Solving Equations A Strategic Approach To Janet's Math Problem

In the realm of mathematics, solving equations is a fundamental skill. The equation presented to Janet, $y+\frac{y^2-5}{y^2-1}=\frac{y^2+y+2}{y+1}$, is a classic example of an algebraic challenge that requires careful manipulation and strategic thinking. The core of solving such equations lies in eliminating denominators to simplify the expression and make it more tractable. This involves identifying the common denominator and multiplying both sides of the equation by it. Let's embark on a detailed exploration of Janet's problem, dissecting the equation, identifying the critical steps, and understanding the underlying mathematical principles.

Unveiling the Equation's Structure

At first glance, the equation $y+\frac{y^2-5}{y^2-1}=\frac{y^2+y+2}{y+1}$ might appear daunting due to the presence of fractions and polynomial expressions. However, a closer inspection reveals a structure that can be systematically simplified. The key lies in recognizing the denominators: $y^2-1$ and $y+1$. The denominator $y^2-1$ is a difference of squares, which can be factored into $(y+1)(y-1)$. This factorization is crucial because it reveals a common factor with the other denominator, $y+1$. Understanding this relationship is the first step towards choosing the appropriate multiplier to clear the fractions.

To further clarify, let's rewrite the equation with the factored denominator: $y+\frac{y^2-5}{(y+1)(y-1)}=\frac{y^2+y+2}{y+1}$. Now, the common denominator is evident: it is $(y+1)(y-1)$. This means that multiplying both sides of the equation by $(y+1)(y-1)$ will eliminate the fractions, making the equation easier to solve. The strategy of identifying and utilizing common denominators is a cornerstone of algebraic manipulation, allowing us to transform complex equations into simpler forms.

The Power of Strategic Multiplication

The question at hand asks what Janet should multiply both sides of the equation by. As we've established, the key to solving this equation efficiently is to eliminate the fractions. This is achieved by multiplying both sides of the equation by the common denominator. In this case, the common denominator is $(y+1)(y-1)$, which is the factored form of $y^2-1$. Multiplying both sides by this expression will clear the fractions, leading to a simpler equation that Janet can solve.

This process is not merely a mechanical step; it's a strategic decision rooted in the fundamental principles of algebra. When we multiply both sides of an equation by the same non-zero expression, we maintain the equality. However, we must be mindful of potential restrictions on the variable $y$. Specifically, we must ensure that the expression we are multiplying by is not equal to zero. In this case, $y$ cannot be equal to 1 or -1, as these values would make the denominator zero, leading to an undefined expression. Therefore, it's important to note these restrictions and check for extraneous solutions at the end of the solving process.

By multiplying both sides by $(y+1)(y-1)$, Janet transforms the equation into a more manageable form, paving the way for further simplification and ultimately, the solution. This strategic multiplication is a powerful tool in the arsenal of any equation solver.

The Step-by-Step Transformation

Let's delve into the mechanics of multiplying both sides of the equation by $(y+1)(y-1)$. This will not only answer the question directly but also illustrate the subsequent simplification process. Starting with the equation $y+\frac{y^2-5}{(y+1)(y-1)}=\frac{y^2+y+2}{y+1}$, we multiply each term on both sides by $(y+1)(y-1)$. This gives us:

$y(y+1)(y-1) + \frac{(y^2-5)(y+1)(y-1)}{(y+1)(y-1)} = \frac{(y^2+y+2)(y+1)(y-1)}{y+1}$.

Now, we can simplify each term. In the second term on the left side, the $(y+1)(y-1)$ terms cancel out, leaving us with $y^2-5$. On the right side, the $(y+1)$ terms cancel out, leaving us with $(y^2+y+2)(y-1)$. The equation now looks like this:

$y(y+1)(y-1) + y^2 - 5 = (y^2+y+2)(y-1)$.

Expanding the terms, we get:

$y(y^2-1) + y^2 - 5 = y^3 - y^2 + y^2 - y + 2y - 2$.

Further simplification yields:

$y^3 - y + y^2 - 5 = y^3 + y - 2$.

This transformation highlights the power of multiplying by the common denominator. The fractions have been eliminated, and we are left with a polynomial equation that is much easier to solve. The next steps would involve collecting like terms, rearranging the equation, and solving for $y$. This step-by-step approach demonstrates how a complex equation can be systematically simplified through strategic algebraic manipulation.

The Importance of Restrictions and Extraneous Solutions

As mentioned earlier, it's crucial to consider restrictions on the variable when solving equations involving fractions. In Janet's equation, the original denominators are $y^2-1$ and $y+1$. These denominators cannot be equal to zero, as division by zero is undefined. Therefore, we must exclude any values of $y$ that would make these denominators zero.

Setting $y^2-1 = 0$, we find that $y = 1$ or $y = -1$. Similarly, setting $y+1 = 0$, we find that $y = -1$. Thus, $y$ cannot be equal to 1 or -1. These are the restrictions on the variable.

After solving the simplified polynomial equation, it's essential to check if any of the solutions are extraneous. Extraneous solutions are values that satisfy the transformed equation but not the original equation. This can occur when we perform operations that are not reversible, such as multiplying both sides of an equation by an expression that could be zero.

If Janet finds solutions $y = 1$ or $y = -1$, she must discard them as extraneous solutions. This step is critical to ensure the validity of the solution set. Ignoring these restrictions can lead to incorrect answers and a misunderstanding of the problem.

Therefore, while multiplying by the common denominator is a powerful technique, it's crucial to be mindful of the restrictions on the variable and to check for extraneous solutions. This careful approach ensures that we arrive at the correct and complete solution to the equation.

Summarizing Janet's Optimal Strategy

In conclusion, to solve the equation $y+\frac{y^2-5}{y^2-1}=\frac{y^2+y+2}{y+1}$, Janet should multiply both sides of the equation by $(y+1)(y-1)$, which is the common denominator. This strategic multiplication will eliminate the fractions, transforming the equation into a simpler polynomial equation. However, it's imperative that Janet remembers to consider the restrictions on the variable $y$, specifically that $y$ cannot be equal to 1 or -1. After solving the simplified equation, she must check for extraneous solutions to ensure the validity of her answer.

This approach exemplifies the importance of strategic algebraic manipulation in solving equations. By identifying the common denominator and multiplying both sides of the equation by it, Janet can effectively simplify the problem and find the solution. This method, combined with a careful consideration of restrictions and extraneous solutions, is a powerful tool for tackling algebraic challenges.

This detailed exploration of Janet's equation not only provides the answer to the immediate question but also highlights the broader principles of equation solving and algebraic manipulation. By understanding these principles, students and mathematicians alike can approach complex problems with confidence and strategic insight.