Solving Linear Equations Finding Equivalent Solutions For 12w-27=-34+14w
In the realm of mathematics, particularly in algebra, solving equations is a fundamental skill. Linear equations, with their straightforward structure, serve as the building blocks for more complex mathematical concepts. Understanding how to manipulate and solve these equations is crucial for various applications, from basic problem-solving to advanced scientific modeling. This article delves into the process of solving a given linear equation and identifying other equations that share the same solution. We will explore the underlying principles, step-by-step methods, and practical considerations involved in this essential mathematical task.
H2: The Given Equation: A Starting Point
Our journey begins with the equation:
12w - 27 = -34 + 14w
This equation is a linear equation in one variable, w. Our primary goal is to isolate w on one side of the equation to determine its value. This involves applying algebraic operations to both sides of the equation while maintaining the equality. Before diving into the solution process, it’s important to recognize the key components of this equation. We have terms involving the variable w (12w and 14w), and constant terms (-27 and -34). The strategy for solving this equation will involve grouping like terms together.
To solve this equation effectively, we'll follow a systematic approach. First, we'll aim to gather all the terms containing the variable w on one side of the equation. This can be achieved by adding or subtracting terms from both sides. In this specific equation, we might choose to subtract 12w from both sides to move the w terms to the right side. Alternatively, we could subtract 14w from both sides to move them to the left. The choice often comes down to personal preference or aiming to minimize negative coefficients. Next, we'll collect the constant terms on the opposite side of the equation. This is done by adding or subtracting constant values from both sides. The goal is to isolate the variable term and the constant term on separate sides of the equation. Once we have the variable term and the constant term isolated, we can determine the value of w by dividing both sides of the equation by the coefficient of w. This final step reveals the solution to the equation. Throughout this process, it's crucial to remember that any operation performed on one side of the equation must also be performed on the other side to maintain balance and ensure the equality remains valid. This principle is the cornerstone of solving algebraic equations.
H2: Solving the Equation Step-by-Step
H3: Step 1: Rearrange the terms
To begin, let's subtract 12w from both sides of the equation:
12w - 27 - 12w = -34 + 14w - 12w
This simplifies to:
-27 = -34 + 2w
H3: Step 2: Isolate the variable term
Next, we add 34 to both sides to isolate the term with w:
-27 + 34 = -34 + 2w + 34
This gives us:
7 = 2w
H3: Step 3: Solve for w
Finally, we divide both sides by 2 to solve for w:
7 / 2 = 2w / 2
Therefore,
w = 7/2
So, the solution to the given equation is w = 7/2 or 3.5. This value is the key to identifying equivalent equations. Any equation that shares this solution is considered equivalent to the original equation. The process of solving for w involved a series of algebraic manipulations, each designed to simplify the equation while preserving its fundamental equality. Subtracting 12w from both sides allowed us to consolidate the w terms on one side. Adding 34 to both sides then isolated the term containing w. Finally, dividing by 2 revealed the value of w. Each step is crucial, and understanding the rationale behind each manipulation is essential for mastering equation-solving skills. This methodical approach not only leads to the correct solution but also builds a strong foundation for tackling more complex algebraic problems.
H2: Identifying Equivalent Equations
Now, we need to determine which of the given options has the same solution, w = 7/2.
H3: Option A
The first option is:
9/7 w = 13/14 w + 2
Let's solve this equation for w.
H4: Step 1: Eliminate fractions
Multiply both sides by the least common multiple (LCM) of 7 and 14, which is 14:
14 * (9/7 w) = 14 * (13/14 w + 2)
This simplifies to:
18w = 13w + 28
H4: Step 2: Isolate the variable term
Subtract 13w from both sides:
18w - 13w = 13w + 28 - 13w
This gives us:
5w = 28
H4: Step 3: Solve for w
Divide both sides by 5:
w = 28/5
Since 28/5 is not equal to 7/2, Option A is not equivalent.
H3: Option B
The second option is:
-5/6 w + 2 = 13/3 - 1/6 w
Let's solve this equation for w.
H4: Step 1: Eliminate fractions
Multiply both sides by the LCM of 6 and 3, which is 6:
6 * (-5/6 w + 2) = 6 * (13/3 - 1/6 w)
This simplifies to:
-5w + 12 = 26 - w
H4: Step 2: Rearrange the terms
Add 5w to both sides:
-5w + 12 + 5w = 26 - w + 5w
This gives us:
12 = 26 + 4w
H4: Step 3: Isolate the variable term
Subtract 26 from both sides:
12 - 26 = 26 + 4w - 26
This gives us:
-14 = 4w
H4: Step 4: Solve for w
Divide both sides by 4:
w = -14/4 = -7/2
Since -7/2 is not equal to 7/2, Option B is not equivalent. The process of solving these equations highlights the importance of careful algebraic manipulation. Each step, from eliminating fractions to isolating the variable, must be executed precisely to arrive at the correct solution. The comparison of the solutions obtained for Options A and B with the solution of the original equation (w = 7/2) demonstrates that neither of these options is equivalent. This underscores the need to solve each equation independently to determine its solution and then compare it with the solution of the given equation. The methodical approach used here, involving LCM multiplication to clear fractions and strategic addition and subtraction to isolate the variable, is a valuable technique for solving a wide range of linear equations. This skill is not only essential for academic success but also for various practical applications where equations need to be solved accurately.
H2: Conclusion
In conclusion, we solved the given equation 12w - 27 = -34 + 14w and found the solution to be w = 7/2. By solving the other equations, we determined that neither Option A nor Option B has the same solution. This exercise demonstrates the importance of accurately solving equations and comparing solutions to identify equivalent equations. Mastering the techniques of solving linear equations is a critical skill in mathematics, laying the foundation for more advanced algebraic concepts and problem-solving scenarios. The ability to manipulate equations, isolate variables, and identify equivalent forms is not only valuable in academic settings but also in various real-world applications where mathematical models are used to represent and solve problems.
