Solving 1/r = 2/s + 3/t Expressing S In Terms Of R And T
Introduction
In the realm of algebra, rearranging equations to solve for a specific variable is a fundamental skill. This article delves into the process of expressing s in terms of r and t, given the equation 1/r = 2/s + 3/t. This type of problem is a common exercise in algebraic manipulation, requiring a solid understanding of fractions, common denominators, and isolating variables. Whether you're a student tackling homework, or simply someone looking to sharpen your math skills, this guide will provide a step-by-step approach to solving this equation and similar problems. Understanding how to manipulate equations is crucial not only in mathematics but also in various scientific and engineering fields, making this a valuable skill to master. Let's embark on this algebraic journey together, breaking down each step to ensure clarity and comprehension. This comprehensive guide will not only provide the solution but also delve into the underlying principles, ensuring a solid grasp of the concepts involved. We'll explore common pitfalls to avoid and offer additional tips for success in similar algebraic challenges. By the end of this article, you'll be equipped with the knowledge and confidence to tackle similar equations with ease.
Understanding the Initial Equation
The equation we're tackling is 1/r = 2/s + 3/t. This equation involves fractions, which immediately suggests we'll need to find common denominators at some point. Our primary goal is to isolate s on one side of the equation. This involves several algebraic manipulations, including combining fractions, cross-multiplication (potentially), and rearranging terms. It's crucial to remember the fundamental principle of algebraic manipulation: whatever operation you perform on one side of the equation, you must perform the same operation on the other side to maintain equality. Before diving into the steps, it's helpful to visualize the equation and identify the operations needed. We have fractions being added, so finding a common denominator for those fractions will be a key step. Then, we'll need to isolate the term containing s before finally solving for s itself. This process often involves a combination of addition, subtraction, multiplication, and division, all applied strategically to achieve our goal. Paying close attention to each step and understanding the reasoning behind it is vital for mastering this type of algebraic problem. By breaking down the equation into smaller, manageable steps, we can approach the solution with clarity and precision. Furthermore, a strong understanding of fraction manipulation is paramount to solving this type of equation.
Step-by-Step Solution
1. Isolating the Term with 's'
Our initial goal is to isolate the term containing s, which is 2/s. To achieve this, we need to subtract 3/t from both sides of the equation:
1/r - 3/t = 2/s + 3/t - 3/t
This simplifies to:
1/r - 3/t = 2/s
This step is crucial as it brings us closer to isolating s. Subtracting 3/t from both sides maintains the equality of the equation while moving the term away from the s term. This process is a fundamental technique in algebra, allowing us to manipulate equations and solve for specific variables. By isolating the term containing s, we've effectively set the stage for the subsequent steps in the solution. This step demonstrates the importance of inverse operations in algebra, where we use the opposite operation to undo an existing one. In this case, we used subtraction to undo addition. Understanding this principle is key to successfully manipulating algebraic equations. The next step will involve combining the fractions on the left-hand side, which will require finding a common denominator. This is a common technique when dealing with fractions in equations and is a crucial skill to develop for algebraic problem-solving.
2. Combining Fractions
Now, we need to combine the fractions on the left side of the equation, which are 1/r and -3/t. To do this, we need to find a common denominator. The common denominator for r and t is their product, rt. So, we rewrite the fractions with the common denominator:
(1/r) * (t/t) - (3/t) * (r/r) = 2/s
This gives us:
t/rt - 3r/rt = 2/s
Now that the fractions have a common denominator, we can combine them:
(t - 3r) / rt = 2/s
This step is essential as it simplifies the left side of the equation into a single fraction, making it easier to work with. Finding a common denominator is a fundamental skill in fraction manipulation, and it's crucial for solving equations involving fractions. By combining the fractions, we've reduced the complexity of the equation and brought us closer to isolating s. The process of finding a common denominator involves multiplying each fraction by a form of 1 that allows them to have the same denominator. This ensures that we're not changing the value of the fractions, only their representation. This step highlights the importance of understanding fraction arithmetic in algebraic problem-solving. The next step will involve dealing with the reciprocals of the fractions to further isolate s.
3. Cross-Multiplication or Taking Reciprocals
At this point, we have the equation (t - 3r) / rt = 2/s. To further isolate s, we can either use cross-multiplication or take the reciprocal of both sides. Let's explore taking the reciprocal, as it can be a more direct approach in this case.
Taking the reciprocal of both sides gives us:
rt / (t - 3r) = s/2
This step is a clever way to flip the fractions and bring s into the numerator. Taking the reciprocal of a fraction simply means inverting it, swapping the numerator and denominator. When we take the reciprocal of both sides of an equation, we maintain the equality, as we're performing the same operation on both sides. This step is particularly useful when the variable we're solving for is in the denominator, as it allows us to bring it into the numerator. Alternatively, we could have used cross-multiplication, which involves multiplying the numerator of the left fraction by the denominator of the right fraction and vice versa. However, taking reciprocals often simplifies the process, especially when dealing with single fractions on each side of the equation. The next and final step will be to isolate s completely by multiplying both sides by 2.
4. Isolating 's'
Finally, to isolate s completely, we need to multiply both sides of the equation rt / (t - 3r) = s/2 by 2:
2 * [rt / (t - 3r)] = 2 * (s/2)
This simplifies to:
2rt / (t - 3r) = s
Therefore, we have successfully expressed s in terms of r and t:
s = 2rt / (t - 3r)
This final step completes the solution by isolating s on one side of the equation. Multiplying both sides by 2 is a straightforward operation that removes the denominator from the s term, leaving us with the solution. This step highlights the importance of understanding inverse operations, as we used multiplication to undo the division by 2. The solution, s = 2rt / (t - 3r), expresses s explicitly in terms of r and t, which was our initial goal. This means that if we know the values of r and t, we can directly calculate the value of s. The entire process demonstrates the power of algebraic manipulation in solving equations and expressing variables in terms of others. This is a fundamental skill in mathematics and has wide applications in various fields.
The Final Solution
After meticulously following the steps, we've successfully expressed s in terms of r and t. The final solution is:
s = 2rt / (t - 3r)
This equation provides a clear and concise relationship between s, r, and t. It allows us to calculate the value of s given the values of r and t. This solution is the culmination of several algebraic manipulations, including isolating the term containing s, combining fractions, and taking reciprocals. Each step was crucial in transforming the original equation into a form where s is explicitly defined in terms of the other variables. This process highlights the power of algebraic techniques in solving equations and expressing relationships between variables. The solution is not just a numerical answer but rather a formula that describes how s changes with respect to r and t. This type of solution is invaluable in various applications where we need to understand and predict the behavior of variables based on their relationships. The ability to solve for a specific variable in an equation is a fundamental skill in mathematics and has wide-ranging applications in science, engineering, and other fields.
Common Pitfalls and How to Avoid Them
When solving algebraic equations, it's easy to make mistakes if you're not careful. One common pitfall is incorrectly applying operations to only one side of the equation. Remember, any operation performed must be applied to both sides to maintain equality. Another common mistake is mishandling fractions, especially when finding common denominators or combining fractions. It's crucial to double-check your work when dealing with fractions to ensure accuracy. Sign errors are also a frequent source of mistakes, especially when dealing with negative numbers. Pay close attention to the signs of each term and ensure you're applying the correct operations. Another pitfall is forgetting the order of operations (PEMDAS/BODMAS). Always perform operations in the correct order: Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right). To avoid these pitfalls, it's helpful to write out each step clearly and double-check your work as you go. Practice is also key to improving your algebraic skills and avoiding mistakes. By being aware of these common pitfalls and taking steps to avoid them, you can significantly increase your accuracy and confidence in solving algebraic equations. It's also beneficial to check your final solution by substituting it back into the original equation to ensure it holds true.
Practice Problems
To solidify your understanding of solving for a variable in an equation, let's consider some practice problems. These problems will help you apply the techniques we've discussed and build your confidence in algebraic manipulation. Try solving the following equations for the specified variable:
- Solve for x: 1/a = 3/x + 2/b
- Solve for y: 5/p = 1/q + 4/y
- Solve for z: 2/m = 3/n - 1/z
These problems are similar to the one we solved in this article, but they involve different variables and constants. The key is to follow the same steps: isolate the term containing the variable you're solving for, combine fractions, take reciprocals or cross-multiply, and then isolate the variable completely. Working through these practice problems will reinforce your understanding of the process and help you develop your problem-solving skills. Remember to show your work and double-check your answers. You can also create your own practice problems by changing the numbers and variables in the original equation. The more you practice, the more comfortable you'll become with algebraic manipulation. These practice problems provide an opportunity to apply the concepts learned and identify any areas where further clarification may be needed. Solving these problems will enhance your understanding and ability to tackle similar algebraic challenges.
Conclusion
In conclusion, expressing s in terms of r and t in the equation 1/r = 2/s + 3/t involves a series of algebraic manipulations. We successfully isolated s by following a step-by-step approach: isolating the term containing s, combining fractions, taking reciprocals, and then solving for s. The final solution is:
s = 2rt / (t - 3r)
This process demonstrates the power of algebraic techniques in solving equations and expressing relationships between variables. Understanding these techniques is crucial for success in mathematics and related fields. By mastering the steps involved, you can confidently tackle similar problems and manipulate equations to solve for any variable. Remember to pay attention to common pitfalls, such as mishandling fractions or making sign errors, and to practice regularly to improve your skills. Algebra is a fundamental building block for higher-level mathematics, and a solid understanding of these concepts will serve you well in your academic and professional pursuits. The ability to manipulate equations and solve for specific variables is a valuable skill that can be applied in various contexts. This article has provided a comprehensive guide to solving this particular equation, and the principles learned can be applied to a wide range of algebraic problems. Keep practicing and exploring the world of algebra, and you'll continue to grow your mathematical abilities.
