Solving For X In The Equation Y = (2/5)(x - 28) - 40
Introduction
In this article, we will delve into the process of solving for the variable x in a given linear equation. Specifically, we are given the equation y = (2/5)(x - 28) - 40 and the value of y is provided as 82. Our goal is to find the corresponding value of x that satisfies this equation. This involves using algebraic manipulation to isolate x on one side of the equation. This article provides a step-by-step guide that simplifies the problem so anyone can understand it, regardless of their math proficiency. Understanding how to solve for variables in linear equations is a fundamental skill in algebra and is essential for various applications in mathematics, science, and engineering. This article will not only walk you through the solution but also explain the underlying principles, ensuring you grasp the concepts thoroughly. Linear equations are the building blocks of more complex mathematical models, and mastering them is crucial for further studies in quantitative fields. Furthermore, the ability to manipulate equations and solve for unknowns is a valuable problem-solving skill that extends beyond the classroom, applicable in real-world scenarios such as budgeting, planning, and decision-making.
Problem Statement
We are given the equation:
y = \frac{2}{5}(x - 28) - 40
We are also given that y = 82. Our task is to find the value of x.
Step-by-Step Solution
Step 1: Substitute the Value of y
First, substitute the given value of y into the equation:
82 = \frac{2}{5}(x - 28) - 40
Substituting the known value is the initial step in solving for the unknown variable. This transforms the equation into a single-variable equation, making it easier to manipulate and solve. By replacing y with its numerical value, we establish a clear relationship between the numerical value and the remaining unknown, x. This substitution is crucial because it allows us to focus on isolating x using algebraic operations. This foundational step is common in solving various types of equations and is a critical skill in mathematical problem-solving. It helps to simplify complex problems into manageable steps, providing a clear path towards the solution. In essence, substitution is about replacing a variable with its equivalent value to simplify the equation and make it solvable.
Step 2: Isolate the Term with x
Next, we want to isolate the term that contains x. To do this, we add 40 to both sides of the equation:
82 + 40 = \frac{2}{5}(x - 28) - 40 + 40
122 = \frac{2}{5}(x - 28)
Isolating the term with x is a crucial step in solving the equation. This process involves moving all other terms to the opposite side of the equation, which sets the stage for solving for x directly. By adding 40 to both sides, we effectively cancel out the -40 on the right side, thereby simplifying the equation. This step is guided by the fundamental principle of algebraic manipulation: maintaining the balance of the equation by performing the same operation on both sides. The goal is to create an equation where the term containing x is alone on one side, making it easier to apply further steps to isolate x itself. This isolation process is a cornerstone of algebraic problem-solving and is used extensively in various mathematical contexts.
Step 3: Multiply by the Reciprocal
To get rid of the fraction, we multiply both sides of the equation by the reciprocal of 2/5, which is 5/2:
122 \cdot \frac{5}{2} = \frac{2}{5}(x - 28) \cdot \frac{5}{2}
\frac{122 \cdot 5}{2} = x - 28
305 = x - 28
Multiplying by the reciprocal is a key technique in solving equations involving fractions. The reciprocal of a fraction is obtained by swapping its numerator and denominator. By multiplying both sides of the equation by the reciprocal of the fraction attached to the x term, we effectively cancel out the fraction, simplifying the equation and bringing us closer to isolating x. This step is based on the multiplicative inverse property, which states that a number multiplied by its reciprocal equals 1. This action helps eliminate the fractional coefficient, making the equation easier to manage and solve. Multiplying by the reciprocal is a common and efficient method in algebra, especially when dealing with equations where a variable is multiplied or divided by a fraction.
Step 4: Isolate x
Now, to isolate x, we add 28 to both sides of the equation:
305 + 28 = x - 28 + 28
333 = x
Isolating x is the ultimate goal in solving the equation, and this step achieves that by undoing any remaining operations attached to x. By adding 28 to both sides, we cancel out the -28 on the right side, leaving x alone. This step is a direct application of the addition property of equality, which ensures that adding the same value to both sides of the equation maintains the balance. Once x is isolated, we have found its value, which satisfies the original equation. This isolation process is a fundamental concept in algebra and is crucial for solving various types of equations. It represents the final step in determining the value of the unknown variable and completing the problem.
Step 5: State the Solution
Thus, the value of x is:
x = 333
Stating the solution clearly is an essential final step in solving any mathematical problem. After performing the algebraic manipulations to isolate x, we arrive at a specific numerical value, which is 333 in this case. Explicitly stating the solution ensures that the answer is clearly communicated and understood. It provides a definitive resolution to the problem, confirming the value of x that satisfies the original equation. This final statement is crucial for completeness and clarity, leaving no ambiguity about the result of the problem-solving process. It also allows for easy verification of the solution by substituting it back into the original equation to ensure it holds true.
Verification
To verify our solution, we substitute x = 333 back into the original equation:
y = \frac{2}{5}(333 - 28) - 40
y = \frac{2}{5}(305) - 40
y = 2 \cdot 61 - 40
y = 122 - 40
y = 82
Since this matches the given value of y, our solution is correct.
Verification is a critical step in the problem-solving process, serving as a safeguard to ensure the accuracy of the solution. By substituting the calculated value of x back into the original equation, we can confirm whether the equation holds true. This process involves replacing x with its numerical value and simplifying the expression to see if it equals the given value of y. If the left-hand side of the equation equals the right-hand side after substitution, it validates that the solution is correct. This verification step not only provides confidence in the answer but also helps identify any potential errors made during the solution process. It reinforces the importance of precision and thoroughness in mathematical problem-solving.
Conclusion
We have successfully solved for x in the equation y = (2/5)(x - 28) - 40 when y = 82. The value of x is 333. This step-by-step solution demonstrates the process of using algebraic manipulation to isolate a variable and solve a linear equation. Understanding these techniques is crucial for further studies in mathematics and related fields.
In conclusion, the process of solving for x in a linear equation involves several key steps: substitution, isolating terms, and applying inverse operations. These steps are fundamental in algebra and are used extensively in solving various types of mathematical problems. The ability to manipulate equations and solve for unknowns is a valuable skill that extends beyond the classroom, applicable in many real-world scenarios. By following a systematic approach and verifying the solution, we can ensure accuracy and confidence in our results. This article has provided a clear and detailed explanation of the solution process, reinforcing the understanding of algebraic principles and problem-solving strategies.
