Solving Rational Equations: Finding Valid Solutions

Hey guys! Today, we're going to dive deep into the world of rational equations and tackle a specific problem that involves finding valid solutions. We'll break down the equation step-by-step, explore potential pitfalls like extraneous solutions, and make sure you understand the process inside and out. So, let's get started!

Understanding the Problem

Our main goal here is to figure out the solutions to this equation: $\frac{2}{x+2}+\frac{1}{10}=\frac{3}{x+3}$. We need to identify the values of 'x' that make this equation true. But, it's not as simple as just plugging in numbers. With rational equations, we need to watch out for extraneous solutions – values that seem to work but actually don't when you plug them back into the original equation. These usually arise from the process of clearing denominators, which can sometimes introduce solutions that make the denominators zero, and we can't have that, right? It's like dividing by zero – a big no-no in the math world! So, buckle up, because we're about to get into the nitty-gritty of solving this equation and verifying our answers.

Initial Equation and Potential Pitfalls

The equation we're working with is $\frac{2}{x+2}+\frac{1}{10}=\frac{3}{x+3}$. The very first thing we need to consider are the values of x that would make the denominators zero. Remember, division by zero is undefined, so these values cannot be solutions. We have two denominators with x in them: x + 2 and x + 3. Setting each of these equal to zero gives us:

• x + 2 = 0 => x = -2
• x + 3 = 0 => x = -3

So, x cannot be -2 or -3. These are our critical values. If we get these as solutions later, we'll know they are extraneous and must be discarded. It's essential to keep these critical values in mind throughout the solving process. Think of them as warning signs on our mathematical journey. They tell us where we might encounter problems and help us avoid incorrect answers. Ignoring these values can lead to selecting incorrect solutions, so this step is crucial for accuracy. Before diving into the algebraic manipulations, it is always good to identify these restrictions on x to avoid mistakes down the road. This proactive approach saves time and ensures a more accurate solution.

Clearing the Denominators

To solve the equation $\frac{2}{x+2}+\frac{1}{10}=\frac{3}{x+3}$, the first major step is to eliminate the fractions. Fractions can make equations look intimidating, but clearing the denominators simplifies the equation significantly. We do this by multiplying both sides of the equation by the least common multiple (LCM) of the denominators. Our denominators are x + 2, 10, and x + 3. The LCM of these is 10(x + 2)(x + 3).

Let's multiply both sides of the equation by this LCM:

10(x + 2)(x + 3) * [\frac{2}{x+2}+\frac{1}{10}] = 10(x + 2)(x + 3) * \frac{3}{x+3}

Distributing on the left side gives us:

10(x + 2)(x + 3) * \frac{2}{x+2} + 10(x + 2)(x + 3) * \frac{1}{10} = 10(x + 2)(x + 3) * \frac{3}{x+3}

Now, we can cancel out common factors:

10(x + 3) * 2 + (x + 2)(x + 3) = 10(x + 2) * 3

This simplifies to:

20(x + 3) + (x + 2)(x + 3) = 30(x + 2)

See how much cleaner the equation looks now? By clearing the denominators, we've transformed a rational equation into a more manageable polynomial equation. This is a standard technique when solving equations with fractions. It allows us to work with whole numbers and polynomials, which are often easier to manipulate algebraically. This step is crucial because it sets the stage for solving the equation using standard algebraic methods, like expanding and combining like terms. The careful multiplication and cancellation are key to getting a simplified equation that we can solve accurately.

Solving the Simplified Equation

Now that we've cleared the denominators, we have a much friendlier equation to work with: 20(x + 3) + (x + 2)(x + 3) = 30(x + 2). Our next task is to expand the expressions and simplify the equation further. This involves distributing and combining like terms to get a standard polynomial equation, which we can then solve using various methods. This stage is all about careful algebraic manipulation to get to the solutions. Let's dive in!

Expanding and Simplifying

First, let's expand the terms:

20x + 60 + (x^2 + 3x + 2x + 6) = 30x + 60

Simplifying the left side, we get:

20x + 60 + x^2 + 5x + 6 = 30x + 60

Combining like terms on the left side gives us:

x^2 + 25x + 66 = 30x + 60

Now, let's move all terms to one side to set the equation to zero:

x^2 + 25x + 66 - 30x - 60 = 0

This simplifies to:

x^2 - 5x + 6 = 0

Great! We now have a quadratic equation in the standard form: ax^2 + bx + c = 0. This is a familiar form, and we have several methods to solve it, including factoring, completing the square, or using the quadratic formula. Factoring is often the quickest way if the quadratic expression can be factored easily. The process of expanding and simplifying is crucial because it transforms the equation into a recognizable form that we can solve using standard techniques. Each step, from distributing to combining like terms, needs to be done carefully to avoid errors that can lead to incorrect solutions. The resulting quadratic equation sets the stage for the final step of finding the values of x that satisfy the original equation.

Factoring the Quadratic

Our quadratic equation is x^2 - 5x + 6 = 0. To solve this, we can try factoring. We're looking for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3. So, we can factor the quadratic as:

(x - 2)(x - 3) = 0

Now, we use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. So, we set each factor equal to zero:

• x - 2 = 0 => x = 2
• x - 3 = 0 => x = 3

So, our potential solutions are x = 2 and x = 3. These values are the candidates for the solutions to our original rational equation, but we're not done yet. We need to check these solutions to make sure they are valid and not extraneous. Remember those critical values we identified at the beginning? This is where they come into play. Factoring the quadratic is a powerful technique for solving quadratic equations because it breaks the equation down into simpler linear equations. The key is to correctly identify the factors, which can sometimes require a bit of trial and error or a solid understanding of factoring patterns. The zero-product property is the linchpin of this method, allowing us to transform the factored equation into individual equations that are easy to solve. However, finding potential solutions is only part of the process; the next critical step is to verify these solutions against the original equation.

Checking for Extraneous Solutions

We've found two potential solutions: x = 2 and x = 3. However, with rational equations, we always need to check for extraneous solutions. These are values that satisfy the transformed equation but not the original equation. This usually happens because clearing denominators can introduce solutions that make the original denominators zero. This is a critical step because including extraneous solutions in our answer would be incorrect. Let's plug each of our potential solutions back into the original equation and see if they work.

Checking x = 2

Original equation: $\frac{2}{x+2}+\frac{1}{10}=\frac{3}{x+3}$

Substitute x = 2:

$\frac{2}{2+2}+\frac{1}{10}=\frac{3}{2+3}$

$\frac{2}{4}+\frac{1}{10}=\frac{3}{5}$

$\frac{1}{2}+\frac{1}{10}=\frac{3}{5}$

To add the fractions on the left, we need a common denominator, which is 10:

$\frac{5}{10}+\frac{1}{10}=\frac{3}{5}$

$\frac{6}{10}=\frac{3}{5}$

Simplifying the left side:

$\frac{3}{5}=\frac{3}{5}$

This is true! So, x = 2 is a valid solution. Checking solutions is like a final exam for each potential answer. We are putting the solutions to the ultimate test: do they satisfy the original equation? If a solution passes this test, we can confidently include it in our answer. If it fails, it's an extraneous solution and must be discarded. The verification process ensures that our solutions are mathematically sound and accurately reflect the solutions to the original problem.

Checking x = 3

Now, let's check x = 3 using the original equation: $\frac{2}{x+2}+\frac{1}{10}=\frac{3}{x+3}$

Substitute x = 3:

$\frac{2}{3+2}+\frac{1}{10}=\frac{3}{3+3}$

$\frac{2}{5}+\frac{1}{10}=\frac{3}{6}$

$\frac{2}{5}+\frac{1}{10}=\frac{1}{2}$

To add the fractions on the left, we need a common denominator, which is 10:

$\frac{4}{10}+\frac{1}{10}=\frac{1}{2}$

$\frac{5}{10}=\frac{1}{2}$

Simplifying the left side:

$\frac{1}{2}=\frac{1}{2}$

This is also true! So, x = 3 is a valid solution. By meticulously checking each potential solution, we ensure that our final answer is accurate and complete. This careful approach is what distinguishes solving rational equations from simpler algebraic problems. The process not only validates our solutions but also reinforces our understanding of the equation and the steps involved in solving it.

Conclusion: Valid Solutions and No Extraneous Solutions

After solving the equation $\frac{2}{x+2}+\frac{1}{10}=\frac{3}{x+3}$ and carefully checking our solutions, we found that both x = 2 and x = 3 are valid solutions. Neither of these values makes any of the original denominators zero, and they both satisfy the original equation when substituted back in. Therefore, we can confidently say that the equation has two valid solutions and no extraneous solutions.

So, the correct answer is A: The equation has two valid solutions and no extraneous solutions.

Solving rational equations can seem tricky at first, but by following a systematic approach – clearing denominators, solving the resulting equation, and, most importantly, checking for extraneous solutions – you can tackle these problems with confidence. Remember, the key is to be careful and methodical in each step, and always double-check your work. Keep practicing, and you'll become a pro at solving rational equations in no time! You got this, guys!