Equation Solution: \frac{1}{c-3}-\frac{1}{c}=\frac{3}{c(c-3)}

Let's dive into solving this algebraic equation! The problem asks for the solution to $\frac{1}{c-3}-\frac{1}{c}=\frac{3}{c(c-3)}$. When we encounter equations with fractions, our primary goal is often to eliminate those fractions. This usually involves finding a common denominator and then multiplying each term by it. Before we start, it's crucial to identify any values of the variable that would make our denominators zero, as these are undefined and cannot be part of our solution. In this equation, the denominators are $(c-3)$, $c$, and $c(c-3)$. Therefore, $c$ cannot be equal to 3 (because $c-3=0$) and $c$ cannot be equal to 0 (because $c=0$). These restrictions, $c \neq 0$ and $c \neq 3$, are fundamental and will help us check our final answer later. Keeping these excluded values in mind, we can proceed with solving the equation. The most efficient way to clear the fractions is to multiply every term in the equation by the least common multiple (LCM) of the denominators. The denominators are $(c-3)$, $c$, and $c(c-3)$. The LCM of these is clearly $c(c-3)$. So, let's multiply each term by $c(c-3)$:

$c(c-3) \times \frac{1}{c-3} - c(c-3) \times \frac{1}{c} = c(c-3) \times \frac{3}{c(c-3)}$

Now, let's simplify each part. For the first term, the $(c-3)$ in the numerator cancels out the $(c-3)$ in the denominator, leaving us with $c \times 1$, which is simply $c$. For the second term, the $c$ in the numerator cancels out the $c$ in the denominator, leaving us with $-(c-3) \times 1$, which simplifies to $-(c-3)$. For the right side of the equation, both $c$ and $(c-3)$ in the numerator cancel out the $c(c-3)$ in the denominator, leaving us with $3$. So, the equation transforms into:

$c - (c-3) = 3$

This is a much simpler linear equation to solve. Let's distribute the negative sign in the second term: $c - c + 3 = 3$. Now, combine like terms on the left side: $(c-c) + 3 = 3$, which gives us $0 + 3 = 3$. This simplifies to $3 = 3$.

What does this result, $3=3$, tell us? It means that the original equation is true for all values of $c$ for which the equation is defined. Remember the restrictions we identified at the beginning? We found that $c$ cannot be 0 and $c$ cannot be 3 because those values would lead to division by zero. Since our simplified equation $3=3$ is always true, the solution set includes all real numbers except for those that are excluded. Therefore, the solution is all real numbers except $c=0$ and $c=3$. This corresponds to option C.

Understanding Extraneous Solutions in Algebraic Equations

When solving equations, especially those involving rational expressions (fractions with variables in the numerator or denominator), it's imperative to understand the concept of extraneous solutions. These are solutions that appear during the solving process but do not satisfy the original equation. They often arise when we multiply both sides of an equation by an expression containing the variable, as this operation can sometimes introduce solutions that were not present in the original form. In our equation, $\frac{1}{c-3}-\frac{1}{c}=\frac{3}{c(c-3)}$, we multiplied by $c(c-3)$. This step is perfectly valid for solving, but it requires us to be vigilant about the values of $c$ that would make $c(c-3)$ equal to zero. As we established earlier, these values are $c=0$ and $c=3$. Any potential solution that results in $c=0$ or $c=3$ must be discarded because it would mean we are attempting to divide by zero in the original equation, which is mathematically impossible. The process of multiplying by the LCD (least common denominator) is a powerful technique because it transforms a complex fractional equation into a simpler polynomial equation. In this particular case, the resulting polynomial equation simplified to an identity ($3=3$). An identity means that the equation holds true for all values of the variable for which the original expression was defined.

Since the equation simplified to $3=3$, it implies that any value of $c$ would satisfy the transformed equation. However, we must always refer back to the original equation and its domain restrictions. The original equation is undefined when $c=0$ or $c=3$. Therefore, while the simplified equation is true for all real numbers, the original equation is only true for all real numbers except $c=0$ and $c=3$. This is why it's so important to state the domain restrictions upfront and check any potential solutions against them. If, for instance, our algebraic manipulation had led to $c=0$ as a solution, we would have had to reject it because it's an extraneous solution. Similarly, if we had arrived at $c=3$, it too would be rejected. The fact that our algebraic simplification did not yield any specific values for $c$ but rather an identity ($3=3$) indicates that all values within the domain of the original equation are solutions. This leads us to the conclusion that the solution set consists of all real numbers except for $0$ and $3$. This understanding is critical for mastering rational equations and avoiding common pitfalls.

Step-by-Step Breakdown of the Equation Solution

Let's reiterate the solution process for $\frac{1}{c-3}-\frac{1}{c}=\frac{3}{c(c-3)}$ in a clear, step-by-step manner to solidify your understanding. The first and perhaps most important step in solving rational equations is to identify the domain restrictions. These are the values of the variable that would make any denominator in the equation equal to zero, thus rendering the expression undefined. For this equation, the denominators are $(c-3)$, $c$, and $c(c-3)$. Setting each denominator to zero, we find:

1. $c-3 = 0 \implies c = 3$
2. $c = 0$
3. $c(c-3) = 0 \implies c = 0$ or $c = 3$

Therefore, the restrictions on $c$ are that $c \neq 0$ and $c \neq 3$. Any solution we find must not be equal to these values.

Next, we want to eliminate the fractions. The most effective way to do this is to multiply every term in the equation by the least common denominator (LCD). The denominators are $(c-3)$, $c$, and $c(c-3)$. The LCD is $c(c-3)$. Multiplying each term by the LCD, we get:

$c(c-3) \times \left(\frac{1}{c-3}\right) - c(c-3) \times \left(\frac{1}{c}\right) = c(c-3) \times \left(\frac{3}{c(c-3)}\right)$

Now, we simplify each part by canceling common factors:

• For the first term, $\frac{c(c-3)}{c-3} = c$. So, $c \times 1 = c$.
• For the second term, $\frac{c(c-3)}{c} = c-3$. So, $-(c-3) \times 1 = -(c-3)$.
• For the third term, $\frac{c(c-3)}{c(c-3)} = 1$. So, $1 \times 3 = 3$.

Substituting these simplified terms back into the equation, we obtain:

$c - (c-3) = 3$

Now, we solve this simplified equation. First, distribute the negative sign to the terms inside the parentheses:

$c - c + 3 = 3$

Combine like terms on the left side of the equation:

$(c - c) + 3 = 3$

$0 + 3 = 3$

$3 = 3$

This result, $3=3$, is an identity. An identity is an equation that is true for all values of the variable for which it is defined. Since our simplified equation is always true, it means that the original equation is true for all values of $c$ that are allowed by the domain restrictions. We previously determined that $c$ cannot be 0 and $c$ cannot be 3. Therefore, the solution set includes all real numbers except for 0 and 3. This precisely matches option C from the given choices.

Conclusion: Mastering Rational Equations

In conclusion, solving the equation $\frac{1}{c-3}-\frac{1}{c}=\frac{3}{c(c-3)}$ leads us to an identity ($3=3$) after clearing the denominators. This signifies that the equation holds true for all values of the variable that do not make any of the original denominators zero. By identifying the domain restrictions early on – namely, $c \neq 0$ and $c \neq 3$ – we can confidently state that the solution set comprises all real numbers except these two excluded values. This approach ensures that we avoid extraneous solutions and arrive at the correct answer. Mastering the techniques of finding common denominators, clearing fractions, and respecting domain restrictions is key to successfully solving rational equations.

For further exploration into solving algebraic equations and understanding mathematical concepts, you might find the following resources helpful:

• Visit Khan Academy for comprehensive lessons and practice exercises on algebra and equations.
• Explore Math is Fun for clear explanations and interactive tools related to various mathematical topics, including fractions and equations.