Solving Rational Equation (r-6)/(1-4r) = 1/r A Step By Step Guide

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Introduction

This article delves into the step-by-step solution of the equation ${ \frac{r-6}{1-4r} = \frac{1}{r} }$. Equations like this, which involve rational expressions, require a careful approach to avoid pitfalls such as division by zero and extraneous solutions. We will explore each step meticulously, providing a clear understanding of the algebraic manipulations involved. Understanding how to solve rational equations is a fundamental skill in algebra, with applications spanning various fields, including physics, engineering, and economics. Mastering these techniques not only aids in solving academic problems but also enhances problem-solving abilities in practical situations. This comprehensive guide aims to equip you with the knowledge and confidence to tackle similar problems effectively. We'll begin by discussing the restrictions on the variable ${ r }$, then proceed with cross-multiplication and simplification to arrive at a quadratic equation. Finally, we will solve the quadratic equation and verify the solutions to ensure they are valid. The goal is to provide a clear, step-by-step explanation that is accessible to students and anyone interested in sharpening their algebraic skills. By the end of this article, you will have a solid understanding of how to approach and solve rational equations of this type. Remember, the key to success in algebra is practice and a thorough understanding of the underlying principles. Let's embark on this mathematical journey together!

Identifying Restrictions

The first critical step in solving any rational equation is to identify any restrictions on the variable. These restrictions arise from the denominators in the equation. A rational expression is undefined when its denominator is equal to zero. In the given equation, ${ \frac{r-6}{1-4r} = \frac{1}{r} }$, we have two denominators: ${ 1-4r }$ and ${ r }$. Setting each denominator to zero will help us find the values of ${ r }$ that must be excluded from the solution set.

1. For the denominator ${ 1-4r }$, we set it equal to zero and solve for ${ r }$: ${ 1 - 4r = 0 }$ ${ 4r = 1 }$ ${ r = \frac{1}{4} }$ Thus, ${ r }$ cannot be equal to ${ \frac{1}{4} }$.
2. For the denominator ${ r }$, the restriction is straightforward: ${ r \neq 0 }$ Therefore, ${ r }$ cannot be equal to 0.

These restrictions are crucial because if we obtain ${ r = \frac{1}{4} }$ or ${ r = 0 }$ as a solution during the solving process, we must discard them as they would lead to division by zero in the original equation. Ignoring these restrictions can lead to extraneous solutions, which are solutions that satisfy the transformed equation but not the original one. Always remember to check the denominators before proceeding with the solution. This step ensures the validity of the final answers. Identifying restrictions is not just a preliminary step; it is a fundamental part of the problem-solving process that safeguards against incorrect solutions. It highlights the importance of understanding the domain of the expressions involved. By clearly defining the restrictions upfront, we set the stage for a more accurate and reliable solution.

Cross-Multiplication

Once the restrictions are identified, the next step in solving the equation ${ \frac{r-6}{1-4r} = \frac{1}{r} }$ is to eliminate the fractions. We accomplish this by cross-multiplication. Cross-multiplication is a technique used to solve equations involving fractions by multiplying the numerator of one fraction by the denominator of the other and setting the results equal to each other. This method is essentially a shortcut for multiplying both sides of the equation by the least common denominator (LCD), which in this case is ${ r(1-4r) }$.

Applying cross-multiplication to the equation ${ \frac{r-6}{1-4r} = \frac{1}{r} }$, we multiply ${ (r-6) }$ by ${ r }$ and ${ 1 }$ by ${ (1-4r) }$. This gives us:

${ r(r-6) = 1(1-4r) }$

This step effectively clears the denominators, transforming the rational equation into a more manageable algebraic equation. Expanding both sides, we get:

${ r^2 - 6r = 1 - 4r }$

The equation is now free of fractions and ready for further simplification. Cross-multiplication is a powerful tool, but it is essential to apply it correctly. It is a direct application of the multiplication property of equality. We are essentially multiplying both sides of the equation by the product of the denominators. This technique is applicable only when we have a proportion, which is an equation stating that two ratios (fractions) are equal. By correctly applying cross-multiplication, we can transform complex rational equations into simpler forms that are easier to solve. This step is a crucial bridge between the original equation and the algebraic manipulations that will lead us to the solution.

Simplifying to a Quadratic Equation

After cross-multiplication, our equation is ${ r^2 - 6r = 1 - 4r }$. The next step is to simplify this equation into a standard quadratic form. A quadratic equation is an equation of the form ${ ax^2 + bx + c = 0 }$, where ${ a }$, ${ b }$, and ${ c }$ are constants, and ${ a \neq 0 }$. To transform our equation into this form, we need to move all terms to one side of the equation, leaving zero on the other side.

First, we add ${ 4r }$ to both sides of the equation:

${ r^2 - 6r + 4r = 1 - 4r + 4r }$

This simplifies to:

${ r^2 - 2r = 1 }$

Next, we subtract 1 from both sides of the equation:

${ r^2 - 2r - 1 = 1 - 1 }$

This results in the quadratic equation:

${ r^2 - 2r - 1 = 0 }$

Now, the equation is in the standard quadratic form, with ${ a = 1 }$, ${ b = -2 }$, and ${ c = -1 }$. This form is essential because it allows us to use standard methods such as factoring, completing the square, or the quadratic formula to find the solutions for ${ r }$. Transforming the equation into a standard form is a critical step in solving quadratic equations. It sets the stage for applying the appropriate solution methods. By arranging the terms in the ${ ax^2 + bx + c = 0 }$ format, we can easily identify the coefficients needed for the quadratic formula or assess whether the equation can be factored. This systematic approach ensures that we can apply the most efficient method to solve the equation. Understanding the process of simplification and transformation is fundamental to mastering quadratic equations. It is a skill that is widely applicable in various mathematical and scientific contexts.

Solving the Quadratic Equation

Now that we have the quadratic equation in the standard form ${ r^2 - 2r - 1 = 0 }$, we can solve for ${ r }$. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, factoring is not straightforward, so we will use the quadratic formula. The quadratic formula is a general method that can be used to solve any quadratic equation, regardless of whether it can be factored easily.

The quadratic formula is given by:

${ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} }$

where ${ a }$, ${ b }$, and ${ c }$ are the coefficients of the quadratic equation ${ ax^2 + bx + c = 0 }$. In our equation, ${ r^2 - 2r - 1 = 0 }$, we have ${ a = 1 }$, ${ b = -2 }$, and ${ c = -1 }$. Substituting these values into the quadratic formula, we get:

${ r = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} }$

Simplifying, we have:

${ r = \frac{2 \pm \sqrt{4 + 4}}{2} }$

${ r = \frac{2 \pm \sqrt{8}}{2} }$

Since ${ \sqrt{8} = 2\sqrt{2} }$, we can further simplify:

${ r = \frac{2 \pm 2\sqrt{2}}{2} }$

Dividing each term in the numerator by 2, we obtain the solutions:

${ r = 1 \pm \sqrt{2} }$

Thus, we have two possible solutions:

${ r_1 = 1 + \sqrt{2} }$

${ r_2 = 1 - \sqrt{2} }$

The quadratic formula is a cornerstone of algebra, providing a reliable method for solving any quadratic equation. Understanding and applying the quadratic formula is a crucial skill. It eliminates the guesswork involved in factoring and offers a direct path to the solutions. By correctly substituting the coefficients and simplifying the expression, we can find the roots of any quadratic equation. The discriminant, ${ b^2 - 4ac }$, within the formula, gives us valuable information about the nature of the roots, indicating whether they are real, distinct, or complex. This step-by-step application of the quadratic formula ensures accuracy and a comprehensive understanding of the solution process.

Verifying the Solutions

After finding the solutions ${ r_1 = 1 + \sqrt{2} }$ and ${ r_2 = 1 - \sqrt{2} }$, it is essential to verify that these solutions are valid. This is particularly important when dealing with rational equations because we identified restrictions on the variable earlier. We need to ensure that neither of our solutions makes the denominator of the original equation equal to zero. The restrictions we found were ${ r \neq 0 }$ and ${ r \neq \frac{1}{4} }$.

Since ${ 1 + \sqrt{2} }$ and ${ 1 - \sqrt{2} }$ are both non-zero and not equal to ${ \frac{1}{4} }$, they do not violate the restrictions. However, to be completely sure, we should substitute each solution back into the original equation ${ \frac{r-6}{1-4r} = \frac{1}{r} }$ to check if the equation holds true.

Let's start with ${ r_1 = 1 + \sqrt{2} }$:

${ \frac{(1 + \sqrt{2}) - 6}{1 - 4(1 + \sqrt{2})} = \frac{1}{1 + \sqrt{2}} }$

Simplifying the left side:

${ \frac{-5 + \sqrt{2}}{1 - 4 - 4\sqrt{2}} = \frac{-5 + \sqrt{2}}{-3 - 4\sqrt{2}} }$

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is ${ -3 + 4\sqrt{2} }$:

${ \frac{(-5 + \sqrt{2})(-3 + 4\sqrt{2})}{(-3 - 4\sqrt{2})(-3 + 4\sqrt{2})} = \frac{15 - 20\sqrt{2} - 3\sqrt{2} + 8}{9 - 16(2)} = \frac{23 - 23\sqrt{2}}{-23} = -1 + \sqrt{2} }$

The right side of the original equation is ${ \frac{1}{1 + \sqrt{2}} }$. To rationalize this denominator, we multiply the numerator and denominator by the conjugate ${ 1 - \sqrt{2} }$:

${ \frac{1}{1 + \sqrt{2}} \cdot \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{1 - \sqrt{2}}{1 - 2} = \frac{1 - \sqrt{2}}{-1} = -1 + \sqrt{2} }$

Since both sides are equal, ${ r_1 = 1 + \sqrt{2} }$ is a valid solution.

Now, let's verify ${ r_2 = 1 - \sqrt{2} }$ in a similar way. This process confirms the accuracy of our solutions and ensures that we have not introduced any extraneous roots during the algebraic manipulations. Verification is a crucial step in solving equations, especially rational equations, where the presence of denominators can lead to potential division-by-zero errors. By substituting the solutions back into the original equation, we can be confident that our answers are correct and consistent with the initial problem statement. This step not only validates the solutions but also reinforces the understanding of the equation and the solution process.

Final Answer

After solving the equation ${ \frac{r-6}{1-4r} = \frac{1}{r} }$, we found two potential solutions: ${ r_1 = 1 + \sqrt{2} }$ and ${ r_2 = 1 - \sqrt{2} }$. We verified that both solutions do not violate the restrictions ${ r \neq 0 }$ and ${ r \neq \frac{1}{4} }$. Furthermore, we substituted each solution back into the original equation and confirmed that they satisfy the equation. Therefore, the solutions to the equation are:

${ r = 1 + \sqrt{2} \text{ and } r = 1 - \sqrt{2} }$

These are the final and valid solutions to the given equation. This comprehensive solution process highlights the importance of each step, from identifying restrictions to verifying the answers. A thorough approach ensures accuracy and a deep understanding of the underlying mathematical principles. By following this structured method, you can confidently solve similar equations and tackle more complex problems in algebra and beyond. The ability to solve rational equations is a valuable skill that can be applied in various fields, making it an essential component of mathematical proficiency.

Conclusion

In summary, solving the equation ${ \frac{r-6}{1-4r} = \frac{1}{r} }$ involved several key steps: identifying restrictions, cross-multiplication, simplifying to a quadratic equation, solving the quadratic equation using the quadratic formula, and verifying the solutions. Each step is crucial to ensure the accuracy and validity of the final answers. The journey through this problem highlights the interconnectedness of algebraic concepts and the importance of a systematic approach. By understanding the restrictions on the variable, we avoided potential pitfalls and extraneous solutions. Cross-multiplication allowed us to transform the rational equation into a simpler form, while the quadratic formula provided a reliable method for solving the resulting quadratic equation. Finally, verifying the solutions confirmed their correctness and completeness.

This problem-solving process is not only applicable to rational equations but also serves as a model for tackling a wide range of mathematical challenges. By mastering these techniques, you can enhance your problem-solving skills and gain a deeper appreciation for the beauty and power of algebra. Remember that practice and a thorough understanding of fundamental principles are the keys to success in mathematics. Embrace the challenges, and continue to explore the fascinating world of algebraic equations and solutions. This detailed exploration of solving rational equations equips you with the tools and knowledge needed to confidently approach similar problems in the future. The ability to break down complex problems into manageable steps and apply appropriate techniques is a valuable skill that extends far beyond the realm of mathematics.