Solving Rational Equations Step-by-Step: Problems 16 And 17

Rational equations, such as the one presented in Problem 16, often appear daunting due to the presence of fractions and variables in the denominators. However, with a systematic approach, these equations can be solved effectively. In this detailed walkthrough, we will dissect the equation $16 \frac{3}{x+1}-\frac{x-2}{2}=\frac{x-2}{x+1}, providing a step-by-step solution and highlighting crucial concepts. Understanding these steps is vital not only for solving this particular equation but also for tackling a wide range of rational equations.

The first critical step in solving any rational equation is to eliminate the fractions. This is achieved by identifying the least common denominator (LCD) of all the fractions present in the equation. In this case, we have denominators ${x+1}$ and ${2}$. The LCD is simply the product of these two, which is ${2(x+1)}$. Multiplying both sides of the equation by the LCD will clear the fractions and transform the equation into a more manageable form. It is essential to perform this multiplication carefully, ensuring that each term is correctly multiplied by the LCD. This process simplifies the equation, making it easier to solve for the unknown variable ${x}$.

Now, let's perform the multiplication. Multiplying both sides of the equation $16 \frac{3}{x+1}-\frac{x-2}{2}=\frac{x-2}{x+1}, by ${2(x+1)}$ gives us:

${2(x+1) \left(\frac{3}{x+1}-\frac{x-2}{2}\right) = 2(x+1) \left(\frac{x-2}{x+1}\right).}$

Distributing ${2(x+1)}$ to each term, we get:

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

Next, we simplify each term by canceling out common factors. In the first term, ${(x+1)}$ cancels out, leaving us with ${2 \cdot 3 = 6}$. In the second term, ${2}$ cancels out, giving us ${(x+1)(x-2)}$. In the third term, ${(x+1)}$ cancels out, leaving ${2(x-2)}$. Thus, the equation simplifies to:

${6 - (x+1)(x-2) = 2(x-2).}$

The next step involves expanding and simplifying the equation further. We need to expand the product ${(x+1)(x-2)}$ and the term ${2(x-2)}$. Expanding ${(x+1)(x-2)}$, we use the distributive property (also known as FOIL) to get:

${x^2 - 2x + x - 2 = x^2 - x - 2.}$

Expanding ${2(x-2)}$ gives us:

${2x - 4.}$

Substituting these expansions back into our equation, we have:

${6 - (x^2 - x - 2) = 2x - 4.}$

Distribute the negative sign on the left side:

${6 - x^2 + x + 2 = 2x - 4.}$

Combine like terms on the left side:

${8 - x^2 + x = 2x - 4.}$

Now, to solve this quadratic equation, we need to rearrange it into the standard form ${ax^2 + bx + c = 0}$. Moving all terms to one side gives us:

${0 = x^2 + x - 12.}$

Thus, our quadratic equation is:

${x^2 + x - 12 = 0.}$

Now that we have a standard quadratic equation, the next step is to solve it for ${x}$. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, the equation can be easily factored. We are looking for two numbers that multiply to ${-12}$ and add up to ${1}$. These numbers are ${4}$ and ${-3}$. Therefore, we can factor the quadratic equation as:

${(x + 4)(x - 3) = 0.}$

Setting each factor equal to zero gives us the possible solutions for ${x}$:

${x + 4 = 0 \quad \text{or} \quad x - 3 = 0.}$

Solving these linear equations, we find:

${x = -4 \quad \text{or} \quad x = 3.}$

These values are our potential solutions. However, in the context of rational equations, it is crucial to check these solutions for extraneous roots. Extraneous roots are solutions that satisfy the transformed equation but not the original equation. This typically occurs when a solution makes one of the original denominators equal to zero.

In the original equation, the denominators are ${x+1}$ and ${2}$. The denominator ${2}$ is never zero. However, ${x+1}$ will be zero when ${x = -1}$. Since neither of our solutions, ${x = -4}$ nor ${x = 3}$, makes the denominator zero, both solutions are valid.

Therefore, the solutions to the equation $16 \frac{3}{x+1}-\frac{x-2}{2}=\frac{x-2}{x+1}, are ${x = -4}$ and ${x = 3}$. This careful consideration of extraneous roots ensures the accuracy of our final answer, a critical step in solving rational equations.

In summary, solving rational equations involves several key steps: identifying the LCD, multiplying to eliminate fractions, simplifying the resulting equation, solving for the variable, and checking for extraneous roots. Each of these steps is important and contributes to the final, accurate solution. By following this methodical approach, complex rational equations can be tackled with confidence.

Problem 17 presents another fascinating rational equation: $1-\frac{3}{b}=\frac{-8 b}{b^2+3 b}. Just like in Problem 16, the presence of fractions requires a strategic approach to find the solution. This problem not only tests our algebraic skills but also our understanding of how to manipulate equations with variables in the denominator. To solve this equation effectively, we will systematically eliminate the fractions, simplify the equation, and solve for the unknown variable ${b}$. This process involves identifying the least common denominator, clearing fractions, and finally, ensuring that our solutions are valid by checking for extraneous roots.

To begin, we identify the least common denominator (LCD) for the fractions in the equation. The denominators are ${b}$ and ${b^2 + 3b}$. Notice that ${b^2 + 3b}$ can be factored as ${b(b + 3)}$. Thus, the denominators are ${b}$ and ${b(b + 3)}$. The LCD is therefore ${b(b + 3)}$. Multiplying every term in the equation by this LCD will eliminate the fractions and simplify the equation into a more workable form. This initial step is crucial because it transforms the rational equation into a standard algebraic equation that we can solve using familiar techniques. Remember, accuracy in this step is vital to prevent errors in the subsequent steps.

Multiplying both sides of the equation $1-\frac{3}{b}=\frac{-8 b}{b^2+3 b}, by the LCD ${b(b + 3)}$ gives us:

${b(b + 3) \left(1 - \frac{3}{b}\right) = b(b + 3) \left(\frac{-8b}{b^2 + 3b}\right).}$

Distribute ${b(b + 3)}$ to each term:

${b(b + 3) \cdot 1 - b(b + 3) \cdot \frac{3}{b} = b(b + 3) \cdot \frac{-8b}{b(b + 3)}.}$

Now, we simplify each term by canceling out common factors. In the first term, we simply have ${b(b + 3)}$. In the second term, ${b}$ cancels out, leaving us with ${-3(b + 3)}$. In the third term, ${b(b + 3)}$ cancels out, leaving us with ${-8b}$. Thus, the equation simplifies to:

${b(b + 3) - 3(b + 3) = -8b.}$

The next step is to expand and simplify the equation. Expanding ${b(b + 3)}$ gives us:

${b^2 + 3b.}$

Expanding ${-3(b + 3)}$ gives us:

${-3b - 9.}$

Substituting these expansions back into our equation, we have:

${b^2 + 3b - 3b - 9 = -8b.}$

Combine like terms on the left side:

${b^2 - 9 = -8b.}$

To solve this quadratic equation, we need to rearrange it into the standard form ${ax^2 + bx + c = 0}$. Adding ${8b}$ to both sides gives us:

${b^2 + 8b - 9 = 0.}$

Now that we have a standard quadratic equation, the next crucial step is to solve for ${b}$. Again, we can use various methods, such as factoring, completing the square, or the quadratic formula. In this case, the equation can be easily factored. We are looking for two numbers that multiply to ${-9}$ and add up to ${8}$. These numbers are ${9}$ and ${-1}$. Therefore, we can factor the quadratic equation as:

${(b + 9)(b - 1) = 0.}$

Setting each factor equal to zero gives us the possible solutions for ${b}$:

${b + 9 = 0 \quad \text{or} \quad b - 1 = 0.}$

Solving these linear equations, we find:

${b = -9 \quad \text{or} \quad b = 1.}$

These values are our potential solutions, but we must check for extraneous roots. In the original equation, the denominators are ${b}$ and ${b^2 + 3b}$, which can be written as ${b(b + 3)}$. The denominator will be zero if ${b = 0}$ or ${b = -3}$. Neither of our solutions, ${b = -9}$ nor ${b = 1}$, makes the denominator zero. Therefore, both solutions are valid.

Thus, the solutions to the equation $1-\frac{3}{b}=\frac{-8 b}{b^2+3 b}, are ${b = -9}$ and ${b = 1}$. This process exemplifies the importance of verifying solutions in rational equations to ensure accuracy.

In conclusion, solving rational equations requires careful attention to detail and a systematic approach. The steps involve identifying the LCD, clearing fractions, simplifying the equation, solving for the variable, and crucially, checking for extraneous roots. By mastering these steps, you can confidently tackle a wide array of rational equations. The methodical approach not only leads to correct solutions but also reinforces your understanding of algebraic manipulations and the fundamental principles of equation solving.

Conclusion

In summary, solving rational equations like Problems 16 and 17 involves a multi-step process that demands a thorough understanding of algebraic principles. From identifying the least common denominator to checking for extraneous solutions, each step plays a crucial role in obtaining the correct answer. By mastering these techniques, students can confidently approach complex equations and develop a deeper appreciation for the elegance and rigor of mathematics. The ability to solve rational equations is a valuable skill that extends beyond the classroom, finding applications in various fields of science, engineering, and beyond. Understanding these concepts builds a strong foundation for advanced mathematical studies and real-world problem-solving.