Solving Compound Inequalities A Step By Step Guide

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In mathematics, compound inequalities represent a combination of two or more inequalities connected by the words "and" or "or." Solving compound inequalities involves finding the values that satisfy all the inequalities simultaneously (in the case of "and") or at least one of the inequalities (in the case of "or"). This guide provides a step-by-step approach to solving compound inequalities, with a focus on expressing the solution in interval notation.

Understanding Compound Inequalities

Before diving into the solution process, it's crucial to understand the different types of compound inequalities:

  • "And" Inequalities: These inequalities require that all conditions be true simultaneously. The solution set includes only the values that satisfy all the individual inequalities.
  • "Or" Inequalities: These inequalities require that at least one condition be true. The solution set includes the values that satisfy any of the individual inequalities.

Solving "And" Inequalities

Let's consider the compound inequality: 3y+6ext≤−9extand2y−3<53y + 6 ext{ ≤ } -9 ext{ and } 2y - 3 < 5. This is an example of an "and" inequality, meaning we need to find the values of y that satisfy both inequalities.

Step 1: Solve Each Inequality Separately

First, we solve each inequality independently:

Inequality 1: 3y+6ext≤−93y + 6 ext{ ≤ } -9

  1. Subtract 6 from both sides: 3yext≤−9−63y ext{ ≤ } -9 - 6 3yext≤−153y ext{ ≤ } -15
  2. Divide both sides by 3: yext≤−5y ext{ ≤ } -5

Inequality 2: 2y−3<52y - 3 < 5

  1. Add 3 to both sides: 2y<5+32y < 5 + 3 2y<82y < 8
  2. Divide both sides by 2: y<4y < 4

Step 2: Determine the Intersection

Since this is an "and" inequality, we need to find the values of y that satisfy both yext≤−5y ext{ ≤ } -5 and y<4y < 4. This means we are looking for the intersection of the two solution sets.

Step 3: Express the Solution in Interval Notation

  • The solution to yext≤−5y ext{ ≤ } -5 is the interval (−∞,−5](-\infty, -5].
  • The solution to y<4y < 4 is the interval (−∞,4)(-\infty, 4).

The intersection of these two intervals is the set of all numbers that are both less than or equal to -5 and less than 4. This can be written as (−∞,−5](-\infty, -5].

Therefore, the solution to the compound inequality 3y+6ext≤−9extand2y−3<53y + 6 ext{ ≤ } -9 ext{ and } 2y - 3 < 5 is (−∞,−5](-\infty, -5].

Interval Notation Explained

Interval notation is a concise way to represent a set of numbers. It uses parentheses and brackets to indicate whether the endpoints are included in the set.

  • Parentheses ( ): Indicate that the endpoint is not included in the set. For example, (a,b)(a, b) represents all numbers between a and b, but not a or b.
  • Brackets [ ]: Indicate that the endpoint is included in the set. For example, [a,b][a, b] represents all numbers between a and b, including a and b.
  • Infinity Symbols (∞,−∞)(\infty, -\infty): These symbols represent unbounded intervals. Infinity is not a number, so it is always enclosed in parentheses.

Examples:

  • x>3x > 3: (3,∞)(3, \infty)
  • xext≤−2x ext{ ≤ } -2: (−∞,−2](-\infty, -2]
  • −1<xext≤5-1 < x ext{ ≤ } 5: (−1,5](-1, 5]

Solving "Or" Inequalities

"Or" inequalities require that at least one of the conditions be true. The solution set includes all values that satisfy either inequality or both.

Step 1: Solve Each Inequality Separately

As with "and" inequalities, the first step is to solve each inequality independently.

Step 2: Determine the Union

Since this is an "or" inequality, we need to find the values that satisfy either one or both of the inequalities. This means we are looking for the union of the two solution sets.

Step 3: Express the Solution in Interval Notation

To find the union, combine the intervals, including all values from both intervals. If the intervals overlap, the union will be a single interval. If they do not overlap, the union will be the combination of two separate intervals.

Example:

Solve the compound inequality: x−2>1extor2x+3ext≤−5x - 2 > 1 ext{ or } 2x + 3 ext{ ≤ } -5

Inequality 1: x−2>1x - 2 > 1

  1. Add 2 to both sides: x>3x > 3
  2. Interval Notation: (3,∞)(3, \infty)

Inequality 2: 2x+3ext≤−52x + 3 ext{ ≤ } -5

  1. Subtract 3 from both sides: 2xext≤−82x ext{ ≤ } -8
  2. Divide both sides by 2: xext≤−4x ext{ ≤ } -4
  3. Interval Notation: (−∞,−4](-\infty, -4]

Union of the Intervals

The solution is the union of (3,∞)(3, \infty) and (−∞,−4](-\infty, -4], which can be written as (−∞,−4]∪(3,∞)(-\infty, -4] \cup (3, \infty).

Special Cases and No Solution

No Solution

Sometimes, a compound inequality may have no solution. This occurs when the inequalities contradict each other.

Example:

Solve: x>5extandx<2x > 5 ext{ and } x < 2

There is no number that is both greater than 5 and less than 2. Therefore, the solution set is empty, denoted by ∅\varnothing.

All Real Numbers

In other cases, the solution to a compound inequality may be all real numbers.

Example:

Solve: x>2extorx<5x > 2 ext{ or } x < 5

Any real number will satisfy at least one of these inequalities. Therefore, the solution set is (−∞,∞)(-\infty, \infty).

Tips for Solving Compound Inequalities

  1. Solve Each Inequality Separately: Always begin by solving each inequality independently.
  2. Determine the Correct Operation (And/Or): Identify whether the compound inequality uses "and" (intersection) or "or" (union).
  3. Graph the Solutions: Graphing the solutions on a number line can help visualize the intersection or union.
  4. Express the Solution in Interval Notation: Write the final solution set using interval notation.
  5. Check for Special Cases: Be mindful of cases where there is no solution or the solution is all real numbers.

Examples with Detailed Solutions

Example 1: "And" Inequality

Solve: −3<2x+1ext≤5-3 < 2x + 1 ext{ ≤ } 5

Step 1: Break into Two Inequalities

This inequality can be rewritten as two separate inequalities:

  • −3<2x+1-3 < 2x + 1
  • 2x+1ext≤52x + 1 ext{ ≤ } 5

Step 2: Solve Each Inequality

Inequality 1: −3<2x+1-3 < 2x + 1
  1. Subtract 1 from both sides: −4<2x-4 < 2x
  2. Divide both sides by 2: −2<x-2 < x
Inequality 2: 2x+1ext≤52x + 1 ext{ ≤ } 5
  1. Subtract 1 from both sides: 2xext≤42x ext{ ≤ } 4
  2. Divide both sides by 2: xext≤2x ext{ ≤ } 2

Step 3: Determine the Intersection

We need to find the values of x that satisfy both −2<x-2 < x and xext≤2x ext{ ≤ } 2.

Step 4: Express the Solution in Interval Notation

  • −2<x-2 < x is the interval (−2,∞)(-2, \infty).
  • xext≤2x ext{ ≤ } 2 is the interval (−∞,2](-\infty, 2].

The intersection of these intervals is the set of all numbers greater than -2 and less than or equal to 2, which can be written as (−2,2](-2, 2].

Solution: (−2,2](-2, 2]

Example 2: "Or" Inequality

Solve: 4x−1<7extor3x+5ext≥144x - 1 < 7 ext{ or } 3x + 5 ext{ ≥ } 14

Step 1: Solve Each Inequality

Inequality 1: 4x−1<74x - 1 < 7
  1. Add 1 to both sides: 4x<84x < 8
  2. Divide both sides by 4: x<2x < 2
Inequality 2: 3x+5ext≥143x + 5 ext{ ≥ } 14
  1. Subtract 5 from both sides: 3xext≥93x ext{ ≥ } 9
  2. Divide both sides by 3: xext≥3x ext{ ≥ } 3

Step 2: Determine the Union

We need to find the values of x that satisfy either x<2x < 2 or xext≥3x ext{ ≥ } 3.

Step 3: Express the Solution in Interval Notation

  • x<2x < 2 is the interval (−∞,2)(-\infty, 2).
  • xext≥3x ext{ ≥ } 3 is the interval [3,∞)[3, \infty).

The union of these intervals is the set of all numbers less than 2 or greater than or equal to 3, which can be written as (−∞,2)∪[3,∞)(-\infty, 2) \cup [3, \infty).

Solution: (−∞,2)∪[3,∞)(-\infty, 2) \cup [3, \infty)

Conclusion

Solving compound inequalities requires a systematic approach, including solving each inequality separately, determining the appropriate operation (and/or), and expressing the solution in interval notation. By following these steps and practicing with various examples, you can master the art of solving compound inequalities. Remember to always double-check your solutions and be mindful of special cases and no solutions. This comprehensive guide aims to provide a solid foundation for understanding and solving compound inequalities effectively.

By mastering these concepts, students can confidently tackle more complex mathematical problems and excel in their mathematical journey.