Solve For P: 6 - 2(q - 3p) = 4p

In the realm of mathematics, equations are the bedrock upon which we build our understanding of relationships between variables. Solving for a specific variable, like $p$ in the equation $6-2(q-3 p)=4 p$, is a fundamental skill that allows us to isolate and understand the influence of that variable. This process involves a series of algebraic manipulations designed to get the target variable all by itself on one side of the equation. It's akin to untangling a knot, where each step carefully loosens the connections until the desired string is free. We'll embark on this journey together, breaking down each step to ensure clarity and build confidence in your algebraic problem-solving abilities. Our goal is not just to find the value of $p$, but to understand the logic behind each transformation, making you a more adept and confident mathematician. Remember, practice is key, and by working through this example, you'll be better equipped to tackle similar problems in the future. So, let's dive in and unravel the mystery of $p$!

Understanding the Equation: $6-2(q-3 p)=4 p$

Before we start manipulating the equation, let's take a moment to understand the structure of the given equation: $6-2(q-3 p)=4 p$. This is a linear equation with two variables, $q$ and $p$. Our primary objective is to isolate $p$. The equation involves constants (6 and 4), a variable $q$, and our target variable $p$. The term $-2(q-3 p)$ indicates multiplication and subtraction, requiring us to use the distributive property to simplify it. The presence of $p$ on both sides of the equation means we'll need to gather all terms containing $p$ on one side and all other terms on the opposite side. This is a standard approach for solving linear equations. The distributive property is a crucial tool here; it states that $a(b+c) = ab + ac$. In our case, $a = -2$, $b = q$, and $c = -3p$. Applying this correctly is the first major step towards isolating $p$. Pay close attention to the signs, as a common pitfall is mishandling the negative sign during distribution. Once distributed, the equation will look simpler, paving the way for further algebraic maneuvers. We are essentially transforming the equation into a more manageable form, step by step, ensuring that we maintain the equality of both sides throughout the process. This methodical approach is what makes algebra powerful and predictable.

Step 1: Distribute the $-2$

The first crucial step in solving for $p$ in the equation $6-2(q-3 p)=4 p$ is to apply the distributive property. This involves multiplying the $-2$ by each term inside the parentheses $(q-3 p)$. Remember that a negative number multiplied by a negative number results in a positive number. So, when we multiply $-2$ by $-3p$, we get $+6p$. The term $-2$ multiplied by $q$ results in $-2q$. Therefore, the equation transforms from $6-2(q-3 p)=4 p$ to $6 - 2q + 6p = 4p$. This expansion removes the parentheses and presents the equation in a more linear form, making it easier to work with. It's vital to be meticulous here; a single error in distribution, especially with the signs, can lead to an incorrect final answer. Think of it as opening up the equation to reveal all its components more clearly. This step is fundamental in simplifying complex algebraic expressions and is a cornerstone of algebraic manipulation. We are systematically breaking down the expression to reveal the underlying linear relationships. The goal is to simplify and organize, making the subsequent steps of isolating $p$ more straightforward. By distributing, we convert a compact form into an expanded form, which is often easier to analyze and manipulate in subsequent steps.

Step 2: Isolate the $p$ terms

After distributing the $-2$, our equation is $6 - 2q + 6p = 4p$. The next logical step is to gather all terms containing $p$ on one side of the equation and all other terms (constants and terms with $q$) on the other side. To do this, we can subtract $6p$ from both sides of the equation. This is a valid algebraic operation because it maintains the equality of the equation. Subtracting $6p$ from the left side gives us $6 - 2q + 6p - 6p$, which simplifies to $6 - 2q$. Subtracting $6p$ from the right side gives us $4p - 6p$, which simplifies to $-2p$. So, the equation now becomes $6 - 2q = -2p$. This step is crucial because it consolidates all the $p$ terms, bringing us closer to isolating $p$. The principle here is that whatever operation you perform on one side of the equation, you must perform the same operation on the other side to keep the equation balanced. This systematic approach ensures that we are not altering the fundamental relationship expressed by the equation. By moving all $p$ terms together, we prepare the equation for the final step of solving for $p$. This consolidation is a key strategy in solving linear equations, making the problem more tractable.

Step 3: Isolate $p$ by dividing

We have now reached the stage where our equation is $6 - 2q = -2p$. To solve for $p$, we need to get $p$ by itself. Currently, $p$ is being multiplied by $-2$. The inverse operation of multiplication is division. Therefore, we need to divide both sides of the equation by $-2$. Dividing the left side, $6 - 2q$, by $-2$ gives us rac{6 - 2q}{-2}. We can simplify this further by dividing each term in the numerator by $-2$: rac{6}{-2} - rac{2q}{-2} = -3 - (-q) = -3 + q. Dividing the right side, $-2p$, by $-2$ gives us rac{-2p}{-2}, which simplifies to $p$. Thus, the equation becomes $-3 + q = p$, or more conventionally, $p = q - 3$. This is our final solution, expressing $p$ in terms of $q$. Each step taken was to maintain the equality of the equation while progressively isolating the variable $p$. This final step, division, is the inverse operation that undoes the multiplication of $p$ by $-2$, leaving $p$ alone on one side. The entire process demonstrates the power of inverse operations in algebra to unravel equations and express relationships between variables. The ability to express one variable in terms of another is a fundamental outcome of solving equations.

Conclusion: The Solution for $p$

Throughout this detailed exploration, we have systematically solved the equation $6-2(q-3 p)=4 p$ for the variable $p$. We began by understanding the structure of the equation and then meticulously applied the distributive property to simplify it. Following that, we gathered all terms involving $p$ onto one side of the equation, using subtraction to consolidate them. Finally, we employed division, the inverse operation of multiplication, to isolate $p$. The culmination of these algebraic steps led us to the solution: $p = q - 3$. This result expresses $p$ as a function of $q$, meaning that the value of $p$ depends directly on the value of $q$. This is a common outcome when solving equations with multiple variables; we express one variable in terms of the others. Mastering these algebraic techniques not only helps in solving specific problems but also strengthens your overall mathematical reasoning and problem-solving skills. Remember to always check your work by substituting your solution back into the original equation to ensure it holds true. For further exploration into the fascinating world of algebra and equation solving, you might find the resources at Khan Academy incredibly helpful. Their comprehensive lessons and practice exercises cover a wide range of mathematical topics, providing a solid foundation for your learning journey.