Hotmath
Show To graph a linear inequality in two variables (say, x and y ), first get y alone on one side. Then consider the related equation obtained by changing the inequality sign to an equals sign. The graph of this equation is a line. If the inequality is strict ( < or > ), graph a dashed line. If the inequality is not strict ( ≤ and ≥ ), graph a solid line. Finally, pick one point not on the line ( ( 0 , 0 ) is usually the easiest) and decide whether these coordinates satisfy the inequality or not. If they do, shade the half-plane containing that point. If they don't, shade the other half-plane. Example: Graph the inequality y ≤ 4 x − 2 . This line is already in slope-intercept form , with y alone on the left side. Its slope is 4 and its y -intercept is − 2 . So it's straightforward to graph it. In this case, we make a solid line since we have a "less than or equal to" inequality.
Now, substitute x = 0 , y = 0 to decide whether ( 0 , 0 ) satisfies the inequality. 0 ≤ ? 4 ( 0 ) − 2 0 ≤ ? − 2 This is false. So, shade the half-plane which does not include the point ( 0 , 0 ) .
Learning Objectives Solutions to Linear InequalitiesWe know that a linear equation with two variables has infinitely many ordered pair solutions that form a line when graphed. A linear inequality with two variables, on the other hand, has a solution set consisting of a region that defines half of the plane. Table \(\PageIndex{1}\)
For the inequality, the line defines one boundary of the region that is shaded. This indicates that any ordered pair that is in the shaded region, including the boundary line, will satisfy the inequality. To see that this is the case, choose a few test points and substitute them into the inequality. Table \(\PageIndex{2}\)
Also, we can see that ordered pairs outside the shaded region do not solve the linear inequality. \(\begin{array} {c|c} {\underline{\color{Cerulean}{Test\:point}}}&{\underline{y\leq\frac{3}{2}x+3}}\\{(-2,3)} &{3\leq\frac{3}{2}(-2)+3}\\{}&{3\leq -3+3}\\{}&{3\leq 0\quad\color{red}{x}} \end{array}\) The graph of the solution set to a linear inequality is always a region. However, the boundary may not always be included in that set. In the previous example, the line was part of the solution set because of the “or equal to” part of the inclusive inequality \(≤\). If we have a strict inequality \(<\), we would then use a dashed line to indicate that those points are not included in the solution set. Table \(\PageIndex{3}\)
Consider the point \((0, 3)\) on the boundary; this ordered pair satisfies the linear equation. It is the “or equal to” part of the inclusive inequality that makes it part of the solution set. \(\begin{array}{c|c}{\underline{y<\frac{3}{2}x+3}}&{\underline{y\leq \frac{3}{2}x+3}}\\{3<\frac{3}{2}(0)+3}&{3\leq\frac{3}{2}(0)+3}\\{3<0+3}&{3\leq 0+3}\\{3<3\quad\color{red}{x}}&{3\leq 3\quad\color{Cerulean}{\checkmark}} \end{array}\) So far, we have seen examples of inequalities that were “less than.” Now consider the following graphs with the same boundary: Table \(\PageIndex{4}\)
Given the graphs above, what might we expect if we use the origin \((0, 0)\) as a test point? \(\begin{array}{c|c}{\underline{y\geq\frac{3}{2}x+3}}&{\underline{y\leq\frac{3}{2}x+3}}\\{0\geq\frac{3}{2}(0)+3}&{0\leq\frac{3}{2}(0)+3}\\{0\geq 0+3}&{0\leq0+3}\\{0\geq 3\quad\color{red}{x}}&{0\leq 3\quad\color{Cerulean}{\checkmark}} \end{array}\) Exercise \(\PageIndex{1}\) Which of the ordered pairs \((−2, −1), (0, 0), (−2, 8), (2, 1),\) and \((4, 2)\) solve the inequality \(y>−\frac{1}{2}x+2\)? Answer\((−2, 8)\) and \((4, 2)\) Graphing Solutions to Linear InequalitiesSolutions to linear inequalities are a shaded half-plane, bounded by a solid line or a dashed line. This boundary is either included in the solution or not, depending on the given inequality. If we are given a strict inequality, we use a dashed line to indicate that the boundary is not included. If we are given an inclusive inequality, we use a solid line to indicate that it is included. The steps for graphing the solution set for an inequality with two variables are outlined in the following example. Example \(\PageIndex{1}\) Graph the solution set: \(y>−3x+1\). Solution: Step 1: Graph the boundary line. In this case, graph a dashed line \(y=−3x+1\) because of the strict inequality. By inspection, we see that the slope is \(m=−3=−\frac{3}{1}=\frac{rise}{run}\) and the \(y\)-intercept is \((0, 1)\). Figure \(\PageIndex{7}\) Step 2: Test a point not on the boundary. A common test point is the origin \((0, 0)\). The test point helps us determine which half of the plane to shade. \(\begin{array}{c|c} {\underline{\color{Cerulean}{Test\:point}}}&{\underline{y>-3x+1}}\\{(0,0)}&{0>-3(0)+1}\\{}&{0>1\quad\color{red}{x}} \end{array}\) Step 3: Shade the region containing the solutions. Since the test point \((0, 0)\) was not a solution, it does not lie in the region containing all the ordered pair solutions. Therefore, shade the half of the plane that does not contain this test point. In this case, shade above the boundary line. Answer: Figure \(\PageIndex{8}\) Consider the problem of shading above or below the boundary line when the inequality is in slope-intercept form. If \(y>mx+b\), then shade above the line. If \(y<mx+b\), then shade below the line. Use this with caution; sometimes the boundary is given in standard form, in which case these rules do not apply. Example \(\PageIndex{2}\) Graph the solution set: \(2x−5y≥−10\). Solution: Here the boundary is defined by the line \(2x−5y=−10\). Since the inequality is inclusive, we graph the boundary using a solid line. In this case, graph the boundary line using intercepts. Figure \(\PageIndex{9}\) Next, test a point; this helps decide which region to shade. \(\begin{array} {c|c} {\underline{\color{Cerulean}{Test\:point}}}&{\underline{2x-5y\geq -10}}\\{(0,0)}&{2(0)-5(0)\geq -10} \\ {}&{0\geq -10\quad\color{Cerulean}{\checkmark}} \end{array}\) Since the test point is in the solution set, shade the half of the plane that contains it. Answer: Figure \(\PageIndex{10}\) In this example, notice that the solution set consists of all the ordered pairs below the boundary line. This may be counterintuitive because of the original \(≥\) in the inequality. This illustrates that it is a best practice to actually test a point. Solve for \(y\) and you see that the shading is correct. In slope-intercept form, you can see that the region below the boundary line should be shaded. An alternate approach is to first express the boundary in slope-intercept form, graph it, and then shade the appropriate region. Example \(\PageIndex{3}\) Graph the solution set: \(y<2\). Solution: First, graph the boundary line \(y=2\) with a dashed line because of the strict inequality. Figure \(\PageIndex{11}\) Now, test a point. \(\begin{array}{c|c}{\underline{\color{Cerulean}{Test\:point}}}&{\underline{y<2}}\\{(0,0)}&{0<2\quad\color{Cerulean}{\checkmark}} \end{array}\) In this case, shade the region that contains the test point. Answer: Figure \(\PageIndex{12}\) Exercise \(\PageIndex{2}\) Graph the solution set: \(5x−y≤10\). AnswerFigure \(\PageIndex{13}\) Key Takeaways
Exercise \(\PageIndex{3}\) Solutions to Linear Inequalities (Two Variables) Is the ordered pair a solution to the given inequality?
1. Yes 3. Yes 5. Yes 7. No 9. Yes Exercise \(\PageIndex{4}\) Graphing Solutions to Linear Inequalities Graph the solution set.
1. Figure \(\PageIndex{14}\) 3. Figure \(\PageIndex{15}\) 5. Figure \(\PageIndex{16}\) 7. Figure \(\PageIndex{17}\) 9. Figure \(\PageIndex{18}\) 11. Figure \(\PageIndex{19}\) 13. Figure \(\PageIndex{20}\) 15. Figure \(\PageIndex{21}\) 17. Figure \(\PageIndex{22}\) 19. Figure \(\PageIndex{23}\) 21. Figure \(\PageIndex{24}\) 23. Figure \(\PageIndex{25}\) 25. Figure \(\PageIndex{26}\) 27. Figure \(\PageIndex{27}\) 29. Figure \(\PageIndex{28}\) 31. Figure \(\PageIndex{29}\) 33. \(y>0\) 35. \(x<0\) 37. \(y≥2\) What is the linear inequality in two variables?A linear inequality in two variables is formed when symbols other than equal to, such as greater than or less than are used to relate two expressions, and two variables are involved. Here are some examples of linear inequalities in two variables: 2x<3y+27x−2y>83x+4y+3≤2y−5y+x≥0.
How do you solve linear equations in two variables?The solution of linear equations in two variables, ax+by = c, is a particular point in the graph, such that when x-coordinate is multiplied by a and y-coordinate is multiplied by b, then the sum of these two values will be equal to c. Basically, for linear equation in two variables, there are infinitely many solutions.
|