Solving Rational Equations Determining The Correct Multiplier For Janet's Equation

When dealing with equations involving rational expressions, the primary goal is to eliminate the fractions. This simplifies the equation and allows us to solve for the variable more easily. Janet's equation, y + (y^2 - 5) / (y^2 - 1) = (y^2 + y + 2) / (y + 1), is a classic example of such an equation. To tackle this, we need to identify the least common denominator (LCD) of all the fractions present. This article delves into the process of finding the LCD and using it to solve rational equations.

To effectively solve Janet's equation, y + (y^2 - 5) / (y^2 - 1) = (y^2 + y + 2) / (y + 1), we first need to understand the importance of the least common denominator (LCD). The LCD is the smallest expression that is divisible by all the denominators in the equation. In this case, the denominators are 1 (since y can be written as y/1), y^2 - 1, and y + 1. Identifying the LCD is the crucial first step in clearing the fractions and simplifying the equation. Factoring the denominators is a vital part of this process. We notice that y^2 - 1 is a difference of squares and can be factored into (y - 1)(y + 1). The other denominator, y + 1, is already in its simplest form. The LCD, therefore, must include all unique factors present in the denominators. Here, the factors are (y - 1) and (y + 1). Thus, the LCD is (y - 1)(y + 1), which is equivalent to y^2 - 1. Multiplying both sides of the equation by the LCD will eliminate the fractions, making the equation easier to solve. This technique is fundamental in dealing with rational equations and allows us to transform a complex-looking equation into a more manageable form. The subsequent steps involve simplifying the equation, combining like terms, and isolating the variable to find its value. Understanding the LCD and its role is paramount in this process.

Identifying the Least Common Denominator (LCD)

To solve the equation, y + (y^2 - 5) / (y^2 - 1) = (y^2 + y + 2) / (y + 1), the first critical step is to identify the least common denominator (LCD). The LCD is the smallest expression that all denominators in the equation divide into evenly. This concept is crucial because multiplying both sides of the equation by the LCD will eliminate the fractions, simplifying the equation significantly. To find the LCD, we first need to factor each denominator completely. In our equation, the denominators are 1 (for the term y), y^2 - 1, and y + 1. The denominator y^2 - 1 is a difference of squares, which can be factored into (y - 1)(y + 1). The denominator y + 1 is already in its simplest form and cannot be factored further. Now, we list all the unique factors present in the denominators: (y - 1) and (y + 1). The LCD is the product of these unique factors, which is (y - 1)(y + 1). This expression is equivalent to y^2 - 1. Therefore, the LCD for the given equation is y^2 - 1. Multiplying both sides of the equation by this LCD will clear the fractions, allowing us to solve for y more easily. Recognizing and correctly identifying the LCD is a fundamental skill in solving rational equations, as it transforms a complex equation into a simpler algebraic form.

Multiplying Both Sides by the LCD

After identifying the LCD as y^2 - 1, the next crucial step in solving Janet's equation, y + (y^2 - 5) / (y^2 - 1) = (y^2 + y + 2) / (y + 1), is to multiply both sides of the equation by this LCD. This process eliminates the fractions, making the equation significantly easier to manipulate and solve. Multiplying both sides ensures that the equation remains balanced, adhering to the fundamental principles of algebra. When we multiply the left side of the equation, [y + (y^2 - 5) / (y^2 - 1)], by y^2 - 1, we distribute the LCD to each term. This gives us y(y^2 - 1) + (y^2 - 5). Notice that when we multiply (y^2 - 5) / (y^2 - 1) by y^2 - 1, the (y^2 - 1) terms cancel out, leaving us with just y^2 - 5. On the right side of the equation, we multiply [(y^2 + y + 2) / (y + 1)] by y^2 - 1, which can be written as (y - 1)(y + 1). Here, the (y + 1) term in the denominator cancels out with the (y + 1) factor in the LCD, leaving us with (y^2 + y + 2)(y - 1). The resulting equation, after multiplying both sides by the LCD, is y(y^2 - 1) + (y^2 - 5) = (y^2 + y + 2)(y - 1). This equation is now free of fractions, making it simpler to solve. The next steps involve expanding the expressions, combining like terms, and solving for y. This method of multiplying by the LCD is a cornerstone technique in solving rational equations and is essential for transforming complex equations into manageable forms.

Solving the Simplified Equation

Once we have multiplied both sides of Janet's equation, y + (y^2 - 5) / (y^2 - 1) = (y^2 + y + 2) / (y + 1), by the LCD (y^2 - 1), we arrive at a simplified equation without fractions. This new equation, y(y^2 - 1) + (y^2 - 5) = (y^2 + y + 2)(y - 1), is much easier to work with. The next step is to expand the expressions on both sides of the equation. On the left side, y(y^2 - 1) expands to y^3 - y, so the left side becomes y^3 - y + y^2 - 5. On the right side, (y^2 + y + 2)(y - 1) requires careful distribution. Multiplying each term in the first expression by each term in the second expression gives us y^3 - y^2 + y^2 - y + 2y - 2, which simplifies to y^3 + y - 2. Now, the equation looks like this: y^3 - y + y^2 - 5 = y^3 + y - 2. The next step is to combine like terms and simplify further. We can subtract y^3 from both sides, which eliminates the cubic term. This leaves us with -y + y^2 - 5 = y - 2. Rearranging the terms to one side to set the equation to zero, we subtract y from both sides and add 2 to both sides, resulting in y^2 - 2y - 3 = 0. Now we have a quadratic equation, which can be solved by factoring, completing the square, or using the quadratic formula. In this case, the quadratic expression y^2 - 2y - 3 can be factored into (y - 3)(y + 1). Setting each factor equal to zero gives us two possible solutions: y - 3 = 0 and y + 1 = 0. Solving these equations yields y = 3 and y = -1. However, it is crucial to check these solutions in the original equation to ensure they are valid and do not result in division by zero, as y = -1 would make the denominators y^2 - 1 and y + 1 equal to zero.

Checking for Extraneous Solutions

After solving the simplified equation, we obtained two potential solutions: y = 3 and y = -1. However, when dealing with rational equations, it is crucial to check for extraneous solutions. Extraneous solutions are values that satisfy the simplified equation but do not satisfy the original equation because they lead to division by zero. In Janet's equation, y + (y^2 - 5) / (y^2 - 1) = (y^2 + y + 2) / (y + 1), the denominators are y^2 - 1 and y + 1. These denominators cannot be equal to zero, as division by zero is undefined. This means that y cannot be equal to 1 or -1, since these values would make the denominators zero. Let's first check y = 3. Substituting y = 3 into the original equation, we get 3 + (3^2 - 5) / (3^2 - 1) = (3^2 + 3 + 2) / (3 + 1), which simplifies to 3 + (9 - 5) / (9 - 1) = (9 + 3 + 2) / 4, further simplifying to 3 + 4 / 8 = 14 / 4. This gives us 3 + 1/2 = 7/2, which is true, as 3.5 = 3.5. Therefore, y = 3 is a valid solution. Now, let's check y = -1. Substituting y = -1 into the original equation, we immediately encounter a problem. The denominators y^2 - 1 and y + 1 become zero, which means that y = -1 is an extraneous solution. It does not satisfy the original equation because it leads to division by zero. Therefore, the only valid solution to Janet's equation is y = 3. This step of checking for extraneous solutions is vital in solving rational equations, as it ensures that the solutions obtained are meaningful and correct within the context of the original problem.

Conclusion

In summary, to solve the equation y + (y^2 - 5) / (y^2 - 1) = (y^2 + y + 2) / (y + 1), Janet should multiply both sides of the equation by the least common denominator (LCD). The LCD, in this case, is y^2 - 1. This step eliminates the fractions, simplifying the equation and making it easier to solve. By multiplying both sides by y^2 - 1, Janet can transform the equation into a more manageable form, which can then be solved using standard algebraic techniques. It is crucial to remember to check for extraneous solutions after solving the simplified equation to ensure the validity of the solutions obtained. The process of identifying the LCD, multiplying both sides by it, simplifying the equation, and checking for extraneous solutions is a fundamental approach to solving rational equations. This method not only simplifies the equation but also helps in avoiding errors caused by division by zero. Therefore, the correct answer to what Janet should multiply both sides of the equation by is y^2 - 1. This comprehensive approach ensures accuracy and a thorough understanding of the solution process.