Writing Inequalities With Variables On Both Sides

Imagine a balancing scale, but instead of perfect equilibrium, one side is heavier or lighter than the other. That's the essence of inequalities with variables on both sides. We're dealing with situations where two expressions, each containing a variable, are not equal. One is either greater than, less than, greater than or equal to, or less than or equal to the other. Mastering these inequalities is a crucial step in algebra, opening doors to problem-solving in diverse fields like finance, engineering, and even everyday decision-making.

Understanding the Basics of Inequalities

Before diving into the intricacies of variables on both sides, let's solidify the foundation. Inequalities are mathematical statements that compare two expressions using symbols like:

• > (greater than)
• < (less than)
• ≥ (greater than or equal to)
• ≤ (less than or equal to)

A simple inequality might look like x > 5, meaning x can be any number larger than 5. The key difference between equations and inequalities lies in the solution: equations typically have a single solution (or a few), while inequalities often have a range of solutions. This range is represented visually on a number line or expressed in interval notation.

When we introduce variables on both sides, we're essentially comparing two algebraic expressions. For example, 2x + 3 < x - 1 is an inequality where x appears on both sides of the '<' symbol. Solving these inequalities involves isolating the variable on one side, similar to solving equations, but with a crucial twist we'll explore shortly.

Step-by-Step Guide to Solving Inequalities with Variables on Both Sides

The process of solving inequalities with variables on both sides mirrors solving equations, but with a critical rule to remember. Here's a breakdown of the steps:

1. Simplify Each Side:

• Distribute: If there are any parentheses, distribute any coefficients. For example, in the inequality 3(x + 2) > 2x - 1, distribute the 3 to get 3x + 6 > 2x - 1.
• Combine Like Terms: On each side of the inequality, combine any like terms. For example, in the inequality 4x + 2 - x < 5x + 7, combine the 4x and -x on the left side to get 3x + 2 < 5x + 7.

2. Isolate the Variable Term:

• Move Variable Terms to One Side: Use addition or subtraction to move all terms containing the variable to one side of the inequality. The goal is to have all the x terms on one side and the constant terms on the other. For instance, in the inequality 3x + 6 > 2x - 1, subtract 2x from both sides to get x + 6 > -1. Alternatively, in the inequality 3x + 2 < 5x + 7, subtract 3x from both sides to get 2 < 2x + 7. The choice depends on which approach minimizes negative coefficients.

3. Isolate the Constant Term:

• Move Constant Terms to the Other Side: Use addition or subtraction to move all constant terms (numbers without variables) to the side opposite the variable term. In our example of x + 6 > -1, subtract 6 from both sides to get x > -7. In the example of 2 < 2x + 7, subtract 7 from both sides to get -5 < 2x.

4. Solve for the Variable:

• Divide by the Coefficient: Divide both sides of the inequality by the coefficient of the variable. This isolates the variable and gives you the solution. In our example of x > -7, the variable is already isolated, so the solution is simply x > -7. In the example of -5 < 2x, divide both sides by 2 to get -5/2 < x, which can also be written as x > -5/2.

5. The Flipping Rule: A Critical Exception

• Multiply or Divide by a Negative Number: If, at any point during the process, you multiply or divide both sides of the inequality by a negative number, you must flip the inequality sign. This is the most crucial difference between solving equations and inequalities.

  • Why does this happen? Multiplying or dividing by a negative number reverses the order of the number line. For example, 2 < 5 is a true statement. If we multiply both sides by -1, we get -2 < -5, which is false. To make it true, we need to flip the sign: -2 > -5.

  • Example: Consider the inequality -2x < 6. To solve for x, we need to divide both sides by -2. This requires flipping the inequality sign: x > -3.

6. Express the Solution:

• Number Line: Represent the solution graphically on a number line. Use an open circle for '>' or '<' (the endpoint is not included in the solution) and a closed circle for '≥' or '≤' (the endpoint is included). Shade the region of the number line that represents the solution set.

• Interval Notation: Express the solution using interval notation. Use parentheses '(' and ')' for endpoints that are not included and brackets '[' and ']' for endpoints that are included. Use '∞' (infinity) and '-∞' (negative infinity) to represent unbounded intervals.

  • Examples:
    • x > 5 is represented as (5, ∞)
    • x ≤ -2 is represented as (-∞, -2]
    • -1 < x ≤ 3 is represented as (-1, 3]

Example Problems with Detailed Solutions

Let's solidify the process with several examples:

Example 1: Solve 4x - 7 > 2x + 3

1. Simplify: Both sides are already simplified.

2. Isolate the variable term: Subtract 2x from both sides: 4x - 2x - 7 > 2x - 2x + 3 2x - 7 > 3

3. Isolate the constant term: Add 7 to both sides: 2x - 7 + 7 > 3 + 7 2x > 10

4. Solve for the variable: Divide both sides by 2: 2x / 2 > 10 / 2 x > 5

5. Express the solution:

   • Number Line: Draw a number line. Place an open circle at 5 and shade the region to the right (representing all numbers greater than 5).
   • Interval Notation: (5, ∞)

Example 2: Solve -3(x + 2) ≤ 6 - x

1. Simplify: Distribute the -3 on the left side: -3x - 6 ≤ 6 - x

2. Isolate the variable term: Add x to both sides: -3x + x - 6 ≤ 6 - x + x -2x - 6 ≤ 6

3. Isolate the constant term: Add 6 to both sides: -2x - 6 + 6 ≤ 6 + 6 -2x ≤ 12

4. Solve for the variable: Divide both sides by -2. Remember to flip the inequality sign! -2x / -2 ≥ 12 / -2 x ≥ -6

5. Express the solution:

   • Number Line: Draw a number line. Place a closed circle at -6 and shade the region to the right (representing all numbers greater than or equal to -6).
   • Interval Notation: [-6, ∞)

Example 3: Solve 5x + 4 < 8x - 2

1. Simplify: Both sides are already simplified.

2. Isolate the variable term: Subtract 5x from both sides: 5x - 5x + 4 < 8x - 5x - 2 4 < 3x - 2

3. Isolate the constant term: Add 2 to both sides: 4 + 2 < 3x - 2 + 2 6 < 3x

4. Solve for the variable: Divide both sides by 3: 6 / 3 < 3x / 3 2 < x (which is the same as x > 2)

5. Express the solution:

   • Number Line: Draw a number line. Place an open circle at 2 and shade the region to the right (representing all numbers greater than 2).
   • Interval Notation: (2, ∞)

Example 4: Solve 2(x - 3) + 5 ≥ 4x - 7

1. Simplify: Distribute the 2 on the left side: 2x - 6 + 5 ≥ 4x - 7 2x - 1 ≥ 4x - 7

2. Isolate the variable term: Subtract 2x from both sides: 2x - 2x - 1 ≥ 4x - 2x - 7 -1 ≥ 2x - 7

3. Isolate the constant term: Add 7 to both sides: -1 + 7 ≥ 2x - 7 + 7 6 ≥ 2x

4. Solve for the variable: Divide both sides by 2: 6 / 2 ≥ 2x / 2 3 ≥ x (which is the same as x ≤ 3)

5. Express the solution:

   • Number Line: Draw a number line. Place a closed circle at 3 and shade the region to the left (representing all numbers less than or equal to 3).
   • Interval Notation: (-∞, 3]

Common Mistakes to Avoid

• Forgetting to Flip the Sign: This is the most common error. Always double-check if you've multiplied or divided by a negative number.
• Incorrect Distribution: Ensure you distribute correctly, especially when dealing with negative signs.
• Combining Unlike Terms: Only combine terms with the same variable and exponent.
• Misinterpreting the Inequality Symbol: Understand the difference between '>', '<', '≥', and '≤'.
• Errors with Negative Numbers: Be extra careful when working with negative numbers, as they can easily lead to sign errors.

Advanced Applications and Problem Solving

Solving inequalities with variables on both sides is not just an abstract algebraic exercise. It's a powerful tool for modeling and solving real-world problems. Here are some examples:

1. Budgeting:

• Problem: You have $100 to spend. You want to buy some books that cost $12 each and a video game that costs $40. How many books can you buy?
• Inequality: Let x be the number of books. The inequality is 12x + 40 ≤ 100.
• Solution:
  1. Subtract 40 from both sides: 12x ≤ 60
  2. Divide both sides by 12: x ≤ 5
• Answer: You can buy at most 5 books.

2. Business and Profit:

• Problem: A company sells widgets for $20 each. The cost to produce each widget is $8, and there are fixed costs of $1200 per month. How many widgets must the company sell to make a profit?
• Inequality: Let x be the number of widgets. Profit is revenue minus cost. The inequality is 20x - (8x + 1200) > 0.
• Solution:
  1. Simplify: 20x - 8x - 1200 > 0
  2. Combine like terms: 12x - 1200 > 0
  3. Add 1200 to both sides: 12x > 1200
  4. Divide both sides by 12: x > 100
• Answer: The company must sell more than 100 widgets to make a profit.

3. Distance, Rate, and Time:

• Problem: Two cars start from the same point and travel in opposite directions. One car travels at 60 mph, and the other travels at 45 mph. How long must they travel so that they are more than 500 miles apart?
• Inequality: Let t be the time in hours. The total distance is the sum of the distances each car travels. The inequality is 60t + 45t > 500.
• Solution:
  1. Combine like terms: 105t > 500
  2. Divide both sides by 105: t > 500/105
  3. Simplify: t > 4.76 (approximately)
• Answer: They must travel for more than 4.76 hours to be more than 500 miles apart.

4. Comparing Options:

• Problem: You are deciding between two cell phone plans. Plan A charges $30 per month plus $0.10 per minute of usage. Plan B charges $50 per month plus $0.05 per minute of usage. For how many minutes of usage is Plan A cheaper than Plan B?
• Inequality: Let m be the number of minutes. The inequality is 30 + 0.10m < 50 + 0.05m.
• Solution:
  1. Subtract 30 from both sides: 0.10m < 20 + 0.05m
  2. Subtract 0.05m from both sides: 0.05m < 20
  3. Divide both sides by 0.05: m < 400
• Answer: Plan A is cheaper than Plan B if you use less than 400 minutes per month.

These examples demonstrate that understanding and solving inequalities with variables on both sides has broad applications in various aspects of life, making it a valuable skill to develop. By carefully applying the steps outlined above and avoiding common pitfalls, you can confidently tackle a wide range of inequality problems. Remember to always double-check your work, especially when multiplying or dividing by negative numbers, to ensure the accuracy of your solutions.