Solving For X: A Step-by-Step Guide To Linear Equations

In the realm of mathematics, solving for x is a fundamental skill, a cornerstone upon which more advanced concepts are built. It's the art of isolating the unknown, of deciphering the hidden value that satisfies a given equation. Today, we embark on a journey to master this art, focusing on a specific type of equation: linear equations. Linear equations, characterized by their straight-line graphs, are ubiquitous in mathematics and its applications. They appear in physics, engineering, economics, and countless other fields, making their mastery essential for anyone seeking to understand the world through a mathematical lens. This guide provides a comprehensive approach to solving linear equations, ensuring that you'll be able to tackle any problem with confidence and skill. In this article, we will delve into the step-by-step process of solving a specific linear equation. We will dissect the equation, isolate the variable x, and ultimately arrive at the solution. Furthermore, we will discuss the underlying principles that govern the solution process, providing you with a solid foundation for tackling more complex equations in the future. So, let's embark on this mathematical journey together and unravel the mysteries of solving for x.

The Equation at Hand

Our mission is to solve the following equation for x:

$$\frac{1}{2}\left(\frac{2}{3} x+10\right)=\frac{3}{5} x+1$$

This equation, while seemingly complex at first glance, is a standard linear equation. It involves fractions, parentheses, and the variable x, but with careful manipulation, we can isolate x and determine its value. The beauty of mathematics lies in its systematic approach to problem-solving. We will employ a series of algebraic operations, each step bringing us closer to the solution. These operations are based on the fundamental principle that we can perform the same operation on both sides of an equation without changing its validity. This principle ensures that the equation remains balanced, maintaining the equality between the left-hand side (LHS) and the right-hand side (RHS). Before we dive into the solution, let's take a moment to appreciate the structure of the equation. We have a fraction multiplying a parenthetical expression on the LHS, and a fraction multiplying x plus a constant on the RHS. Our goal is to simplify both sides, eliminate the fractions, and isolate x. This process requires a combination of algebraic techniques, including the distributive property, combining like terms, and the addition and subtraction properties of equality. With patience and precision, we will navigate through these steps and arrive at the solution.

Step-by-Step Solution

Step 1: Distribute the $\frac{1}{2}$ on the Left-Hand Side

The first step in solving for x is to simplify the equation by distributing the $\frac{1}{2}$ on the left-hand side (LHS). This involves multiplying $\frac{1}{2}$ by each term inside the parentheses. The distributive property, a cornerstone of algebra, allows us to multiply a single term by a group of terms enclosed in parentheses. It states that a(b + c) = ab + ac. In our case, a = $\frac{1}{2}$, b = $\frac{2}{3}x$, and c = 10. Applying the distributive property, we get:

$$\frac{1}{2} \cdot \frac{2}{3} x + \frac{1}{2} \cdot 10 = \frac{3}{5} x + 1$$

Multiplying the fractions, we have:

$$\frac{1}{3} x + 5 = \frac{3}{5} x + 1$$

This simplification eliminates the parentheses and brings us closer to isolating x. The equation now looks cleaner and more manageable. We have two terms involving x, one on each side of the equation, and constant terms on both sides as well. Our next step will involve gathering the x terms on one side and the constant terms on the other. This process utilizes the addition and subtraction properties of equality, ensuring that we maintain the balance of the equation. By performing the same operation on both sides, we can strategically move terms around until x is isolated.

Step 2: Eliminate Fractions by Multiplying by the Least Common Multiple (LCM)

Fractions can often make equations appear more complex than they actually are. To simplify our equation and make it easier to solve for x, the next step is to eliminate the fractions. We achieve this by multiplying both sides of the equation by the least common multiple (LCM) of the denominators. In our equation, the denominators are 3 and 5. The LCM of 3 and 5 is 15. This is the smallest number that is divisible by both 3 and 5. Multiplying both sides of the equation by 15 will clear the fractions, resulting in an equation with only integer coefficients. This makes the subsequent steps much easier to perform. Let's multiply both sides of the equation by 15:

$$15\left(\frac{1}{3} x + 5\right) = 15\left(\frac{3}{5} x + 1\right)$$

Now, distribute the 15 on both sides:

$$15 \cdot \frac{1}{3} x + 15 \cdot 5 = 15 \cdot \frac{3}{5} x + 15 \cdot 1$$

Simplifying, we get:

$$5x + 75 = 9x + 15$$

Notice how the fractions have disappeared, leaving us with a cleaner equation. This step is crucial in simplifying the equation and paving the way for isolating x. By multiplying by the LCM, we have transformed the equation into an equivalent form that is much easier to manipulate. Now, we can proceed with gathering the x terms on one side and the constant terms on the other, using the addition and subtraction properties of equality.

Step 3: Move the x Terms to One Side

Now that we have eliminated the fractions, our equation is:

$$5x + 75 = 9x + 15$$

To isolate x, we need to gather all the terms containing x on one side of the equation. It's generally a good practice to move the term with the smaller coefficient of x to the other side to avoid dealing with negative coefficients. In this case, 5x has a smaller coefficient than 9x. Therefore, we will subtract 5x from both sides of the equation. This is based on the subtraction property of equality, which states that subtracting the same quantity from both sides of an equation maintains the equality. Subtracting 5x from both sides, we get:

$$5x + 75 - 5x = 9x + 15 - 5x$$

Simplifying, we have:

$$75 = 4x + 15$$

Now, all the x terms are on the right-hand side of the equation, and we are closer to isolating x. The next step involves moving the constant term on the right-hand side to the left-hand side, so that we have only the x term remaining on the right. This will further simplify the equation and bring us closer to the final solution. Remember, each step we take is guided by the fundamental principles of algebra, ensuring that we maintain the balance of the equation and arrive at the correct answer.

Step 4: Move the Constant Terms to the Other Side

Our equation now stands as:

$$75 = 4x + 15$$

The next step in solving for x is to isolate the term containing x by moving the constant term to the other side of the equation. In this case, we need to move the +15 from the right-hand side (RHS) to the left-hand side (LHS). We achieve this by subtracting 15 from both sides of the equation. This is another application of the subtraction property of equality, which allows us to subtract the same quantity from both sides without altering the balance of the equation. Subtracting 15 from both sides, we get:

$$75 - 15 = 4x + 15 - 15$$

Simplifying, we have:

$$60 = 4x$$

We are now one step closer to solving for x. The constant term is isolated on the LHS, and we have a single term involving x on the RHS. The final step will involve dividing both sides of the equation by the coefficient of x to isolate x completely. This will reveal the value of x that satisfies the original equation. Each step we have taken has been a deliberate move, guided by the principles of algebra, to systematically isolate the unknown variable.

Step 5: Divide to Isolate x

We've arrived at the equation:

$$60 = 4x$$

The final step in solving for x is to isolate x completely. This is achieved by dividing both sides of the equation by the coefficient of x, which is 4 in this case. This step is based on the division property of equality, which states that dividing both sides of an equation by the same non-zero quantity maintains the equality. Dividing both sides by 4, we get:

$$\frac{60}{4} = \frac{4x}{4}$$

Simplifying, we find:

$$15 = x$$

Therefore, the solution to the equation is:

$$x = 15$$

We have successfully isolated x and determined its value. This final step, dividing by the coefficient of x, is the culmination of all the previous steps. It's the key that unlocks the value of the unknown variable. Now, we have a solution, but it's always a good practice to verify our solution to ensure that it satisfies the original equation. This step provides confidence in our answer and helps to catch any potential errors.

Verification

To ensure the accuracy of our solution, we substitute x = 15 back into the original equation:

$$\frac{1}{2}\left(\frac{2}{3} x+10\right)=\frac{3}{5} x+1$$

Substituting x = 15, we get:

$$\frac{1}{2}\left(\frac{2}{3} (15)+10\right)=\frac{3}{5} (15)+1$$

Simplifying the left-hand side (LHS):

$$\frac{1}{2}\left(10+10\right) = \frac{1}{2}(20) = 10$$

Simplifying the right-hand side (RHS):

$$\frac{3}{5} (15)+1 = 9 + 1 = 10$$

Since the LHS equals the RHS (10 = 10), our solution x = 15 is correct.

Conclusion

In this comprehensive guide, we have successfully solved the linear equation

$$\frac{1}{2}\left(\frac{2}{3} x+10\right)=\frac{3}{5} x+1$$

We systematically navigated through the steps, applying the distributive property, eliminating fractions, and isolating x. The solution we arrived at is x = 15, which we verified by substituting it back into the original equation. Solving for x is a fundamental skill in mathematics, and the techniques we have discussed here are applicable to a wide range of linear equations. The key to success lies in understanding the underlying principles of algebra and applying them methodically. Each step we take should be a deliberate move, guided by the goal of isolating the unknown variable. With practice and perseverance, you can master the art of solving for x and unlock the power of mathematics.

This journey through the solution process has not only provided us with the answer but also reinforced our understanding of the principles of algebra. We have seen how the distributive property, the properties of equality, and the concept of the least common multiple work together to simplify equations and lead us to the solution. These tools are essential for tackling more complex mathematical problems in the future. So, continue to practice, explore, and challenge yourself, and you will find that the world of mathematics becomes increasingly accessible and rewarding.