Solving For X The Equation 2/5x - 17 = 15

This article provides a comprehensive, step-by-step solution to the algebraic equation $\frac{2}{5}x - 17 = 15$. We will explore the fundamental principles of solving linear equations, ensuring a clear understanding of each step involved. This guide is designed to help students, educators, and anyone interested in mathematics to grasp the concepts and techniques necessary to solve similar problems effectively. We will break down the problem, explaining each operation in detail, so you can confidently tackle similar algebraic challenges. Understanding how to solve for variables in equations is a crucial skill in mathematics and many real-world applications.

Understanding the Problem

The given equation is a linear equation in one variable, $x$. Our primary goal is to isolate $x$ on one side of the equation to determine its value. The equation is:

$$\frac{2}{5}x - 17 = 15$$

To solve for $x$, we need to reverse the operations that have been applied to it. This involves adding or subtracting constants and multiplying or dividing by coefficients. The order of operations in reverse – often remembered by the acronym SADMEP (Subtraction, Addition, Division, Multiplication, Exponents, Parentheses) – guides us in undoing the operations. In this specific equation, we first need to address the subtraction of 17 and then the multiplication by the fraction $\frac{2}{5}$. By carefully applying these steps, we will systematically isolate $x$ and find its value. This process is fundamental to solving algebraic equations and is a building block for more advanced mathematical concepts.

Step-by-Step Solution

Step 1: Isolate the Term with $x$

Our initial focus is to isolate the term containing $x$, which is $\frac{2}{5}x$. To achieve this, we need to eliminate the constant term, -17, from the left side of the equation. The opposite of subtraction is addition, so we add 17 to both sides of the equation. This maintains the balance of the equation, a critical principle in algebra. Adding the same value to both sides ensures that the equality remains valid. The operation can be represented as follows:

$$\frac{2}{5}x - 17 + 17 = 15 + 17$$

This simplifies to:

$$\frac{2}{5}x = 32$$

Now, the term with $x$ is isolated on one side, and we have a simpler equation to work with. This step is crucial because it brings us closer to isolating $x$ completely and finding its value. The next step involves dealing with the fractional coefficient.

Step 2: Eliminate the Fraction

Now that we have $\frac{2}{5}x = 32$, our next goal is to eliminate the fraction $\frac{2}{5}$ that is multiplying $x$. To do this, we can multiply both sides of the equation by the reciprocal of $\frac{2}{5}$, which is $\frac{5}{2}$. Multiplying by the reciprocal is equivalent to dividing by the fraction, and it effectively cancels out the fraction on the left side of the equation. This operation will leave $x$ isolated. The operation is as follows:

$$\frac{5}{2} \cdot \frac{2}{5}x = 32 \cdot \frac{5}{2}$$

On the left side, $\frac{5}{2}$ multiplied by $\frac{2}{5}$ equals 1, effectively isolating $x$. On the right side, we multiply 32 by $\frac{5}{2}$. This can be done by first multiplying 32 by 5 and then dividing by 2, or by simplifying first and then multiplying.

Step 3: Simplify and Solve for $x$

After multiplying both sides by the reciprocal, we simplify the equation:

$$1 \cdot x = 32 \cdot \frac{5}{2}$$

$$x = 32 \cdot \frac{5}{2}$$

To simplify the right side, we can multiply 32 by 5, which gives us 160. Then, we divide 160 by 2:

$$x = \frac{160}{2}$$

$$x = 80$$

Therefore, the value of $x$ that satisfies the equation is 80. This means that when we substitute 80 for $x$ in the original equation, the equation holds true. This solution is the final answer, as we have successfully isolated $x$ and determined its value.

Verification

To ensure our solution is correct, we can substitute $x = 80$ back into the original equation and verify that both sides of the equation are equal. This process, known as verification, is an essential step in problem-solving, as it helps to catch any potential errors made during the solution process. Substituting the value back into the original equation ensures that the solution satisfies the given condition.

Substituting $x = 80$ into the original equation:

$$\frac{2}{5}x - 17 = 15$$

$$\frac{2}{5}(80) - 17 = 15$$

First, we multiply $\frac{2}{5}$ by 80. This can be done by multiplying 2 by 80, which gives us 160, and then dividing by 5:

$$\frac{160}{5} - 17 = 15$$

$$32 - 17 = 15$$

Now, we subtract 17 from 32:

$$15 = 15$$

Since both sides of the equation are equal, our solution $x = 80$ is correct. This verification step confirms that the value we found for $x$ indeed satisfies the given equation. This is a crucial step in any mathematical problem-solving process, providing assurance that the answer is accurate.

Conclusion

In conclusion, the value of $x$ that satisfies the equation $\frac{2}{5}x - 17 = 15$ is 80. We arrived at this solution by systematically isolating $x$ through a series of algebraic manipulations, including adding a constant to both sides and multiplying by the reciprocal of a fraction. Each step was carefully executed to maintain the equality of the equation and ensure an accurate result. We also verified our solution by substituting $x = 80$ back into the original equation, confirming that it holds true. This step-by-step approach not only provides the correct answer but also reinforces the fundamental principles of solving linear equations. Mastering these techniques is essential for success in algebra and more advanced mathematical studies. Understanding and practicing these methods will empower you to confidently tackle a wide range of algebraic problems.

The correct answer is A. 80.

This detailed solution illustrates the importance of understanding the underlying principles of algebra and the careful application of mathematical operations. By breaking down the problem into manageable steps, we can solve even complex equations with confidence. This approach is not only valuable in academic settings but also in real-world scenarios where mathematical problem-solving is required. Remember, practice is key to mastering these skills, and each problem solved adds to your mathematical proficiency.