Solving For X When H(x) Is 3 Given H(x) Is -2x+5

In the realm of mathematics, solving for variables is a fundamental skill. It allows us to unravel the relationships between quantities and find unknown values. This article delves into the process of solving for x when given a function h(x) = -2x + 5 and the condition h(x) = 3. This exercise not only reinforces algebraic manipulation but also demonstrates the practical application of functions in determining specific input values for desired outputs. We will break down the steps involved, ensuring a clear understanding of the underlying principles. Let's embark on this mathematical journey together, unlocking the value of x that satisfies the given equation. Understanding how to solve such problems is crucial for building a solid foundation in algebra and calculus.

Understanding the Function h(x) = -2x + 5

Before diving into the solution, let's first understand the given function, h(x) = -2x + 5. This is a linear function, which means that when graphed, it will form a straight line. The function takes an input value x, multiplies it by -2, and then adds 5 to the result. The output of this process is denoted as h(x). Linear functions are a cornerstone of algebra, and understanding their structure is key to solving related problems. In this specific function, the coefficient -2 represents the slope of the line, indicating how steeply the line rises or falls. The constant term 5 represents the y-intercept, the point where the line crosses the vertical axis. To truly grasp the function, consider plugging in various values for x and observing the corresponding h(x) values. For instance, if x = 0, then h(0) = -2(0) + 5 = 5. If x = 1, then h(1) = -2(1) + 5 = 3. This hands-on approach helps to visualize the relationship between x and h(x). Moreover, recognizing the linear nature of the function can guide our approach to solving for x. Linear equations have well-defined properties that can be leveraged to isolate the variable we seek. This understanding of the function's behavior is paramount to effectively solving the problem at hand.

Setting up the Equation: h(x) = 3

The problem states that we need to solve for x when h(x) = 3. This means we are looking for the input value x that, when plugged into the function h(x) = -2x + 5, produces an output of 3. To find this value, we set the function equal to 3, creating the equation -2x + 5 = 3. This equation represents the core of the problem. It is a linear equation in one variable, x, and our goal is to isolate x on one side of the equation. Setting up the equation correctly is a critical step in solving any algebraic problem. It translates the problem's condition into a mathematical statement that can be manipulated and solved. The equation -2x + 5 = 3 captures the essence of the problem, stating that the output of the function h(x) must be equal to 3. From this point forward, our focus shifts to employing algebraic techniques to unravel the value of x. This involves performing operations on both sides of the equation to maintain equality while progressively isolating x. The clarity and precision of this initial equation setup lay the groundwork for a successful solution. In essence, it is the bridge between the problem's description and the mathematical solution process.

Isolating x: Step-by-Step Solution

Now that we have the equation -2x + 5 = 3, we can begin the process of isolating x. The first step is to subtract 5 from both sides of the equation. This maintains the equality and moves the constant term from the left side to the right side. Subtracting 5 from both sides gives us -2x + 5 - 5 = 3 - 5, which simplifies to -2x = -2. This step utilizes the fundamental principle of algebraic manipulation: performing the same operation on both sides of an equation preserves the balance. The next step involves dividing both sides of the equation by -2. This isolates x on the left side and provides us with its value. Dividing both sides by -2 gives us -2x / -2 = -2 / -2, which simplifies to x = 1. Therefore, the solution to the equation is x = 1. Isolating the variable is the central technique in solving equations. It involves systematically undoing the operations that are applied to the variable until it stands alone on one side of the equation. Each step in the process is carefully chosen to move us closer to the solution while preserving the equation's integrity. The result, x = 1, signifies that when the input value x is 1, the function h(x) = -2x + 5 outputs the value 3, satisfying the initial condition of the problem.

Verification of the Solution

To ensure the correctness of our solution, it is crucial to verify it. We can do this by substituting the value we found for x, which is x = 1, back into the original function h(x) = -2x + 5 and checking if it yields the expected output of 3. Substituting x = 1 into the function, we get h(1) = -2(1) + 5. Simplifying this expression, we have h(1) = -2 + 5, which equals h(1) = 3. This confirms that our solution x = 1 is indeed correct. Verification is an essential step in problem-solving, providing confidence in the accuracy of the result. It acts as a safeguard against potential errors in the algebraic manipulation process. By plugging the solution back into the original equation, we ensure that it satisfies the given condition. In this case, the verification process clearly demonstrates that when x = 1, the function h(x) does indeed equal 3. This rigorous check reinforces the validity of our solution and underscores the importance of verification in mathematical problem-solving. It completes the process, leaving no doubt that we have accurately found the value of x that satisfies the given conditions.

Conclusion: The Value of x

In conclusion, we have successfully solved for x when h(x) = 3, given the function h(x) = -2x + 5. By setting up the equation -2x + 5 = 3, isolating x through algebraic manipulation, and verifying our solution, we have determined that the value of x that satisfies the condition is x = 1. This exercise demonstrates the power of algebraic techniques in solving for unknowns and understanding the behavior of functions. Understanding the process of solving equations is not just about finding the answer; it's about developing critical thinking and problem-solving skills that are applicable in various fields. The step-by-step approach we employed, from understanding the function to verifying the solution, highlights the importance of a systematic methodology in mathematics. The result, x = 1, is not merely a number; it represents the specific input value that, when applied to the function h(x), yields the desired output of 3. This connection between input and output is fundamental to the concept of functions and their applications. The ability to solve for variables in equations is a cornerstone of mathematical proficiency, and this example serves as a testament to its significance.