Expressing (2x+1)/(3x+1) In Terms Of A Detailed Solution

Introduction

In this detailed mathematical exploration, we will dive deep into the realm of algebraic manipulation and expression simplification. Our central task is to express the rational function ${\frac{2x+1}{3x+1}}$ in terms of the variable 'a', given the initial relationship ${x = \frac{a+3}{2a} - 1}$. This problem exemplifies a common type of algebraic challenge where substitution and simplification are key techniques. We will meticulously dissect the problem, performing each step with clarity and precision to ensure a comprehensive understanding of the solution.

This exploration is not just about arriving at the final answer; it's about understanding the process of algebraic manipulation. We'll focus on how to strategically simplify expressions, combine like terms, and ultimately rewrite the given expression in the desired form. The skills honed in this process are invaluable for tackling a wide array of mathematical problems, making this a worthwhile exercise for students and enthusiasts alike. So, let's embark on this mathematical journey, armed with patience and a keen eye for detail, as we unravel the intricacies of this expression.

Step-by-Step Solution

1. Simplifying the Expression for x

Our journey begins with the expression for x: ${x = \frac{a+3}{2a} - 1}$. To make this expression more manageable, we need to combine the terms on the right-hand side. This requires finding a common denominator, which in this case is ${2a}$. So, we rewrite 1 as ${\frac{2a}{2a}}$ and proceed with the subtraction:

${ x = \frac{a+3}{2a} - \frac{2a}{2a} }$

Now, we can combine the numerators:

${ x = \frac{a + 3 - 2a}{2a} }$

Simplifying the numerator by combining like terms (a and -2a) gives us:

${ x = \frac{3 - a}{2a} }$

This simplified expression for x is our first major milestone. It lays the foundation for the subsequent steps in our solution. By expressing x in this form, we've effectively reduced the complexity of the initial equation, making it easier to work with in the following substitutions.

2. Substituting x into the Target Expression

Now that we have a simplified expression for x, our next crucial step is to substitute this value into the expression we want to express in terms of a: ${\frac{2x+1}{3x+1}}$. This substitution will replace x with ${\frac{3-a}{2a}}$, effectively bridging the gap between the two expressions.

Let's perform the substitution:

${ \frac{2x+1}{3x+1} = \frac{2(\frac{3-a}{2a})+1}{3(\frac{3-a}{2a})+1} }$

This substitution is a pivotal moment in our solution. It transforms the original expression, which was in terms of x, into an expression that involves a. However, the expression is still quite complex, with fractions within fractions. Our next task is to simplify this compound fraction, which will require careful algebraic manipulation.

3. Simplifying the Compound Fraction

Our expression now looks like this:

${ \frac{2(\frac{3-a}{2a})+1}{3(\frac{3-a}{2a})+1} }$

To simplify this, we first distribute the 2 in the numerator and the 3 in the denominator:

${ \frac{\frac{2(3-a)}{2a}+1}{\frac{3(3-a)}{2a}+1} = \frac{\frac{6-2a}{2a}+1}{\frac{9-3a}{2a}+1} }$

Next, we need to add the 1 in both the numerator and the denominator. To do this, we rewrite 1 as ${\frac{2a}{2a}}$ in both cases:

${ \frac{\frac{6-2a}{2a}+\frac{2a}{2a}}{\frac{9-3a}{2a}+\frac{2a}{2a}} }$

Now we can combine the fractions in the numerator and the denominator:

${ \frac{\frac{6-2a+2a}{2a}}{\frac{9-3a+2a}{2a}} }$

Simplifying the numerators, we get:

${ \frac{\frac{6}{2a}}{\frac{9-a}{2a}} }$

We now have a fraction divided by another fraction. To simplify this, we multiply the numerator by the reciprocal of the denominator:

${ \frac{6}{2a} \cdot \frac{2a}{9-a} }$

4. Final Simplification

In the previous step, we arrived at the expression:

${ \frac{6}{2a} \cdot \frac{2a}{9-a} }$

Now, we can simplify this expression by canceling out the common factor of ${2a}$ in the numerator and denominator:

${ \frac{6}{2a} \cdot \frac{2a}{9-a} = \frac{6 \cdot 2a}{2a \cdot (9-a)} = \frac{6}{9-a} }$

This final simplification brings us to our desired result. We have successfully expressed ${\frac{2x+1}{3x+1}}$ in terms of a.

Final Answer

Therefore, given ${x = \frac{a+3}{2a} - 1}$, the expression ${\frac{2x+1}{3x+1}}$ can be expressed in terms of a as:

${ \frac{2x+1}{3x+1} = \frac{6}{9-a} }$

This completes our step-by-step solution. We have meticulously navigated through the algebraic manipulations, starting from the initial expression for x and culminating in the final expression in terms of a. This journey highlights the power of algebraic techniques in simplifying complex expressions and revealing underlying relationships.

Conclusion

In this comprehensive exploration, we successfully expressed the rational function ${\frac{2x+1}{3x+1}}$ in terms of the variable 'a', given the initial condition ${x = \frac{a+3}{2a} - 1}$. Our journey involved a series of strategic algebraic manipulations, including simplifying the expression for x, substituting it into the target expression, and meticulously simplifying the resulting compound fraction. We arrived at the final answer:

${ \frac{2x+1}{3x+1} = \frac{6}{9-a} }$

This problem serves as a powerful illustration of the importance of algebraic skills in mathematics. The ability to manipulate expressions, combine like terms, and simplify complex fractions is crucial for solving a wide range of mathematical problems. The techniques we employed, such as finding common denominators, distributing terms, and canceling common factors, are fundamental tools in the mathematician's toolkit.

Furthermore, this exercise underscores the value of a step-by-step approach to problem-solving. By breaking down the problem into smaller, more manageable steps, we were able to tackle the complexity with clarity and precision. Each step built upon the previous one, leading us steadily towards the final solution. This methodical approach is a valuable strategy not only in mathematics but also in various other fields.

In conclusion, this exploration has not only provided a solution to a specific algebraic problem but has also reinforced the broader principles of mathematical problem-solving. The skills and techniques demonstrated here are applicable to a wide range of mathematical challenges, making this a valuable learning experience for anyone seeking to deepen their understanding of algebra.