Consider The Equation:${ \frac{2}{x-3} + \frac{1}{x} = \frac{x-1}{x-3} }$The Equation Has $\square$ Valid Solution(s) And $\square$ Extraneous Solution(s). Type The Correct Answer In Each Box. Use Numerals Instead Of Words.

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Introduction

In this article, we will delve into the world of algebra and explore a complex equation involving fractions. The given equation is 2xβˆ’3+1x=xβˆ’1xβˆ’3\frac{2}{x-3} + \frac{1}{x} = \frac{x-1}{x-3}. Our goal is to find the valid solution(s) and extraneous solution(s) for this equation. We will break down the solution process into manageable steps, making it easier to understand and follow along.

Step 1: Multiply Both Sides by the Least Common Denominator (LCD)

To simplify the equation and eliminate the fractions, we need to multiply both sides by the least common denominator (LCD), which is (xβˆ’3)x(x-3)x. This will help us get rid of the fractions and make the equation more manageable.

\frac{2}{x-3} + \frac{1}{x} = \frac{x-1}{x-3}
\\(x-3)x \cdot \left(\frac{2}{x-3} + \frac{1}{x}\right) = (x-3)x \cdot \left(\frac{x-1}{x-3}\right)
\\2x + (x-3) = (x-1)(x)

Step 2: Expand and Simplify the Equation

Now that we have multiplied both sides by the LCD, we can expand and simplify the equation. This will help us get closer to finding the solution(s).

2x + x - 3 = x^2 - x
\\3x - 3 = x^2 - x
\\x^2 - 4x + 3 = 0

Step 3: Factor the Quadratic Equation

The equation we obtained in the previous step is a quadratic equation. We can factor this equation to find the solution(s).

(x - 3)(x - 1) = 0
\\x - 3 = 0 \quad \text{or} \quad x - 1 = 0
\\x = 3 \quad \text{or} \quad x = 1

Step 4: Check for Extraneous Solutions

Now that we have found the solution(s), we need to check if they are extraneous or not. To do this, we will substitute the solution(s) back into the original equation and check if it holds true.

\text{Let's start with } x = 3:
\\ \frac{2}{3-3} + \frac{1}{3} = \frac{3-1}{3-3}
\\ \frac{2}{0} + \frac{1}{3} = \frac{2}{0}
\\ \text{This is undefined, so } x = 3 \text{ is an extraneous solution.}

\textNow, let's try } x = 1
\ \frac{2{1-3} + \frac{1}{1} = \frac{1-1}{1-3}
\ \frac{2}{-2} + \frac{1}{1} = \frac{0}{-2}
\ -1 + 1 = 0
\ 0 = 0
\ \text{This is true, so } x = 1 \text{ is a valid solution.}

Conclusion

In this article, we have solved the equation 2xβˆ’3+1x=xβˆ’1xβˆ’3\frac{2}{x-3} + \frac{1}{x} = \frac{x-1}{x-3} and found that it has 1 valid solution(s) and 1 extraneous solution(s). We have also demonstrated the importance of checking for extraneous solutions when solving equations involving fractions.

Final Answer

The final answer is:

  • Valid solution(s): 1
  • Extraneous solution(s): 1

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Q: What is the least common denominator (LCD) and why is it important?

A: The least common denominator (LCD) is the smallest multiple that all the denominators in an equation have in common. In the given equation, the LCD is (xβˆ’3)x(x-3)x. It is important because multiplying both sides of the equation by the LCD eliminates the fractions and makes the equation more manageable.

Q: How do I know when to multiply both sides of the equation by the LCD?

A: You should multiply both sides of the equation by the LCD when you see fractions in the equation. This will help you simplify the equation and make it easier to solve.

Q: What is the difference between a valid solution and an extraneous solution?

A: A valid solution is a solution that satisfies the original equation, while an extraneous solution is a solution that does not satisfy the original equation. In the given equation, x=3x = 3 is an extraneous solution because it makes the equation undefined, while x=1x = 1 is a valid solution because it satisfies the original equation.

Q: Why is it important to check for extraneous solutions?

A: It is important to check for extraneous solutions because they can occur when you multiply both sides of the equation by the LCD. If you don't check for extraneous solutions, you may end up with a solution that is not valid.

Q: How do I check for extraneous solutions?

A: To check for extraneous solutions, you should substitute the solution back into the original equation and check if it holds true. If the solution makes the equation undefined or false, then it is an extraneous solution.

Q: What if I get a quadratic equation after multiplying both sides by the LCD? How do I solve it?

A: If you get a quadratic equation after multiplying both sides by the LCD, you can solve it by factoring or using the quadratic formula. In the given equation, we factored the quadratic equation to find the solutions.

Q: Can I use the quadratic formula to solve the equation?

A: Yes, you can use the quadratic formula to solve the equation. The quadratic formula is x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. In the given equation, we can use the quadratic formula to find the solutions.

Q: What if I have trouble factoring the quadratic equation?

A: If you have trouble factoring the quadratic equation, you can use the quadratic formula to find the solutions. Alternatively, you can use a calculator or online tool to help you factor the equation.

Q: Can I use this method to solve other equations involving fractions?

A: Yes, you can use this method to solve other equations involving fractions. The key is to multiply both sides of the equation by the LCD and then simplify the equation to find the solutions.

Q: What if I get stuck or have trouble solving the equation?

A: If you get stuck or have trouble solving the equation, don't worry! You can try asking a teacher or tutor for help, or you can use online resources such as video tutorials or practice problems to help you understand the concept better.