Solving 2x + 5 < X + 12 A Step-by-Step Guide

In the realm of mathematics, inequalities play a crucial role in defining ranges and boundaries. Understanding how to solve inequalities is fundamental for various mathematical applications and problem-solving scenarios. This article delves into the process of solving a specific linear inequality, 2x + 5 < x + 12, providing a step-by-step guide and exploring the underlying concepts. Mastering inequalities is essential for anyone delving into algebra and beyond.

Understanding Inequalities

Before diving into the solution, it's important to grasp the concept of inequalities. Unlike equations, which assert equality between two expressions, inequalities express a relationship where one expression is greater than, less than, greater than or equal to, or less than or equal to another. In our case, the inequality 2x + 5 < x + 12 indicates that the expression on the left side, 2x + 5, is strictly less than the expression on the right side, x + 12. The '<' symbol signifies this strict inequality, meaning the two sides cannot be equal. Understanding this fundamental distinction is key to correctly manipulating and solving inequalities.

The solutions to an inequality are not single values, but rather a range of values. For example, while the equation x + 1 = 3 has only one solution (x = 2), the inequality x + 1 < 3 has infinitely many solutions (any value of x less than 2). This is because any number less than 2, when added to 1, will result in a sum less than 3. This concept of a range of solutions is central to understanding and interpreting the results of solving inequalities. When we solve 2x + 5 < x + 12, we are essentially finding the set of all x values that make the inequality true.

The principles of solving inequalities are similar to those used for equations, but with a critical difference: multiplying or dividing by a negative number reverses the direction of the inequality sign. This is a crucial rule to remember to avoid errors. For instance, if we have -x < 2, multiplying both sides by -1 results in x > -2, not x < -2. This reversal is necessary to maintain the truth of the inequality. Consider the example -2 < 1. If we multiply both sides by -1 without flipping the sign, we get 2 < -1, which is clearly false. However, flipping the sign gives us 2 > -1, which is true. Keeping this rule in mind is essential for accurately manipulating and solving inequalities.

Step-by-Step Solution of 2x + 5 < x + 12

Now, let's break down the solution to the inequality 2x + 5 < x + 12 step-by-step. Our goal is to isolate the variable x on one side of the inequality to determine the range of values that satisfy the condition. This process involves algebraic manipulations similar to those used for solving equations, but with the crucial consideration of the inequality sign.

Step 1: Isolate the Variable Terms

The first step in solving the inequality is to group the terms containing the variable x on one side and the constant terms on the other side. This is achieved by adding or subtracting terms from both sides of the inequality. In our case, we want to get all the x terms on the left side. To do this, we subtract x from both sides of the inequality: 2x + 5 - x < x + 12 - x. This simplifies to x + 5 < 12. This step is analogous to moving terms across the equals sign in an equation, but here we maintain the inequality relationship.

Step 2: Isolate the Constant Terms

Next, we need to isolate the x term by moving the constant terms to the right side of the inequality. To do this, we subtract 5 from both sides: x + 5 - 5 < 12 - 5. This simplifies to x < 7. Now, the variable x is isolated on the left side, and we have a clear expression for the range of values that satisfy the inequality. This step mirrors the process of isolating the variable in an equation, ensuring we maintain the balance of the inequality.

Step 3: Interpret the Solution

The solution x < 7 tells us that any value of x that is strictly less than 7 will satisfy the original inequality 2x + 5 < x + 12. This is an infinite set of solutions, as there are infinitely many numbers less than 7. This is a key difference between inequalities and equations, where solutions are often single values or a limited set of values. The solution x < 7 represents a range on the number line, encompassing all numbers to the left of 7, but not including 7 itself. We can visualize this on a number line by shading the region to the left of 7 and using an open circle at 7 to indicate that 7 is not included in the solution set.

Representing the Solution

There are several ways to represent the solution x < 7. We've already touched upon the algebraic representation, but let's explore other methods to gain a more comprehensive understanding.

1. Number Line Representation

A number line is a visual tool that clearly depicts the range of solutions. To represent x < 7, draw a number line and mark the number 7. Since the inequality is strictly less than (not less than or equal to), we use an open circle at 7 to indicate that it's not included in the solution set. Then, shade the region to the left of 7, representing all numbers less than 7. This shaded region visually represents the infinite set of solutions to the inequality. This visual representation makes it easy to grasp the concept of a range of solutions.

2. Interval Notation

Interval notation is a concise way to express a range of numbers. For x < 7, we use parentheses and infinity symbols to denote the range. The solution in interval notation is written as (-∞, 7). The parenthesis indicates that 7 is not included in the interval, and -∞ signifies that the interval extends infinitely to the left. Understanding interval notation is crucial for expressing solutions to inequalities in a standardized mathematical format. This notation is commonly used in higher-level mathematics and provides a compact way to represent solution sets.

3. Set-Builder Notation

Set-builder notation provides a formal way to define a set based on a specific condition. For x < 7, the set-builder notation is {x | x < 7}. This is read as