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| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
Chapter 6 Linear Inequalities (Concepts)
This chapter marks a significant expansion from the world of equations, where we seek specific values that make statements of equality true, into the broader realm of Linear Inequalities. While equations deal with precise balance ($=$), inequalities explore relationships of order – situations where one quantity is less than ($<$), greater than ($>$), less than or equal to ($\le$), or greater than or equal to ($\ge$) another. This extension is crucial because many real-world constraints and conditions are expressed as inequalities rather than exact equalities (e.g., needing *at least* a certain amount, spending *no more than* a budget). We will first focus on linear inequalities involving a single variable, typically of the form $ax + b < c$ or $ax + b \ge c$, where $a, b, c$ are constants and $a \neq 0$. Unlike linear equations in one variable which usually have a single solution, linear inequalities typically have an infinite set of solutions, representing a range of values that satisfy the condition.
Solving these single-variable linear inequalities involves techniques similar to solving linear equations, but with one critical difference regarding the inequality sign. The fundamental rules are:
- Adding or subtracting the same number on both sides of an inequality does not change the direction of the inequality sign.
- Multiplying or dividing both sides by the same positive number does not change the direction of the inequality sign.
- Crucially: Multiplying or dividing both sides by the same negative number reverses the direction of the inequality sign (e.g., $<$ becomes $>$, $\ge$ becomes $\le$).
Mastering this last rule is paramount for accuracy. Once solved, the solution set for a single-variable inequality is typically represented algebraically (e.g., $x > 5$) and also graphically as a portion of the number line – often an infinite ray, possibly including or excluding the endpoint depending on whether the inequality is strict ($<, >$) or inclusive ($\le, \ge$).
The chapter then progresses to linear inequalities in two variables, commonly expressed in forms like $ax + by < c$ or $ax + by \ge c$. Here, $a, b, c$ are real constants, and $a, b$ are not both zero. The solution to such an inequality is no longer just a range of numbers on a line, but rather an entire region within the Cartesian coordinate plane. This region is specifically called a half-plane. The straight line corresponding to the equality $ax + by = c$ acts as a boundary, dividing the plane into two distinct half-planes. The solution set to the inequality consists of all the points $(x, y)$ lying in one of these half-planes.
Graphically representing the solution to a two-variable linear inequality involves a clear procedure:
- First, draw the graph of the boundary line $ax + by = c$. This line should be drawn as a dotted (or dashed) line if the inequality is strict ($<$ or $>$), indicating that points on the line itself are not part of the solution. It should be drawn as a solid line if the inequality is inclusive ($\le$ or $\ge$), signifying that points on the line are included in the solution set.
- Next, choose a convenient test point that does not lie on the boundary line (the origin $(0, 0)$ is often the easiest choice, provided the line doesn't pass through it).
- Substitute the coordinates of the test point into the original inequality.
- If the test point satisfies the inequality, then the entire half-plane containing the test point represents the solution region, and it should be shaded. If the test point does not satisfy the inequality, then the other half-plane (the one not containing the test point) is the solution region and should be shaded.
The primary focus often culminates in solving systems of linear inequalities in two variables graphically. This involves considering two or more linear inequalities simultaneously. The process requires graphing the solution region for each inequality within the system on the same coordinate plane. The overall solution to the system is the region where all the individual shaded half-planes overlap. This common, overlapping area is known as the feasible region. Every point $(x, y)$ within this feasible region (including parts of the boundary lines if the corresponding inequalities are inclusive) satisfies all the inequalities in the system concurrently. Understanding how to identify this feasible region is fundamental for tackling optimization problems in the field of Linear Programming, a crucial topic in operations research, economics, and management science.
Linear Inequalities in One Variable
In mathematics, an inequality is a statement that compares two expressions using symbols other than equality (=). These symbols indicate that one expression is greater than, less than, greater than or equal to, or less than or equal to the other. Building upon our knowledge of linear equations, we now study inequalities that involve a linear expression in a single variable.
Types of Inequalities
Inequalities can be classified based on whether they contain variables and the type of comparison they make.
1. Numerical vs. Literal Inequalities
- A Numerical Inequality is a statement that compares two numbers without any variables. It is either true or false.
Examples: $7 > 3$ (True), $5 < 2$ (False), $-1 \le 0$ (True).
- A Literal Inequality is a statement that contains at least one variable. Its truth depends on the value of the variable.
Examples: $x < 10$, $2y - 3 \ge 7$.
2. Strict vs. Slack Inequalities
- A Strict Inequality uses the symbols $<$ (less than) or $>$ (greater than). The values being compared cannot be equal.
Example: $x > 5$ means $x$ can be any number strictly greater than 5, but not 5 itself.
- A Slack Inequality uses the symbols $\le$ (less than or equal to) or $\ge$ (greater than or equal to). The values being compared can be equal.
Example: $x \le 5$ means $x$ can be 5 or any number less than 5.
Solving Inequalities
Solving an inequality means finding all the possible values of the variable that make the statement true. This collection of values is called the solution set.
- Replacement Set: This is the set of all possible numbers from which the value of the variable can be chosen. Unless specified otherwise, the replacement set is assumed to be the set of all real numbers ($\mathbb{R}$).
- Solution Set: This is the subset of the replacement set containing all the numbers that make the inequality true. For example, if the replacement set is $\{1, 2, 3, 4, 5\}$ and the inequality is $x < 4$, the solution set is $\{1, 2, 3\}$.
Inequality Symbols
The primary symbols used to express inequalities are:
- $<$ : Reads as "less than". Example: $3 < 5$.
- $>$ : Reads as "greater than". Example: $10 > 7$.
- $\le$ : Reads as "less than or equal to". Example: $x \le 4$.
- $\ge$ : Reads as "greater than or equal to". Example: $y \ge -2$.
Inequalities involving $<$ or $>$ are called strict inequalities. Those involving $\le$ or $\ge$ are called slack inequalities.
Linear Inequality in One Variable
A linear inequality in one variable is an inequality that involves a linear expression in a single variable (where the variable's highest power is 1). The general forms are:
$ax + b < 0$
$ax + b > 0$
$ax + b \le 0$
$ax + b \ge 0$
where $a$ and $b$ are real numbers and $a \neq 0$.
Examples of Linear Inequalities in One Variable
1. $2x + 3 < 7$
2. $5y - 1 \ge 9$
3. $\frac{z}{2} + 4 > 1$
Rules for Solving Linear Inequalities
Solving a linear inequality means finding its solution set—all the real numbers that make the inequality true. The process is very similar to solving linear equations, with one critical difference.
Rule 1: Addition and Subtraction
You can add or subtract the same number on both sides of an inequality without changing the direction of the inequality sign. This allows you to move terms across the inequality sign by changing their sign, just like with equations.
If $A < B$, then $A + c < B + c$.
Rule 2: Multiplication and Division by a Positive Number
You can multiply or divide both sides of an inequality by the same positive number without changing the direction of the inequality sign.
If $A < B$ and $c > 0$, then $Ac < Bc$.
Rule 3: Multiplication and Division by a Negative Number
If you multiply or divide both sides of an inequality by the same negative number, you must reverse the direction of the inequality sign.
If $A < B$ and $c < 0$, then $Ac > Bc$.
For example, if $-2x < 6$, dividing by -2 reverses the sign: $x > -3$.
Representation of Solutions
The solution set is usually an interval on the number line, which can be represented in two ways:
1. Interval Notation:
- Parentheses $( , )$ are used for endpoints that are NOT included (for $<$ and $>$).
- Square brackets $[ , ]$ are used for endpoints that ARE included (for $\le$ and $\ge$).
- Infinity symbols, $\infty$ and $-\infty$, always use parentheses.
2. Graphical Representation on a Number Line:
- An open circle ($\circ$) is used for an endpoint that is NOT included.
- A closed circle ($\bullet$) is used for an endpoint that IS included.
- The line is shaded over the range of numbers that are in the solution set.
Examples of Solution Set Representations
| Inequality | Interval Notation | Graph on Number Line |
|---|---|---|
| $x > a$ | $(a, \infty)$ | Open circle at $a$, shade to the right. |
| $x \ge a$ | $[a, \infty)$ | Closed circle at $a$, shade to the right. |
| $x < a$ | $(-\infty, a)$ | Open circle at $a$, shade to the left. |
| $x \le a$ | $(-\infty, a]$ | Closed circle at $a$, shade to the left. |
| $a < x < b$ | $(a, b)$ | Open circles at $a$ and $b$, shade between. |
| $a \le x \le b$ | $[a, b]$ | Closed circles at $a$ and $b$, shade between. |
Example 1. Solve $3(x - 1) \le 2(x - 3)$ and represent the solution on a number line.
Answer:
Given:
The linear inequality $3(x - 1) \le 2(x - 3)$.
To Solve:
Find the solution set for $x$ and represent it on a number line.
Solution:
First, expand both sides:
$3x - 3 \le 2x - 6$
Subtract $2x$ from both sides:
$x - 3 \le -6$
Add 3 to both sides:
$x \le -3$
The solution set is all real numbers less than or equal to -3.
In interval notation, this is $(-\infty, -3]$.
Graphical representation:
The final answer is $x \le -3$, or $(-\infty, -3]$.
Example 2. Solve $-2x + 5 < 11$ and represent the solution on a number line.
Answer:
Given:
The linear inequality $-2x + 5 < 11$.
To Solve:
Find the solution set for $x$ and represent it on a number line.
Solution:
Subtract 5 from both sides:
$-2x < 6$
Divide both sides by -2. Since we are dividing by a negative number, we must reverse the inequality sign:
$\frac{-2x}{-2} > \frac{6}{-2}$
(Sign reversed)
$x > -3$
The solution set is all real numbers greater than -3.
In interval notation, this is $(-3, \infty)$.
Graphical representation:
The final answer is $x > -3$, or $(-3, \infty)$.
Example 3. Solve $\frac{x}{2} - 1 \ge \frac{x}{3} + 1$ and represent the solution on a number line.
Answer:
Given:
The inequality $\frac{x}{2} - 1 \ge \frac{x}{3} + 1$.
To Solve:
Find the solution set for $x$ and represent it on a number line.
Solution:
To eliminate the fractions, we multiply both sides of the inequality by the Least Common Multiple (LCM) of the denominators (2 and 3). The LCM is 6.
Multiply the entire inequality by 6 (a positive number, so the sign does not change):
$6 \left( \frac{x}{2} - 1 \right) \ge 6 \left( \frac{x}{3} + 1 \right)$
Distribute the 6 on both sides:
$6 \cdot \frac{x}{2} - 6 \cdot 1 \ge 6 \cdot \frac{x}{3} + 6 \cdot 1$
$3x - 6 \ge 2x + 6$
Subtract $2x$ from both sides:
$x - 6 \ge 6$
Add 6 to both sides:
$x \ge 12$
The solution set is all real numbers greater than or equal to 12.
In interval notation, this is $[12, \infty)$.
Graphical representation:
The final answer is $x \ge 12$, or $[12, \infty)$.
Example 4. Solve the compound inequality $-5 \le \frac{5 - 3x}{2} \le 8$ and represent the solution on a number line.
Answer:
Given:
The compound inequality $-5 \le \frac{5 - 3x}{2} \le 8$.
To Solve:
Find the solution set for $x$ and represent it on a number line.
Solution:
We need to isolate $x$ in the middle. We perform operations on all three parts of the inequality simultaneously.
Multiply all three parts by 2 to clear the fraction:
$2(-5) \le 2\left(\frac{5 - 3x}{2}\right) \le 2(8)$
$-10 \le 5 - 3x \le 16$
Subtract 5 from all three parts:
$-10 - 5 \le 5 - 3x - 5 \le 16 - 5$
$-15 \le -3x \le 11$
Divide all three parts by -3. Remember to reverse both inequality signs:
$\frac{-15}{-3} \ge \frac{-3x}{-3} \ge \frac{11}{-3}$
(Signs reversed)
$5 \ge x \ge -\frac{11}{3}$
It is conventional to write the interval with the smaller number on the left. So, we rewrite the solution as:
$-\frac{11}{3} \le x \le 5$
In interval notation, this is $[-\frac{11}{3}, 5]$. (Note that $-\frac{11}{3} \approx -3.67$)
Graphical representation:
The final answer is $[-\frac{11}{3}, 5]$.
Solution of a System of Linear Inequalities in One Variable
Often, a situation is defined by multiple conditions at once. When these conditions are expressed as linear inequalities involving the same variable, they form a system of linear inequalities. The solution to such a system is the set of all values for the variable that satisfy every single inequality in the system simultaneously.
Method for Solving a System
To find the solution set of a system of linear inequalities in one variable, we follow these steps:
1. Solve Each Inequality Individually: First, solve each inequality in the system separately to find its individual solution set (e.g., $x < 5$ or $x \ge -2$).
2. Graph Each Solution on a Single Number Line: Draw one number line. On this line, graph the solution set for each inequality. It's helpful to use different colors or draw the graphs slightly above one another to keep them distinct.
3. Find the Intersection (Common Region): The solution to the entire system is the region on the number line where all the individual graphs overlap. This intersection represents the values of the variable that satisfy the "AND" condition—they must fulfill the first inequality AND the second inequality AND so on.
4. Write the Final Solution: Express the common region using interval notation or as a compound inequality. If there is no region where all the graphs overlap, the system has no solution, and the solution set is the empty set ($\emptyset$).
Example 1. Solve the system of linear inequalities:
$2x - 1 < 5$
... (i)
$3x + 4 \ge 10$
... (ii)
Represent the solution on a number line.
Answer:
Given:
A system of two linear inequalities:
Inequality (i): $2x - 1 < 5$
Inequality (ii): $3x + 4 \ge 10$
To Solve:
Find the solution set for the system and represent it on a number line.
Solution:
Step 1: Solve each inequality independently.
First, solve Inequality (i):
$2x - 1 < 5$
(Add 1 to both sides)
$2x < 6$
(Divide by 2)
$x < 3$
The solution set for (i) is all numbers less than 3, which is the interval $(-\infty, 3)$.
Next, solve Inequality (ii):
$3x + 4 \ge 10$
(Subtract 4 from both sides)
$3x \ge 6$
(Divide by 3)
$x \ge 2$
The solution set for (ii) is all numbers greater than or equal to 2, which is the interval $[2, \infty)$.
Step 2 & 3: Find the intersection on a number line.
We need to find the numbers $x$ that are simultaneously less than 3 ($x < 3$) AND greater than or equal to 2 ($x \ge 2$).
The common region where the two solutions overlap is from 2 (inclusive) up to 3 (exclusive).
Step 4: Write the final solution.
The intersection of $(-\infty, 3)$ and $[2, \infty)$ is the interval $[2, 3)$.
The final solution is $2 \le x < 3$, which is $[2, 3)$ in interval notation.
Example 2. Solve the system of linear inequalities: $5 - x \le 3x + 1 \le 7$.
Represent the solution on a number line.
Answer:
Given:
The compound linear inequality $5 - x \le 3x + 1 \le 7$.
To Solve:
Find the solution set for the system and represent it on a number line.
Solution:
A compound inequality of the form $A \le B \le C$ is a compact way of writing a system of two inequalities: $A \le B$ AND $B \le C$. We split it into two parts:
Inequality (i): $5 - x \le 3x + 1$
Inequality (ii): $3x + 1 \le 7$
Step 1: Solve each inequality independently.
First, solve Inequality (i):
$5 - x \le 3x + 1$
(Subtract 1 from both sides)
$4 - x \le 3x$
(Add x to both sides)
$4 \le 4x$
(Divide by 4)
$1 \le x$, which is the same as $x \ge 1$.
Next, solve Inequality (ii):
$3x + 1 \le 7$
(Subtract 1 from both sides)
$3x \le 6$
(Divide by 3)
$x \le 2$
Step 2 & 3: Find the intersection on a number line.
We need the numbers $x$ that are simultaneously greater than or equal to 1 ($x \ge 1$) AND less than or equal to 2 ($x \le 2$).
The common region is the interval from 1 to 2, including both endpoints.
Step 4: Write the final solution.
The intersection of $[1, \infty)$ and $(-\infty, 2]$ is the interval $[1, 2]$.
The final solution is $1 \le x \le 2$, which is $[1, 2]$ in interval notation.
Example 3. Solve the system of inequalities: $3(x-2) < 1$ and $4x + 8 > 0$.
Answer:
Solution:
Step 1: Solve each inequality.
Solve the first inequality:
$3(x-2) < 1$
$3x - 6 < 1$
$3x < 7$
$x < \frac{7}{3}$
Solve the second inequality:
$4x + 8 > 0$
$4x > -8$
$x > -2$
Step 2 & 3: Find the intersection.
We need numbers $x$ such that $x > -2$ and $x < \frac{7}{3}$. (Note that $\frac{7}{3} \approx 2.33$).
The common region is between -2 and $\frac{7}{3}$, excluding the endpoints.
Step 4: Write the final solution.
The solution is $-2 < x < \frac{7}{3}$, which is $(-2, \frac{7}{3})$ in interval notation.
Example 4. Solve the system of inequalities: $x - 4 \ge 0$ and $2x + 1 < 3$.
Answer:
Solution:
Step 1: Solve each inequality.
Solve the first inequality:
$x - 4 \ge 0$
$x \ge 4$
Solve the second inequality:
$2x + 1 < 3$
$2x < 2$
$x < 1$
Step 2 & 3: Find the intersection.
We are looking for numbers $x$ that are simultaneously greater than or equal to 4 ($x \ge 4$) AND less than 1 ($x < 1$).
As we can see from the number line, there is no region where the two graphs overlap. It is impossible for a number to be both greater than or equal to 4 and less than 1 at the same time.
Step 4: Write the final solution.
The system has no solution. The solution set is the empty set, denoted by $\emptyset$.
Graphical Solution of Linear Inequalities
While linear inequalities in one variable represent intervals on a 1D number line, linear inequalities in two variables represent entire regions in the 2D Cartesian plane. A linear inequality in two variables, such as $x$ and $y$, is a statement that can be written in one of the following general forms:
$ax + by < c$
$ax + by > c$
$ax + by \le c$
$ax + by \ge c$
Here, $a, b,$ and $c$ are real numbers, and $a$ and $b$ are not both zero. The solution set of such an inequality is the collection of all ordered pairs $(x, y)$ that make the inequality true. Graphically, this infinite set of points forms a region called a half-plane.
Graphical Solution of Inequalities in One Variable (in 2D)
Before graphing two-variable inequalities, it's useful to see how single-variable inequalities look on a 2D plane. These often form the boundaries in systems of inequalities.
Vertical Boundary Lines
Inequalities like $x > a$ or $x \le a$ involve only the x-variable. The boundary line is the vertical line $x=a$. The solution is the entire half-plane to the left or right of this line.
Horizontal Boundary Lines
Inequalities like $y < b$ or $y \ge b$ involve only the y-variable. The boundary line is the horizontal line $y=b$. The solution is the entire half-plane below or above this line.
Method for Graphing Linear Inequalities in Two Variables
To graph the solution set of an inequality like $ax + by < c$, we follow a systematic process to find the correct half-plane.
Step 1: Graph the Boundary Line
First, imagine the inequality is an equation: $ax + by = c$. This equation represents a straight line which will act as the "fence" or boundary of our solution region. Find two points on this line (e.g., the x and y-intercepts) to draw it.
The type of line you draw is very important:
- Draw a dashed line for strict inequalities ($<$ or $>$). This shows that points on the line are not part of the solution.
- Draw a solid line for slack inequalities ($\le$ or $\ge$). This shows that points on the line are part of the solution.
Step 2: Choose a Test Point
The boundary line divides the entire plane into two half-planes. All the points on one side of the line are solutions, and all the points on the other side are not. To find out which side is correct, we pick a simple "test point" that is not on the line.
- The origin, (0, 0), is the easiest choice, unless the boundary line passes directly through it.
- If the line passes through (0, 0), pick any other simple point like (1, 0) or (0, 1).
Step 3: Test the Point
Substitute the coordinates of your test point into the original inequality.
- If you get a true statement (e.g., $0 \le 4$), your test point is a solution. This means the entire half-plane containing the test point is the solution region.
- If you get a false statement (e.g., $0 > 6$), your test point is not a solution. This means the other half-plane (the one not containing the test point) must be the solution region.
Step 4: Shade the Solution Region
Shade the entire half-plane that you identified in Step 3. This shaded area is the graphical representation of the solution set.
Example 1. Graph the solution set of the linear inequality $2x + 3y > 6$.
Answer:
Given:
The linear inequality in two variables $2x + 3y > 6$.
To Graph:
Represent the solution set graphically in the Cartesian plane.
Solution:
Step 1: Graph the Boundary Line.
Replace $>$ with $=$ to get the boundary line equation: $2x + 3y = 6$.
Find two points on the line:
- Set $x = 0$: $2(0) + 3y = 6 \implies 3y = 6 \implies y = 2$. Point: $(0, 2)$.
- Set $y = 0$: $2x + 3(0) = 6 \implies 2x = 6 \implies x = 3$. Point: $(3, 0)$.
Since the inequality is strict ($>$), we draw a dashed line through these points.
Step 2 & 3: Test a Point.
Choose the origin $(0, 0)$ as the test point. Substitute it into the original inequality $2x + 3y > 6$:
$2(0) + 3(0) > 6$
$0 > 6$
This statement is false.
Step 4: Shade the Solution Region.
Because the test point $(0, 0)$ is not a solution, the solution region is the half-plane that does NOT contain the origin. We shade the region above the dashed line.
The final answer is the graph showing the region above the dashed line 2x + 3y = 6.
Example 2. Graph the solution set of the linear inequality $x - 2y \le 4$.
Answer:
Given:
The linear inequality $x - 2y \le 4$.
To Graph:
Represent the solution set graphically in the Cartesian plane.
Solution:
Step 1: Graph the Boundary Line.
The boundary line is $x - 2y = 4$. Find two points:
- Set $x = 0$: $0 - 2y = 4 \implies y = -2$. Point: $(0, -2)$.
- Set $y = 0$: $x - 0 = 4 \implies x = 4$. Point: $(4, 0)$.
Since the inequality is non-strict ($\le$), we draw a solid line through these points.
Step 2 & 3: Test a Point.
Choose the origin $(0, 0)$ as the test point. Substitute into the original inequality $x - 2y \le 4$:
$0 - 2(0) \le 4$
$0 \le 4$
This statement is true.
Step 4: Shade the Solution Region.
Because the test point $(0, 0)$ is a solution, the solution region is the half-plane that contains the origin. We shade the region above the solid line.
The final answer is the graph showing the region above the solid line x - 2y = 4, including the line itself.
Example 3. Graph the solution set of the linear inequality $y \ge 2x$.
Answer:
Given:
The linear inequality $y \ge 2x$.
To Graph:
Represent the solution set graphically in the Cartesian plane.
Solution:
Step 1: Graph the Boundary Line.
The boundary line is $y = 2x$. Find two points:
- If $x = 0$, then $y = 2(0) = 0$. Point: $(0, 0)$.
- If $x = 1$, then $y = 2(1) = 2$. Point: $(1, 2)$.
Since the inequality is non-strict ($\ge$), we draw a solid line through $(0, 0)$ and $(1, 2)$.
Step 2: Choose a Test Point.
The boundary line passes through the origin $(0, 0)$, so we cannot use it as a test point. We must choose another point not on the line. Let's choose the point $(1, 0)$.
Step 3: Test the Point.
Substitute the test point $(1, 0)$ into the original inequality $y \ge 2x$:
$0 \ge 2(1)$
$0 \ge 2$
This statement is false.
Step 4: Shade the Solution Region.
Since the test point $(1, 0)$ is not a solution, the solution region is the half-plane that does NOT contain the point $(1, 0)$. This is the region above and to the left of the solid line.
The final answer is the graph showing the region above the solid line y = 2x, including the line itself.
Example 4. Graph the solution set of the linear inequality $x < 4$.
Answer:
Given:
The linear inequality $x < 4$. This is an inequality in one variable, but we will graph its solution in the two-variable Cartesian plane.
To Graph:
Represent the solution set graphically in the Cartesian plane.
Solution:
Step 1: Graph the Boundary Line.
The boundary line is $x = 4$. This is a vertical line where every point has an x-coordinate of 4.
Since the inequality is strict ($<$), we draw a dashed line at $x=4$.
Step 2 & 3: Test a Point.
We can use the origin $(0, 0)$ as our test point. Substitute into the original inequality $x < 4$:
The inequality only depends on the x-coordinate.
$0 < 4$
This statement is true.
Step 4: Shade the Solution Region.
Since the test point $(0, 0)$ is a solution, we shade the entire half-plane that contains the origin. This is the region to the left of the dashed line.
The final answer is the graph showing the region to the left of the dashed vertical line x = 4.
Graphical Solution of a System of Linear Inequalities in Two Variables
A system of linear inequalities in two variables is a set of two or more linear inequalities that share the same variables (e.g., $x$ and $y$).
The solution to such a system is the set of all ordered pairs $(x, y)$ that make every single inequality in the system true at the same time. Graphically, this solution is the region where the individual solution areas of all the inequalities overlap.
This common, overlapping area is called the feasible region. Any point within this region is a valid solution to the entire system of inequalities.
Method for Graphical Solution of a System
To find the solution for a system of inequalities, we visually combine the solutions for each inequality on a single graph to find their common ground.
Step 1: Graph Each Inequality on the Same Plane
Treat each inequality in the system one by one. For each inequality:
- Draw its boundary line (solid for $\le, \ge$; dashed for $<, >$).
- Use a test point like (0,0) to determine which half-plane is the solution for that inequality.
- Instead of shading heavily, a good technique is to draw small arrows on the boundary line, pointing into the correct solution half-plane. This keeps the graph from getting messy.
Step 2: Identify the Feasible Region
Once all boundary lines and their solution directions are on the graph, look for the single region where all the arrows are pointing. This area of overlap is the feasible region.
Step 3: Shade the Feasible Region
Clearly shade the region of overlap you found in Step 2. This final shaded area is the graphical solution to the system.
Step 4 (Optional): Find the Vertices
The "corner points" of the feasible region, where the boundary lines intersect, are called vertices. These points are often the most important solutions in real-world optimization problems (e.g., finding a maximum profit). To find a vertex, solve the system of two equations corresponding to the two lines that form that corner.
Example 1. Solve the following system of linear inequalities graphically:
$x + y \le 5$
... (i)
$x - y \ge 3$
... (ii)
Answer:
Given:
A system of two linear inequalities:
(i) $x + y \le 5$
(ii) $x - y \ge 3$
To Solve:
Graph the feasible region that satisfies both inequalities.
Solution:
We will graph each inequality on the same Cartesian plane.
Inequality (i): $x + y \le 5$
- Boundary line: $x + y = 5$.
- Points: If $x=0, y=5 \implies (0, 5)$. If $y=0, x=5 \implies (5, 0)$.
- Line type: Solid line (because of $\le$).
- Test point (0, 0): $0 + 0 \le 5 \implies 0 \le 5$. This is True. The solution region is the half-plane containing the origin (below the line).
Inequality (ii): $x - y \ge 3$
- Boundary line: $x - y = 3$.
- Points: If $x=0, y=-3 \implies (0, -3)$. If $y=0, x=3 \implies (3, 0)$.
- Line type: Solid line (because of $\ge$).
- Test point (0, 0): $0 - 0 \ge 3 \implies 0 \ge 3$. This is False. The solution region is the half-plane NOT containing the origin (below the line).
Graphing and Finding the Feasible Region:
We draw both solid lines and find the region that is below $x+y=5$ AND below $x-y=3$.
The common shaded region is the feasible region for the system.
Finding the Vertex:
The vertex of the feasible region is the intersection point of the two boundary lines. We solve the system of equations:
$\begin{array}{r c l} x + y & = & 5 \\ + \quad x - y & = & 3 \\ \hline 2x \quad & = & 8 \\ \hline \end{array}$
This gives $x = 4$. Substituting into the first equation: $4 + y = 5 \implies y = 1$.
The intersection point (vertex) is $(4, 1)$.
The final answer is The feasible region is the unbounded area below both solid lines $x + y = 5$ and $x - y = 3$, with a vertex at $(4, 1)$.
Example 2. Solve the following system of linear inequalities graphically:
$x + 2y \le 8$
... (i)
$x + y \ge 4$
... (ii)
$x \ge 0$
... (iii)
$y \ge 0$
... (iv)
Answer:
Given:
A system of four linear inequalities.
To Solve:
Graph the feasible region that satisfies all inequalities.
Solution:
We graph the solution for each inequality. The inequalities $x \ge 0$ and $y \ge 0$ are known as non-negativity constraints and they simply restrict our solution to the first quadrant of the Cartesian plane.
Inequality (i): $x + 2y \le 8$
- Boundary line: $x + 2y = 8$. Points: $(0, 4)$ and $(8, 0)$.
- Line type: Solid ($\le$).
- Test point (0, 0): $0 \le 8$. True. The region is on or below this line.
Inequality (ii): $x + y \ge 4$
- Boundary line: $x + y = 4$. Points: $(0, 4)$ and $(4, 0)$.
- Line type: Solid ($\ge$).
- Test point (0, 0): $0 \ge 4$. False. The region is on or above this line.
Graphing and Finding the Feasible Region:
We need the region in the first quadrant that is below the line $x+2y=8$ and above the line $x+y=4$.
The feasible region is the quadrilateral area shown in the graph.
Finding the Vertices:
The vertices are the corners of this shaded region:
- Vertex A: Intersection of $x+y=4$ and the x-axis ($y=0$).
$x+0=4 \implies x=4$. Point: $\boldsymbol{(4, 0)}$.
- Vertex B: Intersection of $x+2y=8$ and the x-axis ($y=0$).
$x+0=8 \implies x=8$. Point: $\boldsymbol{(8, 0)}$.
- Vertex C: Intersection of $x+2y=8$ and the y-axis ($x=0$).
$0+2y=8 \implies y=4$. Point: $\boldsymbol{(0, 4)}$.
- Vertex D: Intersection of $x+y=4$ and the y-axis ($x=0$).
$0+y=4 \implies y=4$. Point: $\boldsymbol{(0, 4)}$.
Note that vertices C and D are the same point, so the feasible region is actually a triangle with vertices at (4,0), (8,0), and (0,4).
The final answer is The feasible region is the triangular area in the first quadrant with vertices at (4, 0), (8, 0), and (0, 4), including the boundary lines.
Example 3. Solve the following system of linear inequalities graphically: $x+y < 2$, $x+y \ge 4$.
Answer:
Given:
A system of two linear inequalities:
(i) $x + y < 2$
(ii) $x + y \ge 4$
To Solve:
Graph the solution set of the system.
Solution:
Inequality (i): $x + y < 2$
- Boundary line: $x + y = 2$. This line is parallel to $x+y=4$.
- Points: $(0, 2)$ and $(2, 0)$.
- Line type: Dashed line (because of $<$).
- Test point (0, 0): $0 + 0 < 2 \implies 0 < 2$. This is True. The solution is the half-plane below the line $x+y=2$.
Inequality (ii): $x + y \ge 4$
- Boundary line: $x + y = 4$.
- Points: $(0, 4)$ and $(4, 0)$.
- Line type: Solid line (because of $\ge$).
- Test point (0, 0): $0 + 0 \ge 4 \implies 0 \ge 4$. This is False. The solution is the half-plane above the line $x+y=4$.
Graphing and Finding the Feasible Region:
As seen on the graph, the solution region for the first inequality is the area below the line $x+y=2$, while the solution for the second is the area above the line $x+y=4$. These two regions are separate and have no points in common.
The final answer is There is no solution to this system of inequalities. The feasible region is the empty set ($\emptyset$).
Example 4. Find the feasible region for the system: $x+2y \le 10$, $3x+y \le 15$, $x \ge 0$, $y \ge 0$.
Answer:
Given:
A system of four linear inequalities.
To Solve:
Graph the feasible region.
Solution:
The conditions $x \ge 0$ and $y \ge 0$ restrict the solution to the first quadrant.
Inequality (i): $x + 2y \le 10$
- Boundary: $x + 2y = 10$. Points: $(0, 5), (10, 0)$.
- Type: Solid.
- Test (0,0): $0 \le 10$ (True). Solution is below the line.
Inequality (ii): $3x + y \le 15$
- Boundary: $3x + y = 15$. Points: $(0, 15), (5, 0)$.
- Type: Solid.
- Test (0,0): $0 \le 15$ (True). Solution is below the line.
Graphing and Finding the Feasible Region:
We need the area in the first quadrant that is below both lines.
The feasible region is the bounded quadrilateral shown.
Finding the Vertices:
The vertices are the corners of the shaded region:
- Vertex A: The origin, $\boldsymbol{(0, 0)}$.
- Vertex B: The x-intercept of $3x+y=15$. Point: $\boldsymbol{(5, 0)}$.
- Vertex C: The y-intercept of $x+2y=10$. Point: $\boldsymbol{(0, 5)}$.
- Vertex D: The intersection of $x + 2y = 10$ and $3x + y = 15$.
From $3x+y=15$, we get $y = 15 - 3x$. Substitute this into the first equation:
$x + 2(15 - 3x) = 10$
$x + 30 - 6x = 10$
$-5x = -20 \implies x = 4$.
Now find y: $y = 15 - 3(4) = 15 - 12 = 3$. Point: $\boldsymbol{(4, 3)}$.
The final answer is The feasible region is the bounded quadrilateral area in the first quadrant with vertices at (0, 0), (5, 0), (4, 3), and (0, 5).