Solving The Equation A Step-by-Step Guide

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Unveiling the Equation: A Detailed Breakdown

\frac{x}{2 x+2}=\frac{-2 x}{4 x+4}+ rac{2 x-3}{x+1}

This equation may appear daunting at first glance, but with a systematic approach, we can break it down and find the solution. The equation involves fractions with variables in the denominators, making it essential to carefully consider the restrictions and potential extraneous solutions. To effectively solve this equation, we will embark on a journey through a series of steps, ensuring a clear understanding of each stage. Our primary goal is to isolate the variable x on one side of the equation, revealing its true value. Along the way, we'll employ algebraic techniques such as simplification, combining like terms, and cross-multiplication, ensuring the solution's accuracy and validity. Let's begin this mathematical expedition with a commitment to clarity and precision.

Step 1: Simplify the Equation - Unveiling the Underlying Structure

To begin, let's simplify the equation by factoring out common factors in the denominators. This process will help us identify potential cancellations and simplify the overall expression. Factoring the denominators, we get:

x2(x+1)=βˆ’2x4(x+1)+2xβˆ’3x+1\frac{x}{2(x+1)} = \frac{-2x}{4(x+1)} + \frac{2x-3}{x+1}

Now, we can simplify the second term on the right side by dividing both the numerator and denominator by 2:

x2(x+1)=βˆ’x2(x+1)+2xβˆ’3x+1\frac{x}{2(x+1)} = \frac{-x}{2(x+1)} + \frac{2x-3}{x+1}

This simplification step is crucial because it reveals the common denominator of 2(x+1), which will be instrumental in the next stage of solving the equation. By factoring and simplifying, we've transformed the equation into a more manageable form, paving the way for further algebraic manipulations.

Step 2: Eliminate the Fractions - Clearing the Path to a Solution

To eliminate the fractions, we need to find the least common denominator (LCD) of all the terms in the equation. In this case, the LCD is 2(x+1). Multiplying both sides of the equation by the LCD will clear the fractions, making the equation easier to solve. Let's multiply both sides by 2(x+1):

2(x+1)βˆ—x2(x+1)=2(x+1)βˆ—(βˆ’x2(x+1)+2xβˆ’3x+1)2(x+1) * \frac{x}{2(x+1)} = 2(x+1) * (\frac{-x}{2(x+1)} + \frac{2x-3}{x+1})

On the left side, the 2(x+1) terms cancel out, leaving us with just x. On the right side, we need to distribute the 2(x+1) term to both fractions:

x=2(x+1)βˆ—βˆ’x2(x+1)+2(x+1)βˆ—2xβˆ’3x+1x = 2(x+1) * \frac{-x}{2(x+1)} + 2(x+1) * \frac{2x-3}{x+1}

Now, we can cancel out the common factors in each term:

x=βˆ’x+2(2xβˆ’3)x = -x + 2(2x-3)

This step of eliminating fractions is a pivotal moment in solving the equation. By multiplying both sides by the LCD, we've transformed the equation from one involving fractions to a simpler linear equation, making it significantly easier to manipulate and solve for x. The path to the solution is becoming clearer as we progress through each step.

Step 3: Simplify and Solve for x - Unveiling the Value of the Unknown

Now that we've eliminated the fractions, let's simplify the equation further by distributing the 2 on the right side:

x=βˆ’x+4xβˆ’6x = -x + 4x - 6

Next, combine the like terms on the right side:

x=3xβˆ’6x = 3x - 6

To isolate the x terms, subtract 3x from both sides of the equation:

xβˆ’3x=3xβˆ’6βˆ’3xx - 3x = 3x - 6 - 3x

This simplifies to:

βˆ’2x=βˆ’6-2x = -6

Finally, divide both sides by -2 to solve for x:

βˆ’2xβˆ’2=βˆ’6βˆ’2\frac{-2x}{-2} = \frac{-6}{-2}

This gives us the solution:

x=3x = 3

This step marks the culmination of our algebraic manipulations. By simplifying the equation, combining like terms, and isolating x, we've successfully determined the value of the unknown. The solution, x = 3, represents the value that satisfies the original equation. However, it's crucial to verify this solution to ensure it's not an extraneous root.

Step 4: Verify the Solution - Ensuring Accuracy and Validity

To ensure that our solution is valid, we must substitute x = 3 back into the original equation and check if it holds true. Let's substitute x = 3 into the original equation:

32(3)+2=βˆ’2(3)4(3)+4+2(3)βˆ’33+1\frac{3}{2(3)+2} = \frac{-2(3)}{4(3)+4} + \frac{2(3)-3}{3+1}

Simplify each term:

38=βˆ’616+34\frac{3}{8} = \frac{-6}{16} + \frac{3}{4}

Now, simplify the second term on the right side:

38=βˆ’38+34\frac{3}{8} = \frac{-3}{8} + \frac{3}{4}

To add the fractions on the right side, we need a common denominator, which is 8. So, we rewrite the last term:

38=βˆ’38+68\frac{3}{8} = \frac{-3}{8} + \frac{6}{8}

Now, add the fractions on the right side:

38=38\frac{3}{8} = \frac{3}{8}

Since the equation holds true, our solution x = 3 is valid and not an extraneous root.

This verification step is a critical safeguard in the problem-solving process. By substituting the solution back into the original equation, we confirm its accuracy and ensure that it satisfies the initial conditions. The fact that the equation holds true validates our solution, providing confidence in our answer.

The Answer

Therefore, the solution to the equation is:

D. x=3x=3

Conclusion: Mastering the Art of Equation Solving

In this comprehensive guide, we've meticulously dissected the process of solving the equation:

x2x+2=βˆ’2x4x+4+2xβˆ’3x+1\frac{x}{2 x+2}=\frac{-2 x}{4 x+4}+\frac{2 x-3}{x+1}

From simplifying the equation to eliminating fractions, solving for x, and verifying the solution, we've covered each step in detail. The solution, x = 3, has been confirmed as the correct answer. By understanding the underlying principles and applying the appropriate techniques, you can confidently tackle a wide range of equations. Remember, practice is key to mastering the art of equation solving. The more you practice, the more proficient you'll become in identifying the most efficient methods and avoiding potential pitfalls. Embrace the challenge of equation solving, and you'll unlock a powerful tool for mathematical exploration and problem-solving.