Solve For 'a' In Algebraic Equations

Understanding the Equation: $7a - 2b = 5a + b$

Welcome, math enthusiasts! Today, we're diving into a fundamental algebraic problem: solving for 'a' in the equation $7a - 2b = 5a + b$. This might seem like a simple equation, but mastering these types of problems is crucial for building a strong foundation in mathematics. We'll break down the steps, explain the reasoning behind each move, and ultimately arrive at the correct solution. Our goal is to isolate the variable 'a' on one side of the equation, making it the subject of our inquiry. Think of it like a detective trying to find the value of 'a' by gathering all the clues (terms) and arranging them logically. The equation $7a - 2b = 5a + b$ presents us with terms involving both 'a' and 'b' on both sides. Our primary objective is to consolidate all the 'a' terms on one side and all the 'b' terms on the other. This process involves using inverse operations, a cornerstone of algebraic manipulation. When we move a term from one side of the equals sign to the other, we must perform the opposite operation. For instance, if a term is being added, we subtract it from both sides; if it's being subtracted, we add it to both sides. This ensures that the equality of the equation remains intact throughout our solving process. Let's begin by focusing on the 'a' terms. We have $7a$ on the left side and $5a$ on the right side. To bring them together, it's often easiest to move the smaller 'a' term to the side with the larger 'a' term, which in this case means moving $5a$ from the right to the left. To do this, we subtract $5a$ from both sides of the equation. This gives us $(7a - 5a) - 2b = (5a - 5a) + b$, which simplifies to $2a - 2b = b$. Now, all our 'a' terms are on the left side. The next step is to deal with the 'b' terms. We have $-2b$ on the left and $+b$ on the right. Our goal is to get all the 'b' terms together on the right side, leaving the 'a' term isolated on the left. To move the $-2b$ from the left to the right, we perform the inverse operation: we add $2b$ to both sides of the equation. This results in $2a - 2b + 2b = b + 2b$. Simplifying this, we get $2a = 3b$. We're almost there! The variable 'a' is currently multiplied by 2. To isolate 'a', we need to perform the inverse operation of multiplication, which is division. We will divide both sides of the equation by 2. This yields rac{2a}{2} = rac{3b}{2}. The final simplification gives us a = rac{3}{2}b. This is our solution, expressing 'a' in terms of 'b'. The process itself reinforces the principles of algebraic manipulation and the importance of maintaining balance in an equation. It’s a systematic approach that, with practice, becomes second nature.

Step-by-Step Solution Process

Let's meticulously walk through the process of solving for 'a' in the given equation $7a - 2b = 5a + b$. Our primary objective is to isolate the variable '$a$' on one side of the equation. This involves a series of algebraic manipulations, primarily using inverse operations to move terms across the equals sign while maintaining the equation's balance. We start with:

$$7a - 2b = 5a + b$$

Step 1: Combine 'a' terms. Our first strategic move is to gather all terms containing '$a$' on one side of the equation. It's generally more efficient to move the smaller 'a' term to the side with the larger 'a' term. In this case, $5a$ on the right side is smaller than $7a$ on the left side. To move $5a$ from the right to the left, we subtract $5a$ from both sides of the equation. This is the inverse operation of adding $5a$.

$$7a - 5a - 2b = 5a - 5a + b$$

Simplifying this, we get:

$$2a - 2b = b$$

Step 2: Combine 'b' terms. Now that our 'a' terms are consolidated, we turn our attention to the 'b' terms. We want to isolate the '$2a$' term on the left side, which means we need to move the '$-2b$' term to the right side. The inverse operation of subtracting $2b$ is adding $2b$. So, we add $2b$ to both sides of the equation:

$$2a - 2b + 2b = b + 2b$$

Simplifying this, we arrive at:

$$2a = 3b$$

Step 3: Isolate 'a'. At this stage, '$a$' is multiplied by 2. To get '$a$' by itself, we need to perform the inverse operation of multiplication, which is division. We divide both sides of the equation by 2:

$$\frac{2a}{2} = \frac{3b}{2}$$

Step 4: Final Solution. After performing the division, we are left with our solution:

$$a = \frac{3}{2}b$$

This result expresses '$a$' in terms of '$b$'. This methodical approach ensures accuracy and deepens our understanding of algebraic manipulation. Each step is a logical progression, building upon the previous one to systematically isolate the desired variable. The key is to remember that whatever operation you perform on one side of the equation, you must perform the identical operation on the other side to maintain the equality. This principle of balance is fundamental to solving any equation, whether it's a simple linear equation like this one or a more complex problem encountered in higher mathematics. The process of moving terms by adding or subtracting their opposites, and isolating variables by multiplying or dividing by their coefficients, are the bedrock skills that unlock a vast array of mathematical concepts.

Analyzing the Options and Verifying the Solution

We have successfully solved for 'a' and found that a = rac{3}{2}b. Now, let's look at the provided options and see which one matches our result. The options are:

A. $a=2b$ B. $a=3b$ C. a=rac{3}{2}b D. a=rac{2}{3}b

Comparing our derived solution, a = rac{3}{2}b, with the given options, it's clear that Option C is the correct answer.

To further solidify our confidence in this solution, we can perform a verification step. Let's substitute our solution a = rac{3}{2}b back into the original equation $7a - 2b = 5a + b$ and check if the equality holds true.

Substitute a = rac{3}{2}b into the left side (LHS) of the equation:

$$LHS = 7a - 2b = 7\left(\frac{3}{2}b\right) - 2b$$

$$LHS = \frac{21}{2}b - 2b$$

To subtract these terms, we need a common denominator. We can rewrite $2b$ as $\frac{4}{2}b$:

$$LHS = \frac{21}{2}b - \frac{4}{2}b = \frac{21 - 4}{2}b = \frac{17}{2}b$$

Now, substitute a = rac{3}{2}b into the right side (RHS) of the equation:

$$RHS = 5a + b = 5\left(\frac{3}{2}b\right) + b$$

$$RHS = \frac{15}{2}b + b$$

Again, we need a common denominator. Rewrite $b$ as $\frac{2}{2}b$:

$$RHS = \frac{15}{2}b + \frac{2}{2}b = \frac{15 + 2}{2}b = \frac{17}{2}b$$

Since the LHS ($\frac{17}{2}b$) is equal to the RHS ($\frac{17}{2}b$), our solution a = rac{3}{2}b is indeed correct. This verification process is a powerful tool in mathematics. It not only confirms the accuracy of our solution but also reinforces the understanding of substitution and the fundamental property of equality. By plugging our derived value back into the original problem, we essentially ask, "Does this value make the statement true?" The fact that both sides balance out confirms that our manipulation of the equation was sound. This methodical check is particularly useful when dealing with more complex equations where algebraic errors might be harder to spot. It transforms a potentially abstract problem into a concrete check of truth. Understanding how to verify solutions is as important as knowing how to find them, as it builds critical thinking and problem-solving skills essential for tackling any mathematical challenge.

Conclusion: Mastering Algebraic Equations

In conclusion, we have systematically tackled the problem of solving for 'a' in the equation $7a - 2b = 5a + b$. Through a clear, step-by-step process involving the consolidation of like terms and the application of inverse operations, we arrived at the definitive solution a = rac{3}{2}b. This aligns perfectly with Option C among the choices provided. The verification step, where we substituted our solution back into the original equation, confirmed the accuracy of our findings, demonstrating that both sides of the equation indeed balance when a = rac{3}{2}b.

Mastering equations like this is a fundamental skill in mathematics, forming the bedrock for more advanced concepts in algebra, calculus, and beyond. The principles of isolating variables, maintaining equality through inverse operations, and verifying solutions are transferable to a wide array of mathematical and scientific problems. We encourage you to practice similar problems to further hone your algebraic skills. Remember, consistent practice is key to building confidence and fluency in mathematics.

For further exploration into algebraic concepts and problem-solving techniques, you can refer to resources like Khan Academy, which offers a wealth of free educational materials and practice exercises on various mathematics topics. Additionally, MathWorld provides an extensive encyclopedia of mathematical definitions and concepts.