%20%2F%203k%20%3D%207 "Solving for k in the Equation (5k - 2) / 3k = 7")
This article provides a step-by-step guide on how to solve for k in the equation (5k - 2) / 3k = 7. We'll cover the fundamental concepts, the algebraic manipulations involved, and offer insights to help you understand the process thoroughly. Whether you're a student tackling algebra or simply brushing up on your math skills, this guide will equip you with the knowledge to solve this type of equation effectively.
Understanding the Equation
The equation (5k - 2) / 3k = 7 is a rational equation, which means it involves a fraction where the numerator and/or denominator contain variables. To solve for k, our goal is to isolate k on one side of the equation. This involves a series of algebraic manipulations that we will explore in detail. Before diving into the steps, let's break down the equation:
- 5k - 2: This is the numerator of the fraction, a linear expression in terms of k.
- 3k: This is the denominator of the fraction, also a linear expression in terms of k.
- = 7: This indicates that the entire fraction is equal to the constant 7.
The key to solving rational equations is to eliminate the fraction. We can do this by multiplying both sides of the equation by the denominator. However, we must also be mindful of any restrictions on the variable k. In this case, k cannot be equal to 0, because that would make the denominator 3k equal to 0, resulting in an undefined expression. This is a crucial point to remember when solving rational equations.
Step-by-Step Solution
Now, let's proceed with the step-by-step solution:
Step 1: Eliminate the Fraction
To get rid of the fraction, we multiply both sides of the equation by the denominator, which is 3k.
(5k - 2) / 3k = 7
Multiply both sides by 3k:
3k * (5k - 2) / 3k = 7 * 3k
On the left side, 3k in the numerator and denominator cancel out, leaving us with:
5k - 2 = 7 * 3k
Step 2: Simplify the Equation
Next, we simplify the right side of the equation by multiplying 7 by 3k:
5k - 2 = 21k
Now we have a linear equation that is easier to work with.
Step 3: Isolate the Variable Terms
To isolate the terms containing k, we subtract 5k from both sides of the equation:
5k - 2 - 5k = 21k - 5k
This simplifies to:
-2 = 16k
Step 4: Solve for k
Finally, to solve for k, we divide both sides of the equation by 16:
-2 / 16 = 16k / 16
This gives us:
k = -2 / 16
We can simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2:
k = -1 / 8
Therefore, the solution to the equation is k = -1/8.
Verification
It's always a good practice to verify your solution by plugging it back into the original equation. Let's substitute k = -1/8 into the equation (5k - 2) / 3k = 7:
(5 * (-1/8) - 2) / (3 * (-1/8)) = 7
First, we simplify the numerator:
5 * (-1/8) = -5/8
-5/8 - 2 = -5/8 - 16/8 = -21/8
Next, we simplify the denominator:
3 * (-1/8) = -3/8
Now, we have:
(-21/8) / (-3/8) = 7
To divide fractions, we multiply by the reciprocal of the divisor:
(-21/8) * (-8/3) = 7
The 8s cancel out, and we have:
(-21) / (-3) = 7
7 = 7
Since the equation holds true, our solution k = -1/8 is correct.
Key Concepts and Considerations
Rational Equations
This problem falls under the category of rational equations, which are equations that contain fractions with variables in the numerator and/or denominator. Solving these equations often involves eliminating the fractions by multiplying both sides by a common denominator. This technique transforms the rational equation into a more manageable linear or polynomial equation.
Extraneous Solutions
When solving rational equations, it's crucial to check for extraneous solutions. These are solutions that satisfy the transformed equation but not the original equation. Extraneous solutions can arise when multiplying both sides of the equation by an expression that can be zero. In our example, we noted that k cannot be 0 because that would make the denominator 3k equal to 0. Fortunately, our solution k = -1/8 is not 0, so it is a valid solution.
Domain Restrictions
Identifying domain restrictions is an essential part of solving rational equations. The domain of a rational expression is the set of all real numbers except those that make the denominator zero. In the equation (5k - 2) / 3k = 7, the denominator 3k cannot be zero, so k cannot be 0. Understanding domain restrictions helps in identifying potential extraneous solutions and ensures that the solution is valid.
Alternative Methods
While we have demonstrated a standard method for solving this equation, let's briefly touch upon alternative approaches.
Cross-Multiplication
Another common technique for solving rational equations is cross-multiplication. This method can be applied when you have a single fraction equal to another single fraction or a constant. In our case, we can rewrite the equation as:
(5k - 2) / 3k = 7/1
Cross-multiplying gives us:
1 * (5k - 2) = 7 * 3k
5k - 2 = 21k
This leads us to the same equation we derived earlier, and the subsequent steps are identical.
Common Mistakes to Avoid
When solving equations like this, it's easy to make mistakes. Here are some common pitfalls to watch out for:
- Forgetting to distribute: When multiplying both sides of the equation by an expression, make sure to distribute the multiplication to every term.
- Incorrectly combining like terms: Ensure that you combine like terms accurately. For example, adding or subtracting terms with variables correctly.
- Ignoring domain restrictions: Always identify and consider domain restrictions to avoid extraneous solutions.
- Arithmetic errors: Double-check your calculations to avoid simple arithmetic mistakes.
Real-World Applications
While solving for variables in equations might seem abstract, it has numerous real-world applications. Rational equations, in particular, are used in various fields, such as:
- Physics: Calculating velocities, distances, and times.
- Engineering: Designing structures and systems.
- Economics: Modeling supply and demand.
- Chemistry: Determining reaction rates and concentrations.
Understanding how to solve these equations is a fundamental skill that can be applied to solve practical problems in these fields.
Conclusion
In this article, we have thoroughly explored how to solve the equation (5k - 2) / 3k = 7. We broke down the problem into manageable steps, from eliminating the fraction to isolating the variable k. We also emphasized the importance of verifying the solution and being aware of domain restrictions and potential extraneous solutions. By understanding these concepts and practicing the techniques outlined in this guide, you'll be well-equipped to tackle similar algebraic problems with confidence. Solving for variables in equations is a crucial skill in mathematics and has wide-ranging applications in various scientific and practical domains. Whether you are a student aiming to excel in algebra or a professional applying these concepts in your field, mastering these techniques will undoubtedly prove beneficial.
Solving for k in the equation (5k - 2) / 3k = 7 involves several steps to isolate the variable k. In this comprehensive guide, we provide a detailed explanation of each step, ensuring you grasp the underlying concepts. The initial equation, (5k - 2) / 3k = 7, presents a fraction where both the numerator and the denominator contain the variable k. To find the value of k that satisfies this equation, we must systematically manipulate the equation using algebraic principles. This article will break down each step, offering insights and clarifications to help you understand the process thoroughly. The primary goal is to rewrite the equation in a form where k is alone on one side, allowing us to determine its value.
To begin, eliminating the fraction is a crucial step when solving for k in the equation (5k - 2) / 3k = 7. This is achieved by multiplying both sides of the equation by the denominator, which in this case is 3k. Multiplying both sides by the same expression maintains the equality and helps simplify the equation. By doing so, the 3k in the denominator on the left side cancels out, leaving us with a more manageable linear equation. This step is fundamental in solving rational equations, as it removes the fractional component that complicates the process. It's important to ensure that this multiplication is done correctly to avoid errors in subsequent steps. Once the fraction is eliminated, the equation becomes easier to manipulate and solve for the unknown variable. This process highlights the importance of understanding how to manipulate equations while maintaining their balance.
After eliminating the fraction, simplify the equation further to solve for k in (5k - 2) / 3k = 7. The equation is now in the form 5k - 2 = 21k. This linear equation can be solved by isolating the variable k on one side. Simplifying at this stage involves combining like terms and rearranging the equation to make it easier to solve. This step is crucial as it transforms the equation into a standard linear form, which is straightforward to solve using basic algebraic operations. Ensuring that each term is correctly placed and combined is vital to the accuracy of the final answer. The simplification process prepares the equation for the final steps of isolating k and finding its value. The goal is to make the equation as clear and concise as possible, minimizing the chances of errors in the subsequent steps.
Isolate the variable terms to effectively solve for k in the equation (5k - 2) / 3k = 7. This step involves moving all terms containing k to one side of the equation and the constant terms to the other side. By subtracting 5k from both sides of the equation 5k - 2 = 21k, we achieve this isolation. This process results in a simpler equation where the variable terms are grouped together, making it easier to determine the value of k. Proper isolation of variables is a fundamental technique in algebra and is essential for solving various types of equations. This step sets the stage for the final calculation, where we divide by the coefficient of k to find its value. The careful and accurate isolation of variable terms ensures the correct solution is obtained.
The final step is to explicitly solve for k in the equation (5k - 2) / 3k = 7. After isolating the variable terms, we are left with -2 = 16k. To find the value of k, we divide both sides of the equation by the coefficient of k, which is 16. This gives us k = -2 / 16, which can be simplified to k = -1 / 8. This is the solution to the original equation. Always remember to simplify fractions to their lowest terms for a clean and accurate result. This final calculation provides the numerical value of k that satisfies the equation. Verifying this solution by substituting it back into the original equation is a good practice to ensure accuracy. This step completes the process of solving for k, demonstrating the application of algebraic principles to find the unknown value.
To ensure accuracy, verifying the solution is a crucial step after solving for k in the equation (5k - 2) / 3k = 7. This involves substituting the calculated value of k, which is -1/8, back into the original equation to check if it holds true. By replacing k with -1/8, we can evaluate both sides of the equation and see if they are equal. This process helps to identify any errors made during the solving process and confirms the correctness of the solution. Verification is particularly important in rational equations, where extraneous solutions can arise. A thorough check ensures that the value obtained for k satisfies the initial conditions of the equation. This step reinforces the integrity of the solution and builds confidence in the result.
An alternative approach to solving for k in (5k - 2) / 3k = 7 involves using cross-multiplication. This technique is particularly useful when dealing with proportions or equations where one fraction is equal to another. By cross-multiplying, we eliminate the fractions and obtain a linear equation that is easier to solve. This method involves multiplying the numerator of the first fraction by the denominator of the second fraction, and vice versa. In this case, cross-multiplying (5k - 2) / 3k = 7/1 gives us 1 * (5k - 2) = 7 * 3k, which simplifies to 5k - 2 = 21k. From this point, the steps to isolate k are similar to those described earlier. Cross-multiplication provides a straightforward way to transform rational equations into simpler forms, making them more accessible to solve.
While solving equations like (5k - 2) / 3k = 7, common mistakes can easily occur. One such mistake is not distributing correctly when multiplying both sides of the equation by a term. Another frequent error is mishandling negative signs, which can lead to incorrect solutions. It’s also crucial to remember the order of operations (PEMDAS/BODMAS) to avoid calculation errors. Neglecting to check for extraneous solutions is another common pitfall, especially in rational equations. Always double-check each step and verify the final solution in the original equation to minimize these mistakes. Recognizing these potential errors can significantly improve your accuracy in solving algebraic problems.
Real-world applications of solving equations like (5k - 2) / 3k = 7 are numerous and span various fields. Algebraic equations are fundamental in physics for calculating velocities, distances, and forces. In engineering, they are used to design structures and systems. Economics relies on such equations to model supply and demand, while chemistry employs them to determine reaction rates and concentrations. These equations also appear in computer science, financial analysis, and many other areas. Understanding how to solve these equations is not just an academic exercise; it is a practical skill that is essential in many professions and daily life situations. The ability to manipulate and solve algebraic equations is a cornerstone of problem-solving across various domains.
In conclusion, solving for k in the equation (5k - 2) / 3k = 7 involves a series of algebraic steps aimed at isolating the variable. This comprehensive guide has detailed each step, from eliminating the fraction to verifying the solution, ensuring a thorough understanding of the process. We began by eliminating the fraction through multiplication, simplified the equation, isolated the variable terms, and finally, solved for k. Verification of the solution and awareness of alternative methods and common mistakes have also been emphasized. The real-world applications of such equations highlight their importance in various fields. Mastering these techniques not only enhances your algebraic skills but also provides a solid foundation for tackling more complex problems in mathematics and beyond.
