3.4 Linear Inequalities
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Learning Objectives
- Understand what linear inequalities are and their notation
- Solve linear inequalities in one variable
- Represent solutions on a number line
- Solve systems of linear inequalities
- Apply linear inequalities to solve real-world problems
Key Concepts
Introduction to Linear Inequalities
Linear inequalities are mathematical statements that compare expressions using inequality symbols: < (less than), > (greater than), ≤ (less than or equal to), or ≥ (greater than or equal to).
Unlike equations which have specific solutions, inequalities have a range of solutions that satisfy the given condition.
Linear inequalities contain variables with a maximum power of 1, such as 2x + 3 < 7 or 5 - x ≥ 2x + 1.
Solving Linear Inequalities in One Variable
Solving linear inequalities follows similar steps to solving linear equations, with one important difference: when multiplying or dividing both sides by a negative number, the inequality sign must be reversed.
The goal is to isolate the variable on one side of the inequality.
The solution to a linear inequality is typically expressed as an interval or represented on a number line.
Example 1
Solve the inequality: 3x - 5 < 7
Solution:
Step 1: Isolate the variable term. 3x - 5 < 7 3x < 12 Step 2: Divide both sides by 3 (positive, so inequality sign stays the same). x < 4 The solution is x < 4, which means all real numbers less than 4 satisfy the original inequality.
Example 2
Solve the inequality: 2 - 4x ≥ 10
Solution:
Step 1: Isolate the variable term. 2 - 4x ≥ 10 -4x ≥ 8 Step 2: Divide both sides by -4 (negative, so inequality sign must be reversed). x ≤ -2 The solution is x ≤ -2, which means all real numbers less than or equal to -2 satisfy the original inequality.
Representing Solutions on a Number Line
Solutions to linear inequalities can be visualized on a number line.
For strict inequalities (< or >), use an open circle at the boundary point to indicate it's not included in the solution.
For non-strict inequalities (≤ or ≥), use a closed circle (or filled dot) at the boundary point to indicate it's included in the solution.
Example 1
Represent the solution to x < 4 on a number line.
Solution:
To represent x < 4 on a number line: 1. Locate the point 4 on the number line. 2. Draw an open circle at 4 to indicate that 4 itself is not part of the solution. 3. Draw an arrow extending to the left from 4 to indicate all values less than 4. The solution includes all real numbers to the left of 4, but not including 4 itself.
Example 2
Represent the solution to x ≥ -2 on a number line.
Solution:
To represent x ≥ -2 on a number line: 1. Locate the point -2 on the number line. 2. Draw a closed circle (filled dot) at -2 to indicate that -2 itself is part of the solution. 3. Draw an arrow extending to the right from -2 to indicate all values greater than -2. The solution includes -2 and all real numbers to the right of -2.
Solving Compound Inequalities
Compound inequalities involve two or more inequalities joined by 'and' or 'or'.
For 'and' compound inequalities (intersection), both conditions must be satisfied simultaneously.
For 'or' compound inequalities (union), at least one of the conditions must be satisfied.
Example 1
Solve the compound inequality: 2 < 3x - 1 < 8
Solution:
Step 1: This is a three-part inequality that means 2 < 3x - 1 AND 3x - 1 < 8. We can solve it in one go. 2 < 3x - 1 < 8 Step 2: Add 1 to all parts to isolate the term with x. 3 < 3x < 9 Step 3: Divide all parts by 3 to isolate x. 1 < x < 3 The solution is 1 < x < 3, which means all real numbers between 1 and 3, not including 1 and 3 themselves.
Example 2
Solve the compound inequality: x - 4 ≤ -2 or 2x + 1 > 7
Solution:
Step 1: Solve each inequality separately. For x - 4 ≤ -2: x - 4 ≤ -2 x ≤ 2 For 2x + 1 > 7: 2x + 1 > 7 2x > 6 x > 3 Step 2: Combine the solutions with 'or'. x ≤ 2 or x > 3 This means all real numbers that are either less than or equal to 2, or greater than 3, satisfy the original compound inequality.
Solving Systems of Linear Inequalities
A system of linear inequalities consists of two or more inequalities that must be satisfied simultaneously.
To solve a system of linear inequalities, find the solution set for each inequality and then determine their intersection.
The solution to a system of linear inequalities is the set of all points that satisfy all the inequalities in the system.
Example 1
Solve the system of inequalities: x + 2y ≤ 6 x - y > 1
Solution:
Step 1: Solve each inequality for y to graph them. For x + 2y ≤ 6: 2y ≤ 6 - x y ≤ (6 - x)/2 y ≤ 3 - x/2 For x - y > 1: -y > 1 - x y < x - 1 Step 2: The solution is the region where both inequalities are satisfied. For y ≤ 3 - x/2, the region is below the line y = 3 - x/2. For y < x - 1, the region is below the line y = x - 1. The solution is the intersection of these two regions, which is the region below both lines.
Example 2
Solve the system of inequalities: 2x + y ≥ 4 3x - 2y ≤ 6 x ≥ 0 y ≥ 0
Solution:
Step 1: Solve each inequality for y when possible. For 2x + y ≥ 4: y ≥ 4 - 2x For 3x - 2y ≤ 6: -2y ≤ 6 - 3x y ≥ (3x - 6)/2 For x ≥ 0: This is a vertical boundary at x = 0. For y ≥ 0: This is a horizontal boundary at y = 0. Step 2: The solution is the region where all four inequalities are satisfied. This is the region above the line y = 4 - 2x, above the line y = (3x - 6)/2, to the right of the y-axis (x ≥ 0), and above the x-axis (y ≥ 0). The solution is the intersection of these four regions.
Applications of Linear Inequalities
Linear inequalities are used to model many real-world scenarios where variables are constrained but not exactly determined.
Common applications include resource allocation, budgeting, production planning, and scheduling problems.
In these applications, inequalities often represent constraints or limitations on resources.
Example 1
A company manufactures two types of products: A and B. Each product A requires 2 hours of labor and 3 units of material, while each product B requires 3 hours of labor and 2 units of material. The company has available 12 hours of labor and 12 units of material per day. Let x be the number of product A and y be the number of product B manufactured per day. Set up the inequalities that represent the constraints.
Solution:
Step 1: Identify the constraints based on the available resources. Labor constraint: The total labor hours used cannot exceed 12 hours. 2x + 3y ≤ 12 Material constraint: The total units of material used cannot exceed 12 units. 3x + 2y ≤ 12 Non-negativity constraints: The number of products cannot be negative. x ≥ 0 y ≥ 0 Step 2: The complete system of inequalities is: 2x + 3y ≤ 12 3x + 2y ≤ 12 x ≥ 0 y ≥ 0 These inequalities define the feasible region for the company's production.
Example 2
A student needs to score at least 70% to get a B grade in a course. The course has two tests worth 30% each and a final exam worth 40%. The student scored 75% and 80% on the two tests. What minimum percentage must the student score on the final exam to get a B grade?
Solution:
Step 1: Let x be the percentage the student scores on the final exam. Step 2: Set up the inequality based on the weighted average. 0.3(75) + 0.3(80) + 0.4x ≥ 70 Step 3: Simplify and solve for x. 22.5 + 24 + 0.4x ≥ 70 46.5 + 0.4x ≥ 70 0.4x ≥ 23.5 x ≥ 58.75 The student must score at least 58.75% on the final exam to get a B grade.
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