Equation Solver: Is The Student Right?

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Hey everyone! Today, we're diving into a math problem that many students encounter: solving algebraic equations. We'll be looking at a specific equation and checking if a student's solution is correct. This is a common type of problem, and understanding how to solve it is key to mastering algebra. So, let's get started and see if our student nailed it! Remember, the goal here isn't just to find the answer but to understand why the answer is what it is. This is all about the journey of learning, you know?

The Equation and the Student's Claim

Alright, let's break down the problem. A student tackled the equation: 3a+2−6aa2−4=1a−2\frac{3}{a+2}-\frac{6 a}{a^2-4}=\frac{1}{a-2}. They claimed that the solution to this equation is a = -2. Our task? To figure out if they're right! This means we need to meticulously solve the equation ourselves, step-by-step, and then compare our findings with the student's answer. This process of verification is super important in math; it helps us catch mistakes and builds a deeper understanding of the concepts. Keep in mind that when we're solving equations like this, we're essentially looking for the value(s) of the variable (in this case, a) that make the equation true. It's like finding a secret code that unlocks the solution. We'll need to use our knowledge of algebra to manipulate the equation, simplify it, and isolate the variable to find the correct value(s). It's all about following the rules and being patient. Are you ready to dive in, guys?

Step-by-Step Solution: Unveiling the Truth

Let's meticulously solve the equation 3a+2−6aa2−4=1a−2\frac{3}{a+2}-\frac{6 a}{a^2-4}=\frac{1}{a-2}.

  1. Factoring the Denominator: The first thing we want to do is to factor the denominator of the second term. Notice that a² - 4 is a difference of squares, and it can be factored into (a + 2)(a - 2). So, our equation becomes: 3a+2−6a(a+2)(a−2)=1a−2\frac{3}{a+2}-\frac{6 a}{(a+2)(a-2)}=\frac{1}{a-2}.
  2. Finding a Common Denominator: To combine the fractions, we need a common denominator. In this case, the least common denominator (LCD) is (a + 2)(a - 2). We will multiply each term to achieve the LCD. The first term must be multiplied by (a - 2) / (a - 2). The second term already has the LCD. The third term must be multiplied by (a + 2) / (a + 2). This gives us: 3(a−2)(a+2)(a−2)−6a(a+2)(a−2)=1(a+2)(a+2)(a−2)\frac{3(a-2)}{(a+2)(a-2)}-\frac{6 a}{(a+2)(a-2)}=\frac{1(a+2)}{(a+2)(a-2)}.
  3. Simplifying and Combining: Now that all the terms have the same denominator, we can combine the numerators: 3(a−2)−6a(a+2)(a−2)=a+2(a+2)(a−2)\frac{3(a-2)-6a}{(a+2)(a-2)}=\frac{a+2}{(a+2)(a-2)}. Then simplify the numerator on the left side: 3(a - 2) - 6a = 3a - 6 - 6a = -3a - 6. So we have: −3a−6(a+2)(a−2)=a+2(a+2)(a−2)\frac{-3a-6}{(a+2)(a-2)}=\frac{a+2}{(a+2)(a-2)}.
  4. Multiplying by the Denominator: Now, we can eliminate the denominators by multiplying both sides of the equation by (a + 2)(a - 2), which leaves us with: -3a - 6 = a + 2.
  5. Isolating the Variable: Next, let's gather all the terms with 'a' on one side and the constants on the other side. Add 3a to both sides and subtract 2 from both sides, which gives us: -6 - 2 = a + 3a, which simplifies to: -8 = 4a.
  6. Solving for a: To solve for a, we divide both sides by 4: a = -8 / 4, so a = -2.

The Plot Thickens: A Potential Roadblock

Now, here's where things get really interesting. We've solved the equation, and we also got a = -2! But wait a second... Remember, the original equation had terms with (a + 2) and (a - 2) in the denominators. If a = -2, then a + 2 = 0. And, you can't divide by zero! Dividing by zero is undefined in mathematics; it breaks all the rules. This means that a = -2 is an extraneous solution. An extraneous solution is a solution that we obtain through our algebraic manipulation, but it doesn't actually work in the original equation because it creates a division by zero error. The original equation is not defined when a = -2. So, what's our conclusion, guys?

The Verdict: Was the Student Correct?

So, was the student correct? The short answer is no. While we arrived at a = -2 as a potential solution through our calculations, this value leads to division by zero in the original equation. That means it's not a valid solution. Therefore, the original equation has no solution because a cannot equal -2. It's crucial to always check your solutions against the original equation, especially when dealing with fractions and variables in the denominator. This step helps us identify and avoid extraneous solutions. It is like being a detective; you must examine all the evidence to get to the truth. Always be careful and critical of your answers, and you will do great.

Importance of Checking Solutions

This example highlights the importance of checking our solutions. It's not enough to simply solve the equation and get an answer; we need to verify that the answer is valid within the context of the original problem. This is a critical step in algebra and in many other areas of mathematics. Checking our solutions helps us to catch any errors that might arise from algebraic manipulations, such as the introduction of extraneous solutions or overlooking domain restrictions. By plugging our solution back into the original equation, we can ensure that it satisfies all the conditions and doesn't lead to any mathematical impossibilities, such as division by zero. This is a crucial practice for ensuring accuracy and understanding the limitations of our solutions. Checking also allows us to solidify our understanding and improve our ability to solve problems confidently and effectively.

Tips for Solving Similar Equations

To become better equation solvers, here are some tips:

  • Always Factor: Always factor denominators to identify common factors and the LCD.
  • Find LCD: Identify and use the least common denominator.
  • Combine Terms: Combine like terms and simplify the equation as much as possible.
  • Check Your Work: After solving, always check your answer. Plug your answer into the equation.
  • Watch Out for Domain Restrictions: Be mindful of values that make the denominator zero.

By following these steps and paying close attention to detail, you will be well on your way to mastering algebraic equations.

Conclusion: Mastering the Equation

In conclusion, this equation serves as a fantastic illustration of the need for careful problem-solving and critical thinking in algebra. We discovered that although our calculations led us to a = -2, this solution was invalid because it would cause division by zero in the original equation. This emphasizes the vital importance of checking our solutions and being mindful of domain restrictions. Remember, mathematics is not just about finding answers; it's about understanding the underlying principles and ensuring that our solutions are accurate and meaningful. So, keep practicing, keep questioning, and keep exploring the wonderful world of mathematics!