greater than thaty-coordinate.
The graphic below shows the regionX>jand some ordered pairs in the coordinate plane. Look at each pair ordered. Is heX-Coordinate greater thany-coordinate? Is the ordered pair inside or outside the shaded area?
The ordered pairs (4, 0) and (0, −3) fall within the shaded area. In these ordered pairs thatX-coordinate is greater than thaty-coordinate. These ordered pairs are in the solution set of the equationX>j.
The ordered pairs (−3, 3) and (2, 3) are outside the shaded area. In these ordered pairs thatX-coordinate iskleinerthey-coordinate, so they are not included in the solution set for the inequality.
The ordered pair (−2, −2) lies on the boundary line. It is not a solution since −2 is not greater than −2. However, if the inequality had beenX≥j(read as "Xis greater than or equal toj"), then (−2, −2) would have been included (and the line would have been represented by a solid line, not a dashed line).
Let's look at another example: Inequality 3X+ 2j≤ 6. The graph below shows the range of values that makes this inequality true (shaded in red), the limit line 3X+ 2j= 6, as well as a handful of ordered pairs. The boundary line is solid this time because the points on the boundary line are 3X+ 2j= 6 makes the inequality 3X+ 2j≤ 6 true.
As in the previous example, you can replace theX-jy-Values in each of the (x, y) ordered pairs in the inequality to find solutions. While you could possibly do this in your head due to the disparityX>j, sometimes creating a table of values is useful for more complicated inequalities.
| orderly pair | makes the inequality 3X+ 2j ≤ 6 a true statement | makes the inequality 3X+ 2j ≤ 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 |
When replacing (x, y) produces a true statement in the inequality, then the ordered pair is a solution of the inequality, and the point is drawn inside the shaded areaor 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 is drawn outside the shaded area., or the point becomes part of a dashed boundary line.
| example | ||
| Problem | Use the diagram to determine which ordered pairs shown below are solutions to the inequality X–j< 3. | |
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| The solutions are in the shaded area. Because this is a "less than" problem, the ordered pairs on the boundary line are not included in the solution set. | ||
| (−1, 1) (−2,−2) | These values are in the shaded area, as are the solutions. (When substituting into the inequalityX–j< 3, generate true statements.) | |
| (1,−2) (3,−2) (4, 0) | These values are not in the shaded area, so they are not solutions. (When substituting into the inequalityX–j< 3, produce false statements.) | |
| responder | (−1, 1), (−2,−2) | |
| example | ||
| Problem | es (2,−3) a solution of the inequality | |
| j<−3X+ 1 | and 2,−3) is a solution, then it gives a true statement when plugged into the inequality j<−3X+ 1. | |
| −3 <−3(2) + 1 | ErsatzX= 2 Jj=−3 in inequality. | |
| −3 <−6 + 1 | Evaluate. | |
| −3 <−5 | This statement isnotrue, then the ordered pair (2,−3) esnoa solution. | |
| responder | (2,−3) is not a solution. | |
Which ordered pair is a solution of inequality 2j- 5X< 2?
A) (−5, 1)
B) (−3, 3)
C) (1, 5)
D) (3, 3)
Show/hide answer
So how do you get the algebraic form of an inequality such asj> 3X+ 1, to a graph of this inequality? Graphing inequalities is pretty easy if you follow a few steps.
graphic inequalities
To graph an inequality:
Ö Draw the associated limit line. Replace the sign <, >, ≤, or ≥ in the inequality with = to find the equation of the limit line.
Ö Identify at least one ordered pair on each side of the boundary line and replace them (x, y) values in the inequality. Shade the area containing the ordered pairs that make the inequality true.
Ö If the points on the limit line are solutions, use a solid line to draw the limit line. This happens for inequalities ≤ or ≥.
Ö If the points on the limit line are not solutions, use a dotted line for the limit line. This happens for the inequalities < or >.
Let's graph the inequalityX+ 4j≤ 4.
To graph the limit line, find at least two values that lie on the lineX+ 4j= 4. You can use theX- andj-Sections for this equation by replacing 0 in forXfirst and find the value ofj; then substitute 0 forjand findX.
| X | j |
| 1 | |
| 4 |
Draw the points (0, 1) and (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.
The next step is to find the region that contains the solutions. Are you above or below the dividing line? To identify the region where the inequality holds, you can try a pair of ordered pairs, one on each side of the boundary line.
If you replace (−1, 3) inX+ 4j≤ 4:
| −1 + 4(3) ≤ 4 |
| −1 + 12 ≤ 4 |
| 11 ≤ 4 |
This is an incorrect statement because 11 is not less than or equal to 4.
On the other hand, if you replace (2, 0) inX+ 4j≤ 4:
| 2 + 4(0) ≤ 4 |
| 2 + 0 ≤ 4 |
| 2 ≤ 4 |
This is true! The area containing (2, 0) should be shaded since this is the area of solutions.
And there you have it: the graph of the solution set forX+ 4j≤ 4.
| example | ||||||||
| Problem | Draw the inequality 2j> 4x – 6. | |||||||
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| Resolverj. | |||||||
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| Create a table of values to find two points on the line Enter the points and draw the line. The line is dashed because the sign in the inequality is >, not≥and hence the points on the line are not solutions of the inequality. | |||||||
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| 2j> 4X– 6 Exam 1: (−3, 1) 2(1) > 4(−3) – 6 2 >–12 – 6 2 >−18 Right! Test 2: (4, 1) 2(1) > 4(4) – 6 2 > 16 – 6 2 > 10 Wrong! | Find an ordered pair on each side of the boundary line. paste theX- andj-Values in inequality As (−3, 1) yields a true statement including the region (−3, 1) should be shaded. | |||||||
| responder | The inequality graph 2j> 4X– 6 es:
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A quick note on the above issue. Note that you can use the points (0, −3) and (2, 1) to draw the boundary line, but these points are not included in the solution domain because the domain does not contain the boundary line!
What does the graph mean when drawn on a coordinate plane?j≥Xlooks?
A)
B)
C)
D)
Show/hide answer
summary
When inequalities are graphed on a coordinate plane, the solutions reside in an area of the coordinate plane, represented as a shaded area on the plane. The boundary line of the inequality is drawn as a solid line when points on the line itself satisfy the inequality, as in the cases of ≤ and ≥. It is drawn as a dashed line when the points on the line do not satisfy the inequality, as in the cases of < and >. You can determine which region to shade by testing some points in the inequality. Using a coordinate plane is particularly useful for visualizing the solution space for two-variable inequalities.