Solving Equations: A Step-by-Step Guide

Nov 13, 2025 by Alex Johnson 40 views

Hey there, math enthusiasts! Today, we're diving into a fun little equation that Janet's trying to crack. Equations can sometimes seem a bit intimidating, but trust me, with the right approach, they're totally manageable. We're going to break down how to tackle this specific problem and, more importantly, learn some general strategies that you can apply to many similar equations. So, let's get started and help Janet solve this equation!

Understanding the Problem: The Equation in Question

Let's take a look at the equation Janet is working with: $y+\frac{y^2-5}{y^2-1}=\frac{y^2+y+2}{y+1}$ . Our main goal is to figure out what Janet should multiply both sides of the equation by to simplify things and make it easier to solve. This is a common strategy when dealing with equations that have fractions – getting rid of those pesky denominators is often the first and most crucial step.

Why Clearing Denominators Matters

Why bother getting rid of denominators? Well, fractions can complicate things. They can make it harder to see the underlying structure of the equation and can be prone to errors. By multiplying both sides of an equation by a common multiple of the denominators, we can "clear" the fractions, transforming the equation into something much more manageable. This usually involves multiplying both sides by the least common multiple (LCM) of the denominators to simplify the equation effectively. This process avoids introducing more complex fractions and makes algebraic manipulation easier and less prone to errors. Remember that the ultimate goal is to isolate the variable, and clearing denominators is a powerful tool in achieving that goal.

Finding the Right Multiplier

Now, let's get to the heart of the matter: what should Janet multiply by? We need to look at the denominators in the equation to find the common multiple that will eliminate the fractions. Let's analyze the denominators individually to understand why.

Identifying the Denominators

First, we have $y^2 - 1$ and $y + 1$ . Recognizing these denominators is essential. The expression $y^2 - 1$ is a difference of squares, and it can be factored into $(y - 1)(y + 1)$ . The other denominator is simply $(y + 1)$ . This factoring is a crucial step; it helps us identify the least common multiple (LCM) needed to clear the fractions. The denominators guide us in selecting the right expression to multiply both sides of the equation by, ensuring we can simplify and solve the equation efficiently.

Choosing the Correct Multiplier

Looking at the options, we must determine which one, when multiplied, will eliminate all denominators. Here’s a breakdown:

A. $y$ : Multiplying by $y$ won't eliminate either denominator.
B. $y^2 - 1$ : This is equivalent to $(y-1)(y+1)$ . If we multiply both sides of the equation by this, we'll eliminate the denominator in the first fraction. Since $(y + 1)$ is a factor of $y^2 - 1$ , it will also eliminate the denominator in the second fraction on the right side of the equation. This is our best bet.
C. $y + 1$ : While this will eliminate the denominator of the right side, it won't get rid of the denominator $y^2 - 1$ . Additionally, it would leave us with a more complex expression.
D. $y^2 + y + 2$ : This doesn't share any factors with the existing denominators, making it ineffective for simplifying the equation. It will only make it more complicated.

Therefore, the correct answer is B. $y^2 - 1$ because it will eliminate all denominators in the equation.

Step-by-Step: Solving the Equation (Illustrative Example)

Let's walk through what happens when we multiply both sides by $y^2 - 1$ , to visualize how the equation changes. Remember, this is not a comprehensive solution to the equation, but it illustrates the first step:

Original Equation: $y+\frac{y^2-5}{y^2-1}=\frac{y^2+y+2}{y+1}$ .
Multiply Both Sides by $(y^2 - 1)$ : $(y^2-1) \cdot \left(y+\frac{y^2-5}{y^2-1}\right)=(y^2-1) \cdot \left(\frac{y^2+y+2}{y+1}\right)$ .
Distribute and Simplify: This becomes $y(y^2 - 1) + (y^2 - 5) = (y - 1)(y^2 + y + 2)$ .

Notice how the fractions are gone? Now the equation is much easier to work with. We can expand, collect like terms, and solve for y.

Expanding and Simplifying

To continue solving, we would:

Expand: $y^3 - y + y^2 - 5 = y^3 + y^2 + 2y - y^2 - y - 2$ .
Collect like terms: Combine terms with the same powers of y.
Isolate y: Move all terms involving y to one side and constants to the other. Be careful with signs!
Solve: Once you have a simplified equation, solve for y.

This is the general approach, which varies depending on the initial equation; the key is to perform each step with care, ensuring you maintain the equation's balance and accuracy.

Why This Approach Works

This method works because it transforms an equation involving fractions into a simpler one, which is easier to solve. Multiplying by the least common multiple (LCM) of the denominators is a fundamental algebraic technique. This approach effectively removes the fractions and streamlines the solving process. The efficiency comes from focusing on the structure of the equation and using the rules of algebra to systematically simplify it.

General Tips for Solving Equations with Fractions

Always Factor: Factor denominators whenever possible. This helps identify the LCM.
Check Your Work: After solving, plug your solution back into the original equation to verify that it is correct. Also, ensure your solution doesn't make any original denominators equal to zero (which would make the fraction undefined).
Stay Organized: Keep your work neat and clearly show each step. This will help you avoid errors.

Conclusion: Mastering Equation Solving

So, to recap, to solve the equation, Janet should multiply both sides by $y^2 - 1$ . This clears the fractions and simplifies the equation, making it easier to solve. Remember, mastering equation-solving is all about understanding the underlying principles and practicing consistently. The more you practice, the more comfortable and confident you'll become! Don't hesitate to ask for help or look for further resources if you're stuck. Math can be tricky, but it's also incredibly rewarding when you finally get the hang of it!

I hope this guide has been helpful! Keep practicing, and you'll be solving equations like a pro in no time.

For additional help with algebra, you can explore resources on websites such as:

Khan Academy - It offers a comprehensive course on algebra and other mathematical topics.