Solving For W In The Equation (2x-1)/y = (w+2)/(2z) A Step-by-Step Guide

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Hey there, math enthusiasts! Today, we're diving into a fun algebraic problem where we need to isolate a specific variable. Specifically, we are going to focus on solving for w in the equation $\frac{2x-1}{y} = \frac{w+2}{2z}$. This type of problem is a classic example of manipulating equations, a fundamental skill in algebra and beyond. Understanding how to rearrange equations not only helps in solving mathematical problems but also sharpens your logical thinking and problem-solving abilities in general. So, grab your thinking caps, and let's get started!

Understanding the Equation

Before we jump into the steps, let's take a moment to understand what the equation is telling us. We have a fraction on the left side, $\ rac{2x-1}{y}$, and another fraction on the right side, $\ rac{w+2}{2z}$. The equals sign tells us that these two fractions are equivalent. Our goal is to get w all by itself on one side of the equation. To do this, we'll use a series of algebraic manipulations, ensuring that we maintain the balance of the equation at every step. Think of it like a balancing scale – whatever we do to one side, we must also do to the other.

Key Concepts: Inverse Operations

The core idea behind solving equations is using inverse operations. Inverse operations are operations that undo each other. For example, addition and subtraction are inverse operations, and multiplication and division are inverse operations. When we want to isolate a variable, we use inverse operations to "undo" the operations that are being applied to it. For instance, if w is being added to 2, we'll subtract 2 to isolate w. Similarly, if w is being divided by a number, we'll multiply by that number to isolate w. Mastering these inverse operations is key to confidently maneuvering through algebraic equations.

Variables and Constants

In this equation, we have several variables: x, y, w, and z. Variables are symbols that represent unknown values. We also have constants, which are fixed values (like the numbers 1 and 2 in our equation). When we solve for w, we want to express w in terms of the other variables. This means our final answer will likely involve x, y, and z. Keeping track of which variables we're solving for and which ones we're treating as knowns is crucial in avoiding confusion and ensuring accuracy.

Step-by-Step Solution

Alright, let's break down the solution into manageable steps. We'll go through each step in detail, explaining the reasoning behind it. Remember, the key is to maintain the balance of the equation throughout the process. Let's go!

Step 1: Multiply Both Sides by 2z

Our first goal is to get rid of the fraction on the right side of the equation. To do this, we'll multiply both sides of the equation by 2z. This is an application of the multiplication property of equality, which states that if you multiply both sides of an equation by the same non-zero value, the equation remains balanced. So, we have:

2x−1y∗2z=w+22z∗2z \frac{2x-1}{y} * 2z = \frac{w+2}{2z} * 2z

On the right side, the 2z in the numerator and denominator cancel each other out, leaving us with:

2z(2x−1)y=w+2 \frac{2z(2x-1)}{y} = w + 2

This step is crucial because it eliminates the denominator on the side with w, making it easier to isolate.

Step 2: Simplify the Left Side

Now, let's simplify the left side of the equation. We can distribute the 2z in the numerator:

4xz−2zy=w+2 \frac{4xz - 2z}{y} = w + 2

This step just tidies up the equation, making it visually simpler and easier to work with. It's always a good practice to simplify expressions whenever possible, as it reduces the chances of making errors in subsequent steps.

Step 3: Subtract 2 from Both Sides

The last step to isolate w is to get rid of the +2 on the right side. We can do this by subtracting 2 from both sides of the equation. This is an application of the subtraction property of equality, which is similar to the multiplication property we used earlier:

4xz−2zy−2=w+2−2 \frac{4xz - 2z}{y} - 2 = w + 2 - 2

On the right side, the +2 and -2 cancel each other out, leaving us with just w:

4xz−2zy−2=w \frac{4xz - 2z}{y} - 2 = w

Step 4: (Optional) Combine Terms on the Left Side

We've successfully isolated w, but we can optionally combine the terms on the left side to express the answer as a single fraction. To do this, we need to rewrite 2 as a fraction with a denominator of y:

2=2yy 2 = \frac{2y}{y}

Now we can subtract the fractions:

4xz−2zy−2yy=w \frac{4xz - 2z}{y} - \frac{2y}{y} = w

Combine the numerators:

4xz−2z−2yy=w \frac{4xz - 2z - 2y}{y} = w

So, our final answer is:

w=4xz−2z−2yy w = \frac{4xz - 2z - 2y}{y}

This step is optional, as the previous form of the answer ( $\ rac{4xz - 2z}{y} - 2 = w$ ) is also perfectly correct. However, expressing the answer as a single fraction can sometimes be useful in further calculations or when comparing results.

Final Answer

Therefore, the solution for w is:

w=4xz−2z−2yy w = \frac{4xz - 2z - 2y}{y}

Or, equivalently:

w=4xz−2zy−2 w = \frac{4xz - 2z}{y} - 2

Common Mistakes to Avoid

When solving equations like this, there are a few common mistakes that students often make. Being aware of these pitfalls can help you avoid them and ensure you get the correct answer. Here are some key things to watch out for:

Forgetting to Apply Operations to Both Sides

One of the most frequent errors is only applying an operation to one side of the equation. Remember, the equals sign represents a balance, so whatever you do to one side, you must also do to the other. For example, if you multiply the left side by 2z, you must also multiply the right side by 2z. Failing to do so will throw off the balance and lead to an incorrect solution.

Incorrectly Cancelling Terms

Another common mistake is incorrectly canceling terms. Cancellation is a powerful tool, but it must be used correctly. You can only cancel factors that are multiplied, not terms that are added or subtracted. For example, in the expression $\ rac{2z(2x-1)}{y}$, you can't cancel the 2z with the y because y is not a factor of the entire numerator. Similarly, in the step where we subtracted 2 from both sides, it's crucial to remember that we are subtracting the entire quantity, not just a part of it.

Distributing Negatives Incorrectly

When dealing with negative signs, it's easy to make mistakes. If you have a negative sign in front of a parenthesis, remember to distribute it to every term inside the parenthesis. For instance, if you had an expression like $-(a + b)$, it's important to distribute the negative sign to both a and b, resulting in $-a - b$. Neglecting to do this can lead to sign errors, which can significantly alter the solution.

Combining Unlike Terms

A fundamental rule of algebra is that you can only combine like terms. Like terms are terms that have the same variable raised to the same power. For example, 3x and 5x are like terms, but 3x and 3x² are not. Trying to combine unlike terms is a common mistake that can lead to incorrect simplifications. In our equation, we made sure to only combine terms that had the same variables and exponents.

Order of Operations (PEMDAS/BODMAS)

Following the correct order of operations is crucial in simplifying expressions. Remember the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction) or BODMAS (Brackets, Orders, Division and Multiplication, Addition and Subtraction). These acronyms remind you of the order in which operations should be performed. For example, multiplication and division should be done before addition and subtraction. Failing to follow the order of operations can lead to incorrect results.

Not Simplifying Completely

It's always a good practice to simplify your answer as much as possible. This means combining like terms, reducing fractions, and removing any unnecessary parentheses. A fully simplified answer is not only cleaner but also easier to work with in further calculations. In our solution, we took the optional step of combining the terms on the left side to express the answer as a single fraction, which is a form of simplification.

Real-World Applications

While solving for variables in equations might seem like an abstract mathematical exercise, it has numerous real-world applications. The ability to manipulate equations is essential in various fields, including physics, engineering, economics, and computer science. Let's explore a few examples:

Physics

In physics, many formulas relate different physical quantities. For example, the formula for the force (F) acting on an object is given by F = ma, where m is the mass of the object and a is its acceleration. If you know the force and the mass, you can solve for the acceleration by rearranging the equation to a = F/m. This skill is crucial for analyzing motion, calculating trajectories, and understanding the behavior of objects under various forces.

Engineering

Engineers frequently use equations to design structures, circuits, and systems. For instance, in electrical engineering, Ohm's Law states that the voltage (V) across a conductor is equal to the current (I) flowing through it times the resistance (R), or V = IR. If you need to determine the resistance required for a specific voltage and current, you would solve for R by rearranging the equation to R = V/I. This type of algebraic manipulation is essential for designing safe and efficient electrical systems.

Economics

Economics involves many mathematical models that describe relationships between economic variables. For example, the quantity demanded (Qd) of a product might depend on its price (P), income (I), and other factors. If you have an equation that models this relationship, you might need to solve for the price that would lead to a specific quantity demanded. Rearranging equations is a fundamental skill for economists analyzing markets, forecasting trends, and making policy recommendations.

Computer Science

In computer science, equations are used in algorithm design, data analysis, and machine learning. For example, in machine learning, models are often trained by minimizing a cost function, which is an equation that measures the error of the model's predictions. Solving for the parameters that minimize the cost function often involves rearranging equations and using optimization techniques. The ability to manipulate equations is crucial for developing effective machine learning algorithms.

Everyday Life

Even in everyday life, we often use algebraic thinking to solve problems. For instance, if you're planning a road trip and want to calculate how long it will take to reach your destination, you might use the formula time = distance / speed. If you know the distance and your desired travel time, you can solve for the required speed by rearranging the equation to speed = distance / time. This type of problem-solving ability is valuable in many practical situations.

Practice Problems

To solidify your understanding of solving for variables in equations, let's try a few practice problems. These problems will give you the opportunity to apply the steps we've discussed and build your confidence. Remember, practice makes perfect!

Practice Problem 1

Solve for a in the equation:

3a+25=b−12 \frac{3a + 2}{5} = \frac{b - 1}{2}

Practice Problem 2

Solve for y in the equation:

x=2y−3z4 x = \frac{2y - 3z}{4}

Practice Problem 3

Solve for r in the equation:

A=Ï€r2 A = \pi r^2

(Assume A and $\\pi$ are known constants.)

Take your time to work through these problems, and don't hesitate to refer back to the steps we discussed earlier. The more you practice, the more comfortable you'll become with rearranging equations and solving for variables.

Conclusion

So there you have it, guys! We've successfully solved for w in the equation $\frac{2x-1}{y} = \frac{w+2}{2z}$. We've also explored the importance of inverse operations, common mistakes to avoid, real-world applications, and some practice problems to help you hone your skills. Solving for variables is a fundamental skill in algebra, and mastering it will open doors to more advanced mathematical concepts and applications. Keep practicing, and you'll become a pro at rearranging equations in no time!