Solving (r-6)/(1-4r) = 1/r A Step-by-Step Guide

This article delves into the step-by-step process of solving the equation (r-6)/(1-4r) = 1/r. This equation falls under the category of rational equations, where the variable appears in the denominator. Mastering the techniques to solve such equations is crucial for anyone studying algebra or related fields. We will not only provide the solution but also explain the underlying concepts and potential pitfalls to avoid.

Understanding Rational Equations

Before diving into the specifics of this equation, let's establish a foundational understanding of rational equations. Rational equations are equations that contain one or more fractions where the variable is in the denominator. Solving these equations requires careful attention to avoid extraneous solutions, which are solutions obtained through the algebraic process but do not satisfy the original equation. These extraneous solutions often arise due to the multiplication by expressions that can be zero. Therefore, it's always necessary to check the solutions obtained against the original equation's restrictions. One fundamental principle in solving rational equations is to eliminate the denominators. This is typically achieved by multiplying both sides of the equation by the least common multiple (LCM) of the denominators. However, it is important to identify any values of the variable that would make the denominator zero, as these values are excluded from the solution set. In our given equation, the denominators are 1-4r and r. We must ensure that 1-4r ≠ 0 and r ≠ 0, which implies r ≠ 1/4 and r ≠ 0. These restrictions are crucial and will be revisited when we verify our solutions. The process of solving rational equations often involves multiple steps, including finding the LCM, simplifying expressions, solving polynomial equations (linear, quadratic, etc.), and finally, checking for extraneous solutions. Each of these steps requires careful manipulation and attention to detail to arrive at the correct solution. Solving rational equations is not just an exercise in algebraic manipulation; it is a critical skill that has applications in various fields, including physics, engineering, and economics. These equations often model real-world scenarios, such as rates of work, mixing problems, and electrical circuits.

Step-by-Step Solution

Now, let's tackle the given equation step-by-step:

1. Identify Restrictions

As mentioned earlier, the first crucial step is to identify any restrictions on the variable r. We need to ensure that the denominators in the original equation, (r-6)/(1-4r) = 1/r, are not equal to zero. This gives us two conditions:

• 1 - 4r ≠ 0
• r ≠ 0

Solving the first inequality, we get:

1 - 4r ≠ 0 -4r ≠ -1 r ≠ 1/4

So, we have two restrictions: r ≠ 0 and r ≠ 1/4. These values cannot be solutions to the equation. These restrictions arise because division by zero is undefined in mathematics. If we were to substitute either 0 or 1/4 into the original equation, we would end up with a denominator of zero, making the expression undefined. Therefore, it's imperative to identify these restrictions at the outset to avoid extraneous solutions later on. Failing to identify these restrictions can lead to incorrect solutions and a misunderstanding of the problem. The restrictions also define the domain of the rational expressions involved. The domain is the set of all possible values of the variable that make the expression defined. In this case, the domain is all real numbers except for 0 and 1/4. Understanding the domain is crucial for interpreting the solutions in the context of a real-world problem, where certain values may not be physically meaningful.

2. Eliminate Denominators

The next step is to eliminate the denominators to simplify the equation. To do this, we multiply both sides of the equation by the least common multiple (LCM) of the denominators. In this case, the denominators are 1 - 4r and r. Their LCM is simply their product, r(1 - 4r). Multiplying both sides of the equation by this LCM, we get:

r(1 - 4r) * (r - 6) / (1 - 4r) = r(1 - 4r) * (1 / r)

This simplifies to:

r(r - 6) = (1 - 4r)

Multiplying by the LCM effectively clears the fractions from the equation, transforming it into a simpler polynomial equation. This step is crucial because polynomial equations are generally easier to solve than rational equations. However, as mentioned earlier, this multiplication can sometimes introduce extraneous solutions, which is why checking the solutions against the original equation is so important. The LCM approach is a standard technique for solving rational equations, and it can be applied to equations with multiple fractions. The key is to correctly identify the LCM and then multiply both sides of the equation by it. This will eliminate all the denominators and leave you with a polynomial equation that you can solve using standard algebraic techniques.

3. Simplify and Rearrange

Now, we expand and simplify the equation:

r^2 - 6r = 1 - 4r

To solve this quadratic equation, we need to rearrange it into the standard form ax^2 + bx + c = 0. Adding 4r and subtracting 1 from both sides, we get:

r^2 - 6r + 4r - 1 = 0 r^2 - 2r - 1 = 0

The process of simplifying and rearranging the equation is essential for identifying the type of equation we are dealing with. In this case, we have a quadratic equation, which can be solved using various methods such as factoring, completing the square, or the quadratic formula. Bringing all the terms to one side and setting the equation equal to zero allows us to apply these standard solution techniques. The coefficients of the quadratic equation play a critical role in determining the nature of the solutions. The discriminant, which is the part of the quadratic formula under the square root sign (b^2 - 4ac), tells us whether the solutions are real or complex and how many distinct solutions there are. If the discriminant is positive, there are two distinct real solutions; if it is zero, there is one real solution (a repeated root); and if it is negative, there are two complex solutions. Understanding the properties of quadratic equations is crucial for solving a wide range of mathematical problems in various fields.

4. Solve the Quadratic Equation

The equation r^2 - 2r - 1 = 0 is a quadratic equation that doesn't factor easily, so we use the quadratic formula:

r = [-b ± √(b^2 - 4ac)] / 2a

In this case, a = 1, b = -2, and c = -1. Plugging these values into the quadratic formula, we get:

r = [2 ± √((-2)^2 - 4 * 1 * -1)] / (2 * 1) r = [2 ± √(4 + 4)] / 2 r = [2 ± √8] / 2 r = [2 ± 2√2] / 2 r = 1 ± √2

So, we have two potential solutions: r = 1 + √2 and r = 1 - √2. The quadratic formula is a powerful tool for solving quadratic equations, especially those that cannot be easily factored. It provides a direct method for finding the solutions, regardless of the complexity of the coefficients. The formula is derived by completing the square on the general quadratic equation ax^2 + bx + c = 0. The ± sign in the formula indicates that there are two possible solutions, corresponding to the two roots of the quadratic equation. The square root term in the formula, the discriminant, plays a crucial role in determining the nature of the solutions. The quadratic formula is not just a mathematical tool; it has applications in various fields, such as physics, engineering, and computer science, where quadratic equations often arise in modeling real-world phenomena.

5. Check for Extraneous Solutions

Finally, we must check whether these solutions satisfy the original equation and are not extraneous. Recall that we had restrictions r ≠ 0 and r ≠ 1/4. Both 1 + √2 and 1 - √2 are real numbers and are not equal to 0 or 1/4. Now, we need to substitute each potential solution back into the original equation to verify that they indeed satisfy the equation.

Checking r = 1 + √2:

[(1 + √2) - 6] / [1 - 4(1 + √2)] = 1 / (1 + √2) (-5 + √2) / (-3 - 4√2) = 1 / (1 + √2)

To verify this, we can cross-multiply:

(-5 + √2)(1 + √2) = 1(-3 - 4√2) -5 - 5√2 + √2 + 2 = -3 - 4√2 -3 - 4√2 = -3 - 4√2

This holds true, so r = 1 + √2 is a valid solution.

Checking r = 1 - √2:

[(1 - √2) - 6] / [1 - 4(1 - √2)] = 1 / (1 - √2) (-5 - √2) / (-3 + 4√2) = 1 / (1 - √2)

Cross-multiply:

(-5 - √2)(1 - √2) = 1(-3 + 4√2) -5 + 5√2 - √2 + 2 = -3 + 4√2 -3 + 4√2 = -3 + 4√2

This also holds true, so r = 1 - √2 is a valid solution.

The step of checking for extraneous solutions is absolutely crucial in solving rational equations. As we mentioned earlier, multiplying both sides of the equation by the LCM can sometimes introduce solutions that do not satisfy the original equation. These extraneous solutions arise because the LCM can be zero for certain values of the variable, and multiplying by zero is not a reversible operation. Therefore, substituting the potential solutions back into the original equation is the only way to ensure that they are valid. This process involves careful substitution and simplification to verify that both sides of the equation are equal. If a potential solution does not satisfy the original equation, it must be discarded. In this case, both of our potential solutions, 1 + √2 and 1 - √2, passed the check and were confirmed to be valid solutions. This rigorous checking process ensures the accuracy and validity of the solutions obtained.

Conclusion

Therefore, the solutions to the equation (r-6)/(1-4r) = 1/r are r = 1 + √2 and r = 1 - √2. Remember the crucial steps: identify restrictions, eliminate denominators, solve the resulting equation, and always check for extraneous solutions. Understanding these steps thoroughly will enable you to solve a wide range of rational equations confidently. Rational equations, while sometimes challenging, are a fundamental part of algebra and have applications in various fields. By mastering the techniques for solving them, you will enhance your problem-solving skills and gain a deeper understanding of mathematical concepts.

Key takeaways: Solving rational equations requires careful attention to detail and a systematic approach. Identifying restrictions, eliminating denominators, solving the resulting polynomial equation, and checking for extraneous solutions are the core steps in this process. Remember that extraneous solutions can arise due to the multiplication by expressions that can be zero, so always verify your solutions against the original equation. With practice and a solid understanding of the underlying principles, you can confidently tackle rational equations and apply them to real-world problems.