Solving For U: A Step-by-Step Guide To U - 15/u = 2

Hey math enthusiasts! Ever stumbled upon an equation that looks a bit intimidating at first glance? Well, today we're diving into one that might seem tricky but is totally manageable with the right approach. We're going to break down how to solve for u in the equation u - 15/u = 2. Don't worry, we'll take it step by step, so you'll be a pro in no time. Let's get started!

Understanding the Equation

Before we jump into the solution, let's make sure we understand what we're dealing with. The equation u - 15/u = 2 is a rational equation, meaning it involves a variable in the denominator. These types of equations often require a bit of algebraic maneuvering to solve, but nothing we can't handle! The key here is to eliminate the fraction, which will make the equation much easier to work with. We're essentially trying to find the value(s) of u that make this equation true. Think of it like a puzzle – we have all the pieces, now we just need to arrange them correctly. The variable u appears in two places, once as a standalone term and once in the denominator of a fraction. This is what makes the equation a little different from a simple linear equation. To solve for u, our main goal is to get rid of that fraction and rearrange the equation into a more familiar form, like a quadratic equation. Remember, quadratic equations are those that can be written in the form ax² + bx + c = 0, and we have some cool tools to solve them, like factoring or the quadratic formula. So, our initial strategy will be to multiply both sides of the equation by u. This will clear the fraction and set us up for the next steps. This process is a common technique when dealing with rational equations. It helps simplify the equation and makes it easier to isolate the variable we're trying to solve for. Once we have a clearer picture, we can apply other algebraic techniques to find the value(s) of u.

Step-by-Step Solution

Okay, let's get down to the nitty-gritty and solve this equation. We'll break it down into clear steps so you can follow along easily. Grab your pencil and paper, and let's do this!

Step 1: Eliminate the Fraction

The first thing we want to do is get rid of that fraction. To do this, we'll multiply both sides of the equation by u. Remember, whatever you do to one side of the equation, you have to do to the other to keep things balanced.

So, we have:

u - 15/u = 2

Multiply both sides by u:

u(u - 15/u) = 2u

This simplifies to:

u² - 15 = 2u

See how the fraction is gone? Awesome! Now we have a much cleaner equation to work with. This step is crucial because it transforms the equation from a rational one into a quadratic one, which is something we know how to handle. By multiplying by u, we've effectively cleared the denominator and made the equation more manageable. However, it's important to note that we've also introduced a potential restriction. Since we multiplied by u, we need to remember that u cannot be equal to 0. If u were 0, the original equation would be undefined due to the division by u. We'll need to keep this in mind and check our solutions later to make sure they don't violate this restriction. This is a common practice when solving rational equations – always be mindful of any values that would make the denominator zero.

Step 2: Rearrange into a Quadratic Equation

Now, let's rearrange the equation into the standard form of a quadratic equation, which is ax² + bx + c = 0. This will make it easier to identify the coefficients and apply our solving methods.

We have:

u² - 15 = 2u

Subtract 2u from both sides:

u² - 2u - 15 = 0

Great! We now have a quadratic equation in the standard form. This form is super helpful because it allows us to easily identify the values of a, b, and c, which we'll need for methods like the quadratic formula. In this case, a = 1, b = -2, and c = -15. Recognizing this standard form is a key step in solving quadratic equations. It sets us up for using various techniques like factoring, completing the square, or the quadratic formula. By getting the equation into this form, we've essentially translated it into a language that we can easily understand and manipulate. From here, we can choose the most efficient method to find the solutions for u. This rearrangement is a fundamental skill in algebra, and it's used in many different types of problem-solving scenarios.

Step 3: Solve the Quadratic Equation

There are a couple of ways we can solve this quadratic equation: factoring or using the quadratic formula. Let's try factoring first, as it's often the quickest method if it works.

We're looking for two numbers that multiply to -15 and add up to -2. Those numbers are -5 and 3.

So, we can factor the equation as:

(u - 5)(u + 3) = 0

Now, set each factor equal to zero and solve for u:

u - 5 = 0 or u + 3 = 0

u = 5 or u = -3

Awesome! We've found two potential solutions: u = 5 and u = -3. Factoring is a powerful technique for solving quadratic equations because it breaks down the problem into simpler parts. The idea behind factoring is that if the product of two factors is zero, then at least one of the factors must be zero. This allows us to split the quadratic equation into two linear equations, which are much easier to solve. In this case, we found the factors (u - 5) and (u + 3). Setting each of these equal to zero gave us the two possible solutions for u. If factoring doesn't come easily, don't worry! The quadratic formula is a reliable backup method that always works. However, when factoring is possible, it's often the most efficient way to solve the equation. This step demonstrates the importance of recognizing patterns and choosing the appropriate method to solve a problem.

Step 4: Check for Extraneous Solutions

Remember how we said u couldn't be 0? We need to check our solutions to make sure they don't violate this condition and that they actually work in the original equation.

Let's check u = 5:

5 - 15/5 = 2

5 - 3 = 2

2 = 2 (This is true!)

Now let's check u = -3:

-3 - 15/(-3) = 2

-3 + 5 = 2

2 = 2 (This is also true!)

Both solutions work! Hooray! This step is crucial in solving rational equations because it helps us identify and eliminate any extraneous solutions. Extraneous solutions are values that we get through the solving process but don't actually satisfy the original equation. They often arise when we perform operations like multiplying both sides of the equation by a variable expression, as we did in Step 1. By plugging our solutions back into the original equation, we can verify whether they are valid. In this case, both u = 5 and u = -3 checked out, meaning they are both legitimate solutions. However, it's important to be aware that this isn't always the case, and sometimes we might find solutions that don't work, which we would then discard. This step emphasizes the importance of being thorough and careful when solving mathematical problems.

Final Answer

So, the solutions for u in the equation u - 15/u = 2 are:

u = 5 and u = -3

We did it! We successfully solved for u. High five! Remember, the key to tackling these types of equations is to break them down into manageable steps. By eliminating the fraction, rearranging into a quadratic equation, solving for the variable, and checking for extraneous solutions, you can conquer even the trickiest-looking problems. Math can be like a fun puzzle if you approach it with the right strategies and a little bit of confidence. And there you have it, guys! You've now got another tool in your math belt. Solving rational equations might seem daunting at first, but with a systematic approach, you can totally nail it. Remember to always check your solutions, and don't be afraid to break down problems into smaller, more manageable steps. Keep practicing, and you'll become a math whiz in no time!