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 equality 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 ( ≤ or ≥ ), graph a solid line.
Finally, pick one point that is not on either 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.
Graph each of the inequalities in the system in a similar way. The solution of the system of inequalities is the intersection region of all the solutions in the system.
Example 1:
Solve the system of inequalities by graphing:
y ≤ x − 2 y > − 3 x + 5
First, graph the inequality y ≤ x − 2 . The related equation is y = x − 2 .
Since the inequality is ≤ , not a strict one, the border line is solid.
Graph the straight line.
Consider a point that is not on the line - say, ( 0 , 0 ) - and substitute in the inequality y ≤ x − 2 .
0 ≤ 0 − 2 0 ≤ − 2
This is false. So, the solution does not contain the point ( 0 , 0 ) . Shade the lower half of the line.
Similarly, draw a dashed line for the related equation of the second inequality y > − 3 x + 5 which has a strict inequality. The point ( 0 , 0 ) does not satisfy the inequality, so shade the half that does not contain the point ( 0 , 0 ) .
The solution of the system of inequalities is the intersection region of the solutions of the two inequalities.
Example 2:
Solve the system of inequalities by graphing:
2 x + 3 y ≥ 12 8 x − 4 y > 1 x < 4
Rewrite the first two inequalities with y alone on one side.
3 y ≥ − 2 x + 12 y ≥ − 2 3 x + 4 − 4 y > − 8 x + 1 y < 2 x − 1 4
Now, graph the inequality y ≥ − 2 3 x + 4 . The related equation is y = − 2 3 x + 4 .
Since the inequality is ≥ , not a strict one, the border line is solid.
Graph the straight line.
Consider a point that is not on the line - say, ( 0 , 0 ) - and substitute in the inequality.
0 ≥ − 2 3 ( 0 ) + 4 0 ≥ 4
This is false. So, the solution does not contain the point ( 0 , 0 ) . Shade upper half of the line.
Similarly, draw a dashed line of related equation of the second inequality y < 2 x − 1 4 which has a strict inequality. The point ( 0 , 0 ) does not satisfy the inequality, so shade the half that does not contain the point ( 0 , 0 ) .
Draw a dashed vertical line x = 4 which is the related equation of the third inequality.
Here point ( 0 , 0 ) satisfies the inequality, so shade the half that contains the point.
The solution of the system of inequalities is the intersection region of the solutions of the three inequalities.
Learning Objectives
Solutions to Systems of Linear Inequalities
A system of linear inequalities consists of a set of two or more linear inequalities with the same variables. The inequalities define the conditions that are to be considered simultaneously. For example,
We know that each inequality in the set contains infinitely many ordered pair solutions defined by a region in a rectangular coordinate plane. When considering two of these inequalities together, the intersection of these sets defines the set of simultaneous ordered pair solutions. When we graph each of the above inequalities separately, we have
\(y>x-2\) \(y\leq 2x+2\)When graphed on the same set of axes, the intersection can be determined.
The intersection is shaded darker and the final graph of the solution set is presented as follows:
The graph suggests that \((3, 2)\) is a solution because it is in the intersection. To verify this, show that it solves both of the original inequalities:
\(\color{Cerulean}{Check:}\:\:\color{black}{(3,2)}\)
\(\begin{array}{c|c}{Inequality\:1:\quad y>x-2}&{Inequality\:2:\quad y/leq 2x+2}\\{2>3-2}&{2\leq 2(3)+2}\\{2>1\quad\color{Cerulean}{\checkmark}}&{2\leq 8\quad\color{Cerulean}{\checkmark}} \end{array}\)
Points on the solid boundary are included in the set of simultaneous solutions and points on the dashed boundary are not. Consider the point \((−1, 0)\) on the solid boundary defined by \(y=2x+2\) and verify that it solves the original system:
\(\color{Cerulean}{Check:}\:\:\color{black}{(-1,0)}\)
\(\begin{array}{c|c}{Inequality\:1:\quad y>x-2}&{Inequality\:2:\quad y\leq 2x+2}\\{0>-1-2}&{0\leq 2(-1)+2}\\{0>-3\quad\color{Cerulean}{\checkmark}}&{0\leq 0\quad\color{Cerulean}{\checkmark}} \end{array}\)
Notice that this point satisfies both inequalities and thus is included in the solution set. Now consider the point \((2, 0)\) on the dashed boundary defined by \(y=x−2\) and verify that it does not solve the original system:
\(\color{Cerulean}{Check:}\:\:\color{black}{(2,0)}\)
\(\begin{array}{c|c}{Inequality\:1:\quad y>x-2}&{Inequality\:2:\quad y\leq 2x+2}\\{0>2-2}&{0\leq 2(2)+2}\\{0>0\quad\color{red}{x}}&{0\leq\quad\color{Cerulean}{\checkmark}} \end{array}\)
This point does not satisfy both inequalities and thus is not included in the solution set.
Solving Systems of Linear Inequalities
Solutions to a system of linear inequalities are the ordered pairs that solve all the inequalities in the system. Therefore, to solve these systems, graph the solution sets of the inequalities on the same set of axes and determine where they intersect. This intersection, or overlap, defines the region of common ordered pair solutions.
Example \(\PageIndex{1}\)
Graph the solution set:
\(\left\{\begin{aligned} −2x+y&>−4 \\ 3x−6y& ≥6 \end{aligned}\right.\).
Solution:
To facilitate the graphing process, we first solve for \(y\).
2x-4 \\ y&\leq \frac{1}{2}x-1 \end{aligned}\right.\)
For the first inequality, we use a dashed boundary defined by \(y=2x−4\) and shade all points above the line. For the second inequality, we use a solid boundary defined by \(y=\frac{1}{2}x−1\) and shade all points below. The intersection is darkened.
Figure \(\PageIndex{5}\)Now we present the solution with only the intersection shaded.
Answer:
Example \(\PageIndex{2}\)
Graph the solution set:
\(\left\{\begin{aligned} −2x+3y&>6 \\ 4x−6y&>12 \end{aligned}\right.\).
Solution:
Begin by solving both inequalities for \(y\).
Use a dashed line for each boundary. For the first inequality, shade all points above the boundary. For the second inequality, shade all points below the boundary.
Figure \(\PageIndex{7}\)As you can see, there is no intersection of these two shaded regions. Therefore, there are no simultaneous solutions.
Answer:
No solution, \(∅\)
Example \(\PageIndex{3}\)
Graph the solution set:
Solution:
Figure \(\PageIndex{8}\)After graphing all three inequalities on the same set of axes, we determine that the intersection lies in the triangular region pictured.
Answer:
The graphic suggests that \((−1, 1)\) is a common point. As a check, substitute that point into the inequalities and verify that it solves all three conditions.
\(\color{Cerulean}{Check:}\:\:\color{black}{(-1,1)}\)
\(\begin{array}{c|c|c} {Inequality\:1:}&{Inequality\:2:}&{Inequality\:3:}\\{y\geq -4}&{y<x+3}&{y/leq -3x+3}\\{1\geq -4\quad\color{Cerulean}{\checkmark}}&{1<-1+3}&{1\leq -3(-1)+3}\\{}&{1<2\quad\color{Cerulean}{\checkmark}}&{1\leq 3+3}\\{}&{}&{1\leq 6\quad\color{Cerulean}{\checkmark}} \end{array}\)
Key Takeaways
- To solve systems of linear inequalities, graph the solution sets of each inequality on the same set of axes and determine where they intersect.
Exercise \(\PageIndex{1}\) Solving Systems of Linear Inequalities
Determine whether the given point is a solution to the given system of linear equations.
- \((3, 2)\); \(\left\{\begin{aligned} y&≤x+3\\y&≥−x+3\end{aligned}\right.\)
- \((−3, −2)\); \(\left\{\begin{aligned} y&<-3x+4\\y&\geq 2x-1\end{aligned}\right.\)
- \((5,0)\); \(\left\{\begin{aligned} y&>−x+5\\y&≤\frac{3}{4}x−2\end{aligned}\right.\)
- \((0, 1)\); \(\left\{\begin{aligned} y&<\frac{2}{3}x+1\\y&≥\frac{5}{2}x−2\end{aligned}\right.\)
- \((−1, \frac{8}{3})\); \(\left\{\begin{aligned}−4x+3y&≥−12\\2x+3y&<6 \end{aligned}\right.\)
- \((−1, −2)\); \(\left\{\begin{aligned}−x+y&<0\\x+y&<0\\x+y&<−2 \end{aligned}\right.\)
1. Yes
3. No
5. No
Exercise \(\PageIndex{2}\) Solving Systems of Linear Inequalities
Graph the solution set.
- \(\left\{\begin{aligned} y&\leq x+3 \\ y&\geq -x+3\end{aligned}\right.\)
- \(\left\{\begin{aligned} y&<-3x+4 \\ y&\geq 2x-1\end{aligned}\right.\)
- \(\left\{\begin{aligned} y&>x \\ y&<-1 \end{aligned}\right.\)
- \(\left\{\begin{aligned} y&<\frac{2}{3}x+1\\ y& ≥\frac{5}{2}x−2 \end{aligned}\right.\)
- \(\left\{\begin{aligned} y&>−x+5\\y&≤\frac{3}{4}x−2 \end{aligned}\right.\)
- \(\left\{\begin{aligned} y&>\frac{3}{5}x+3\\y&<\frac{3}{5}x−3 \end{aligned}\right.\)
- \(\left\{\begin{aligned} x+4y&<12\\−3x+12y&≥−12 \end{aligned}\right.\)
- \(\left\{\begin{aligned} −x+y&≤6\\2x+y&≥1 \end{aligned}\right.\)
- \(\left\{\begin{aligned} −2x+3y&>3\\4x−3y&<15 \end{aligned}\right.\)
- \(\left\{\begin{aligned} −4x+3y&≥−12\\2x+3y&<6 \end{aligned}\right.\)
- \(\left\{\begin{aligned} 5x+y&≤4\\−4x+3y&<−6 \end{aligned}\right.\)
- \(\left\{\begin{aligned} 3x+5y&<15\\−x+2y&≤0\end{aligned}\right.\)
- \(\left\{\begin{aligned} x&≥0\\5x+y&>5\end{aligned}\right.\)
- \(\left\{\begin{aligned} x&≥−2\\y&≥1\end{aligned}\right.\)
- \(\left\{\begin{aligned} x−3&<0\\y+2&≥0\end{aligned}\right.\)
- \(\left\{\begin{aligned} 5y&≥2x+5\\−2x&<−5y−5\end{aligned}\right.\)
- \(\left\{\begin{aligned} x−y&≥0\\−x+y&<1\end{aligned}\right.\)
- \(\left\{\begin{aligned} −x+y&≥0\\y−x&<1\end{aligned}\right.\)
- \(\left\{\begin{aligned} x&>−2\\x&≤2\end{aligned}\right.\)
- \(\left\{\begin{aligned} y&>−1\\y&<2\end{aligned}\right.\)
- \(\left\{\begin{aligned} −x+2y&>8\\3x−6y&≥18\end{aligned}\right.\)
- \(\left\{\begin{aligned} −3x+4y&≤4\\6x−8y&>−8\end{aligned}\right.\)
- \(\left\{\begin{aligned} 2x+y&<3\\−x&≤12y\end{aligned}\right.\)
- \(\left\{\begin{aligned} 2x+6y&≤6\\−13x−y&≤3\end{aligned}\right.\)
- \(\left\{\begin{aligned} y&<3\\y&>x\\x&>-4\end{aligned}\right.\)
- \(\left\{\begin{aligned} y&<1\\y&\geq x-1\\y&<-3x+3\end{aligned}\right.\)
- \(\left\{\begin{aligned} -4x+3y&>-12\\y&\geq 2\\2x+3y&>6\end{aligned}\right.\)
- \(\left\{\begin{aligned} -x+y&<0\\x+y&\leq 0\\x+y&>-2\end{aligned}\right.\)
- \(\left\{\begin{aligned} x+y&<2\\x&<3\\-x+y&\leq 2\end{aligned}\right.\)
- \(\left\{\begin{aligned} y+4&\geq 0\\ \frac{1}{2}x+\frac{1}{3}y&\leq 1\\-\frac{1}{2}x+\frac{1}{3}y&\leq 1\end{aligned}\right.\)
- Construct a system of linear inequalities that describes all points in the first quadrant.
- Construct a system of linear inequalities that describes all points in the second quadrant.
- Construct a system of linear inequalities that describes all points in the third quadrant.
- Construct a system of linear inequalities that describes all points in the fourth quadrant.
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21. No solution, \(∅\)
23.
25.
27.
29.
31. \(\left\{\begin{aligned} x&>0 \\ y&>0 \end{aligned}\right.\)
33. \(\left\{\begin{aligned} x&<0 \\y&<0 \end{aligned}\right.\)