Solving For X: A Step-by-Step Guide To The Equation

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Hey everyone! Today, we're diving into a classic algebra problem: finding the value of x in the equation 3xβˆ’24βˆ’2x+13=1βˆ’xβˆ’12\frac{3x-2}{4} -\frac{2x+1}{3} =1-\frac{x-1}{2}. Don't worry, it might look a little intimidating at first glance, but trust me, we'll break it down into easy-to-follow steps. This guide is all about making math approachable, so grab a pen and paper, and let's get started. We'll walk through each step, explaining the 'why' behind the 'how,' to ensure you not only get the right answer but also understand the process. By the end of this, you'll feel confident tackling similar equations. So, let's unlock the mystery of x together!

Understanding the Equation and Our Goal

Before we jump into the calculations, let's take a moment to understand what we're dealing with. The equation 3xβˆ’24βˆ’2x+13=1βˆ’xβˆ’12\frac{3x-2}{4} -\frac{2x+1}{3} =1-\frac{x-1}{2} is a linear equation. Our goal is to isolate x on one side of the equation, meaning we want to manipulate the equation until we get something like x = [a number]. The key to solving these types of equations is to perform the same operations on both sides to maintain the balance. Think of it like a seesaw; whatever you do on one side, you must do on the other to keep it level. The initial challenge in this equation is the presence of fractions. Our first objective will be to eliminate these fractions to make the equation easier to handle. This involves finding a common denominator and multiplying each term accordingly. Once the fractions are gone, we can then focus on grouping the x terms on one side and the constant numbers on the other side. This process will systematically move us closer to finding the value of x. Remember, each step we take is designed to simplify the equation, bringing us closer to our ultimate goal: finding what x equals.

The Importance of Order of Operations

When solving equations, especially those with multiple operations, following the correct order of operations is crucial. You might remember the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). While our equation doesn't have exponents, or direct multiplications, we should keep in mind that the subtraction signs also act as multipliers of the terms after it. Applying these rules correctly ensures that each step is performed in the right order, which is essential to reaching the correct solution. For instance, when dealing with the terms in the numerator (like 3xβˆ’23x - 2), we need to make sure we treat the subtraction correctly – it's crucial for keeping the equation balanced and accurate. If you are not familiar with the order of operations, do a quick review, since this is a fundamental concept in algebra.

Step-by-Step Solution: Eliminating Fractions

Alright, let's get down to business and start solving this equation. The first, and often most important, step is to eliminate those pesky fractions. We can do this by finding the least common multiple (LCM) of the denominators: 4, 3, and 2. The LCM of 4, 3, and 2 is 12. This means we need to multiply every term in the equation by 12. Doing so will clear out the fractions, making the equation much cleaner and easier to work with. Let’s do it step by step:

  • Multiply each term by 12: 12Γ—3xβˆ’24βˆ’12Γ—2x+13=12Γ—1βˆ’12Γ—xβˆ’1212 \times \frac{3x-2}{4} - 12 \times \frac{2x+1}{3} = 12 \times 1 - 12 \times \frac{x-1}{2}

  • Simplify each term: 3(3xβˆ’2)βˆ’4(2x+1)=12βˆ’6(xβˆ’1)3(3x-2) - 4(2x+1) = 12 - 6(x-1)

  • Now, we've successfully cleared the fractions! The equation is now free of fractions and is simpler than before. From here on, our work will be focused on isolating x. You might be asking, β€œWhy are we doing this?” Because fractions complicate calculations. By removing the fractions, we simplify our equation, which in turn reduces the chances of errors and makes the equation friendlier to work with. Remember that we must apply this multiplication to every term in the equation to maintain balance.

Expanding and Simplifying the Equation

With the fractions out of the way, the next step is to expand the parentheses and simplify the equation. This involves applying the distributive property, multiplying the numbers outside the parentheses by each term inside. This is a common operation in algebra, and it's essential for simplifying expressions and moving closer to the solution. This process systematically removes the parentheses, which groups together the x terms and constants on either side of the equation. Watch carefully as we make this happen.

  • Expand the parentheses: 9xβˆ’6βˆ’8xβˆ’4=12βˆ’6x+69x - 6 - 8x - 4 = 12 - 6x + 6

  • Combine like terms on each side: (9xβˆ’8x)+(βˆ’6βˆ’4)=12+6βˆ’6x(9x - 8x) + (-6 - 4) = 12 + 6 - 6x xβˆ’10=18βˆ’6xx - 10 = 18 - 6x

  • At this stage, we have transformed our equation into a much simpler form. The key is to be meticulous with the distribution and to make sure that each term is handled correctly. If you're unsure about this step, go back and double-check your work; a minor mistake here can throw off the entire equation.

Isolating the Variable

Now that we've simplified the equation, the next step is to isolate the variable x. This means getting all the x terms on one side of the equation and the constants (numbers without x) on the other side. This is achieved by performing inverse operations. We'll add or subtract terms from both sides to move them. We're getting closer to solving this equation and determining the value of x. Let's see how this works:

  • Add 6x6x to both sides: xβˆ’10+6x=18βˆ’6x+6xx - 10 + 6x = 18 - 6x + 6x 7xβˆ’10=187x - 10 = 18

  • Add 10 to both sides: 7xβˆ’10+10=18+107x - 10 + 10 = 18 + 10 7x=287x = 28

  • Notice how, step by step, we're bringing x closer to being on its own. Adding and subtracting terms to both sides of the equation maintain the equality, and we're consistently working to isolate x. Don’t worry; we are almost there!

Solving for x: The Final Step

We are now on the final leg of our journey! The equation is now at its simplest form, with x almost entirely isolated. Now, all that's left to do is to solve for x directly. This involves performing one final operation to get x by itself, and we'll then have our answer. The last step here is often the easiest, but it's crucial to get it right. Let’s do it:

  • Divide both sides by 7: 7x7=287\frac{7x}{7} = \frac{28}{7} x=4x = 4

  • And there we have it! x = 4. We've solved the equation! This is the value of x that satisfies the original equation. It's a great feeling to solve an equation from start to finish. Always take a moment to celebrate these milestones, and congratulate yourself on a job well done!

Verification of the Solution

Before we call it a day, let's verify our solution. The most important step to ensure the validity of the solution we've found is to substitute the value back into the original equation and check if both sides are equal. This helps in confirming that no mistakes were made during any step of the process. This step is about solidifying your understanding and building confidence. Let’s confirm the answer!

  • Substitute x = 4 into the original equation: 3(4)βˆ’24βˆ’2(4)+13=1βˆ’4βˆ’12\frac{3(4)-2}{4} -\frac{2(4)+1}{3} =1-\frac{4-1}{2}

  • Simplify both sides: 12βˆ’24βˆ’8+13=1βˆ’32\frac{12-2}{4} - \frac{8+1}{3} = 1 - \frac{3}{2} 104βˆ’93=1βˆ’32\frac{10}{4} - \frac{9}{3} = 1 - \frac{3}{2} 52βˆ’3=22βˆ’32\frac{5}{2} - 3 = \frac{2}{2} - \frac{3}{2} 52βˆ’62=βˆ’12\frac{5}{2} - \frac{6}{2} = -\frac{1}{2} βˆ’12=βˆ’12-\frac{1}{2} = -\frac{1}{2}

  • Both sides of the equation are equal! This confirms that our solution x = 4 is correct. Congrats, you successfully solved the equation!

Conclusion: Mastering the Art of Solving Equations

Well, guys, we made it! We successfully found the value of x for our equation. Solving equations like these is a fundamental skill in algebra, and it opens the door to more complex mathematical concepts. Remember, the key is to break down the problem into manageable steps, understanding each operation and why it's being performed. Practice is paramount. The more you solve equations, the more confident and proficient you will become. Keep practicing, and don't be discouraged by initial challenges. If you ever get stuck, revisit the steps, and remember the principles of maintaining balance within the equation. You've got this!

Additional Tips for Success

  • Practice Regularly: Consistent practice is the best way to improve your skills. Solve a variety of equations to reinforce your understanding.
  • Review the Basics: Make sure you're comfortable with fractions, the order of operations, and the distributive property.
  • Seek Help When Needed: Don't hesitate to ask for help from teachers, tutors, or classmates when you're struggling. Explaining the problem to someone else can often clarify the solution for you.
  • Use Visual Aids: Diagrams and visual representations can make complex concepts easier to understand. Try drawing the steps out to visualize the problem.
  • Take Your Time: Rushing can lead to mistakes. Work through each step carefully and double-check your work.

Now, go out there and tackle some more equations! You are now well-equipped with the knowledge and skills to conquer similar algebraic challenges. Keep learning, keep practicing, and enjoy the journey of mastering mathematics. You got this!