Solving Equations: A Step-by-Step Guide

Hey math enthusiasts! Today, we're diving into the world of equation solving. Specifically, we're tackling the equation $-\frac{18}{v^2-4 v+3}=-\frac{3 v}{v-3}$. Don't worry if it looks a bit intimidating at first; we'll break it down step by step and make sure you understand every bit of the process. Our goal here is to find the value(s) of 'v' that make this equation true. Now, let's get started!

Understanding the Problem: The Equation Unpacked

First things first, let's take a closer look at our equation. We have $-\frac{18}{v^2-4 v+3}=-\frac{3 v}{v-3}$. This is a rational equation, meaning it involves fractions where the variables are in the denominators. The presence of variables in the denominator is a key indicator that we need to be extra cautious. Why? Because we can't divide by zero! So, before we jump into solving, we need to identify any values of 'v' that would make our denominators equal to zero. These values are called excluded values, and they're super important to keep in mind throughout the solving process.

Looking at the denominators, we have two expressions to consider: $v^2 - 4v + 3$ and $v - 3$. Let's start with $v - 3$. If $v - 3 = 0$, then $v = 3$. This means that 'v' cannot be equal to 3, as it would make the denominator zero, and our equation undefined. Now, let's analyze the other denominator, $v^2 - 4v + 3$. This is a quadratic expression, and we can factor it to make it easier to work with. Factoring $v^2 - 4v + 3$, we get $(v - 3)(v - 1)$. Setting this equal to zero, $(v - 3)(v - 1) = 0$, we find that $v = 3$ and $v = 1$ are potential excluded values. However, since $v=3$ is already identified, it is the only one. Therefore, the excluded values for our equation are $v = 3$ and $v = 1$. It's crucial that we remember these excluded values, because if we find a solution that matches one of them, then that's not a valid solution.

Now, armed with this understanding, we can go ahead and begin solving the equation. Remember, always be mindful of excluded values throughout the entire process. This is so that at the end of solving the equation you can identify the extraneous solutions.

Step-by-Step Solution: Finding the Value of 'v'

Now, let's dive into the core of the problem: solving for 'v'. Our main goal here is to isolate 'v' on one side of the equation. We'll achieve this by performing a series of algebraic operations. Our equation is $-\frac{18}{v^2-4 v+3}=-\frac{3 v}{v-3}$. To start, it's often helpful to get rid of the fractions. To do this, we'll multiply both sides of the equation by the least common denominator (LCD). In this case, since we already factored $v^2 - 4v + 3$ into $(v - 3)(v - 1)$, our LCD is $(v - 3)(v - 1)$.

Multiplying both sides by $(v - 3)(v - 1)$, we get:

$(v - 3)(v - 1) * \frac{-18}{(v - 3)(v - 1)} = (v - 3)(v - 1) * \frac{-3v}{v - 3}$

Notice that on the left side, $(v - 3)(v - 1)$ cancels out with the denominator, leaving us with -18. On the right side, $(v - 3)$ cancels out, leaving us with $-3v(v - 1)$. This simplifies our equation to:

$-18 = -3v(v - 1)$

Now, let's simplify further by dividing both sides by -3. This gives us:

$6 = v(v - 1)$

Next, we need to expand the right side to get a quadratic equation. This means multiplying v by both terms inside the parentheses:

$6 = v^2 - v$

To solve this, we want to set the equation to zero. So, we'll subtract 6 from both sides:

$0 = v^2 - v - 6$

Now we have a quadratic equation in the standard form. We can solve this by factoring. We are looking for two numbers that multiply to -6 and add up to -1. Those numbers are -3 and 2. Therefore, our factored equation is:

$0 = (v - 3)(v + 2)$

This gives us two potential solutions: $v = 3$ and $v = -2$. But wait! Remember those excluded values we found earlier? One of our potential solutions, $v = 3$, is an excluded value. That means it cannot be a valid solution because it would make our original equation undefined (division by zero). Therefore, we discard $v = 3$. This leaves us with just one solution to consider: $v = -2$.

Verification and Conclusion

Alright, guys, we've arrived at a potential solution: $v = -2$. However, we're not quite done yet. It's always a good practice to verify your solution. Verification is our final check to make sure our value of 'v' actually works in the original equation. Let's substitute $v = -2$ back into our original equation: $-\frac{18}{v^2-4 v+3}=-\frac{3 v}{v-3}$.

Substituting $v = -2$, we get:

$-\frac{18}{(-2)^2-4(-2)+3}=-\frac{3(-2)}{-2-3}$

Let's simplify both sides of the equation. On the left side, we get:

$-\frac{18}{4 + 8 + 3} = -\frac{18}{15} = -\frac{6}{5}$

On the right side, we get:

$-\frac{-6}{-5} = -\frac{6}{5}$

Since both sides are equal, our solution $v = -2$ is valid! We can confidently say that the solution to the equation $-\frac{18}{v^2-4 v+3}=-\frac{3 v}{v-3}$ is $v = -2$. This marks the end of our journey through this equation. You've now seen a practical example of solving rational equations, which involves factoring, finding the LCD, and the all important excluded values. Remember that practice is key, so keep practicing and you'll become more confident in solving similar problems.

In conclusion, through a methodical approach, we successfully solved the equation. We factored, simplified, and verified our solution, always keeping an eye on those important excluded values. Solving equations might seem like a complex task at first, but by breaking it down into smaller, manageable steps, you can achieve amazing results. So, the next time you encounter an equation, remember the steps we've covered today, and approach it with confidence. You've got this!