Solving [x+2] + [x-3] - [x+4] = 3: A Step-by-Step Guide

Oct 14, 2025 by ADMIN 56 views

Hey guys! Today, we're diving into a really cool math problem: finding the real numbers x that satisfy the equation [x+2] + [x-3] - [x+4] = 3. This involves understanding the floor function, which might seem a bit tricky at first, but trust me, we'll break it down together. So, grab your pencils, and let's get started!

Understanding the Floor Function

Before we jump into solving the equation, let's make sure we're all on the same page about the floor function. The floor function, denoted by [x], gives you the greatest integer less than or equal to x. Think of it as rounding down to the nearest whole number. For example, [3.14] = 3, [5] = 5, and [-2.7] = -3. It's super important to grasp this concept because it's the key to unlocking our equation.

The floor function, often denoted as [x], plays a crucial role in various mathematical contexts, including number theory, real analysis, and computer science. To truly master solving equations involving the floor function, it's essential to develop a deep understanding of its properties and behavior. The floor function essentially maps a real number to the greatest integer that is less than or equal to that number. For instance, if we consider the number 4.7, the floor function [4.7] would return 4 because 4 is the greatest integer less than or equal to 4.7. Similarly, for a negative number like -2.3, the floor function [-2.3] would return -3, as -3 is the greatest integer less than or equal to -2.3. Understanding how the floor function handles both positive and negative numbers is crucial for accurately solving equations. When dealing with equations containing floor functions, it's often beneficial to consider different intervals for x and analyze how the floor function behaves within each interval. This approach allows us to break down the problem into smaller, more manageable parts, making it easier to identify the solutions. Furthermore, recognizing the properties of the floor function, such as its behavior with inequalities and its relationship with integer values, can significantly simplify the solution process. For instance, if we know that [x] = n, where n is an integer, then we can deduce that n ≤ x < n + 1. This type of deduction is invaluable when solving equations and inequalities involving floor functions. By thoroughly understanding these aspects, you'll be well-equipped to tackle a wide range of problems involving the floor function and enhance your problem-solving skills in mathematics.

Breaking Down the Equation

Now, let's tackle the equation: [x+2] + [x-3] - [x+4] = 3. The trick here is to realize that the floor function changes its value only at integer points. So, we need to consider intervals where the expressions inside the floor functions (x+2, x-3, and x+4) cross integer values. This means we'll be looking at intervals defined by integers related to -2, 3, and -4. Think of these as critical points that will help us break down the problem into smaller, more manageable chunks.

To effectively solve the equation [x+2] + [x-3] - [x+4] = 3, we need to consider the intervals where the expressions inside the floor functions, namely x+2, x-3, and x+4, transition between integer values. These transition points are crucial because the value of the floor function changes only at integer boundaries. In our equation, the critical points are derived from the integers related to -2, 3, and -4, which are the constants added to x within the floor functions. These critical points help us partition the real number line into distinct intervals, each of which we can analyze separately to determine the behavior of the floor functions. For instance, if we consider the expression x+2, it becomes an integer when x is an integer minus 2. Similarly, x-3 becomes an integer when x is an integer plus 3, and x+4 becomes an integer when x is an integer minus 4. By identifying these critical points, we can create intervals such as (-∞, -4), [-4, -3), [-3, -2), and so on. Within each of these intervals, the floor functions [x+2], [x-3], and [x+4] will have constant integer values, which simplifies the equation and allows us to solve for x. This approach of dividing the problem into intervals based on the critical points of the floor functions is a fundamental technique for solving equations of this type. It enables us to transform a complex equation involving floor functions into a series of simpler equations that are much easier to handle. By systematically analyzing each interval, we can identify all possible solutions for x that satisfy the original equation. This method not only helps in finding the solutions but also provides a clear understanding of how the floor functions behave across different ranges of x. Mastering this technique is invaluable for tackling more advanced problems involving floor functions and other piecewise-defined functions.

Defining the Intervals

Based on these critical points, we can define the following intervals: (-∞, -4), [-4, -3), [-3, -2), [-2, -1), [-1, 0), [0, 1), [1, 2), [2, 3), [3, 4), [4, ∞). We'll analyze the equation in each of these intervals to see where it holds true. This might seem like a lot of work, but it's a systematic way to ensure we don't miss any solutions. Remember, math is all about being thorough and precise!

To systematically solve the equation [x+2] + [x-3] - [x+4] = 3, defining the appropriate intervals is a crucial step. These intervals are determined by the points where the expressions inside the floor functions (x+2, x-3, and x+4) become integers, as the floor function's value changes only at integer values. By identifying these critical points, we can partition the real number line into segments within which the floor functions behave predictably. For our equation, the relevant expressions are x+2, x-3, and x+4. Setting each of these equal to integers will give us the boundaries of our intervals. Specifically, we need to consider the integers related to -2, 3, and -4, which are the constants added to x within the floor functions. This leads us to consider intervals based on integers from -4 onwards. For example, we start with the interval (-∞, -4), where x is less than -4. Then we move to [-4, -3), where x is greater than or equal to -4 but less than -3, and so on. We continue this process, creating intervals that cover the entire range of real numbers. This systematic approach ensures that we account for every possible value of x and that we don't miss any potential solutions. The intervals we define are: (-∞, -4), [-4, -3), [-3, -2), [-2, -1), [-1, 0), [0, 1), [1, 2), [2, 3), [3, 4), and [4, ∞). Analyzing the equation within each of these intervals allows us to replace the floor functions with constant integer values, transforming the original equation into a series of simpler equations. This step-by-step method is essential for solving equations involving floor functions accurately and efficiently. By carefully defining and analyzing each interval, we can identify all solutions for x that satisfy the given equation.

Solving in Each Interval

Now comes the fun part: solving the equation in each interval. Let's start with the interval [-4, -3). In this interval, we have:

[x+2] = -2
[x-3] = -7
[x+4] = 0

Plugging these values into the equation, we get:

-2 + (-7) - 0 = 3

-9 = 3

This is not true, so there are no solutions in this interval. We'll repeat this process for each interval, carefully calculating the floor values and checking if the equation holds. It's like detective work, but with numbers!

Solving the equation [x+2] + [x-3] - [x+4] = 3 within each interval is a methodical process that involves substituting the floor functions with their corresponding integer values and then checking if the resulting equation is satisfied. This approach transforms the original equation into a series of simpler equations that are much easier to solve. Let's illustrate this process by examining the interval [-4, -3). In this interval, x is greater than or equal to -4 but less than -3. We need to determine the values of [x+2], [x-3], and [x+4] for any x within this interval. For [x+2], since x is between -4 and -3, x+2 will be between -2 and -1. Therefore, [x+2] = -2. For [x-3], x being between -4 and -3 means x-3 will be between -7 and -6. Thus, [x-3] = -7. Finally, for [x+4], x being between -4 and -3 means x+4 will be between 0 and 1. So, [x+4] = 0. Now, we substitute these values back into the original equation: [x+2] + [x-3] - [x+4] = 3 becomes -2 + (-7) - 0 = 3, which simplifies to -9 = 3. This is clearly false, indicating that there are no solutions for x within the interval [-4, -3). We repeat this process for each of the intervals we defined earlier. By systematically analyzing each interval, we can determine whether the equation holds true for any x within that interval. If the equation does not hold true, as we saw in the interval [-4, -3), then there are no solutions in that interval. If the equation does hold true for a certain interval, we have found a range of values for x that satisfy the original equation. This step-by-step method ensures that we exhaustively search for all possible solutions and provides a clear and organized way to solve equations involving floor functions.

Finding the Solution

After going through all the intervals (and I won't bore you with the calculations for each one, but you should definitely try it yourself!), you'll find that the equation holds true in the interval [3, 4). In this interval:

[x+2] = 5
[x-3] = 0
[x+4] = 7

Plugging these in:

5 + 0 - 7 = 3

This simplifies to -2 = 3, which is not correct, I made a mistake here, let's calculate again the correct values for the interval [3,4):

[x+2] = 5
[x-3] = 0
[x+4] = 7

So the equation becomes:

5 + 0 - 7 = -2, and it is not equal to 3, so there is no solution in the interval [3,4) as well.

Let's check the interval [4,5):

[x+2] = 6
[x-3] = 1
[x+4] = 8

6 + 1 - 8 = -1 != 3, so it is not the correct solution.

Let's analyze the interval where x is within [2, 3):

[x + 2] will be [2 + 2, 3 + 2) which is [4, 5), so [x + 2] = 4
[x - 3] will be [2 - 3, 3 - 3) which is [-1, 0), so [x - 3] = -1
[x + 4] will be [2 + 4, 3 + 4) which is [6, 7), so [x + 4] = 6