Solving 1/(c-3) - 1/c = 3/(c(c-3))

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematics to tackle a specific algebraic equation that might have left some of you scratching your heads. We're talking about solving $\frac{1}{c-3}-\frac{1}{c}=\frac{3}{c(c-3)}$. Now, before we get too bogged down in the technicalities, let's remember that mathematics is all about logic and problem-solving, skills that are super useful in every aspect of life, not just in the classroom. So, grab your favorite beverage, get comfy, and let's break this down step-by-step.

Understanding the Equation and Potential Pitfalls

First off, let's look at the equation itself: $\frac{1}{c-3}-\frac{1}{c}=\frac{3}{c(c-3)}$. Our main goal here is to find the value or values of 'c' that make this statement true. However, as with many equations involving fractions, there's a crucial detail we need to pay close attention to: the denominators. We absolutely cannot have a denominator equal to zero, because division by zero is undefined. This means we need to identify any values of 'c' that would make any of our denominators zero and exclude them from our potential solutions right from the start. Looking at the denominators in our equation, we have $(c-3)$, $c$, and $c(c-3)$. Setting each of these to zero gives us:

• $c-3 = 0 \implies c = 3$
• $c = 0$
• $c(c-3) = 0 \implies c = 0$ or $c = 3$

So, right off the bat, we know that $c=0$ and $c=3$ are not possible solutions to this equation. If we arrive at either of these values as a potential answer later on, we'll have to discard them because they would make parts of the original equation meaningless. This is a super important concept in algebra, often referred to as identifying the 'excluded values' or 'restrictions' on the variable. It's like setting the ground rules before you start playing the game – if you don't have these rules, the game can break down.

The Strategy: Clearing the Denominators

Now that we've identified our excluded values, let's move on to solving the equation. The most common and effective strategy when dealing with equations involving multiple fractions is to eliminate the denominators altogether. We can do this by multiplying every term in the equation by the Least Common Denominator (LCD). The LCD is the smallest expression that is a multiple of all the denominators present. In our equation, the denominators are $(c-3)$, $c$, and $c(c-3)$. The LCD is clearly $c(c-3)$ because it contains all the factors present in the other denominators.

So, let's multiply each term by $c(c-3)$:

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

Now, let's simplify each part. In the first term, the $(c-3)$ in the numerator cancels out the $(c-3)$ in the denominator, leaving us with just $c \times 1$, which is $c$.

$\quad c \quad - \quad ...$

In the second term, the $c$ in the numerator cancels out the $c$ in the denominator, leaving us with $(c-3) \times 1$, which is $(c-3)$. Remember to keep that minus sign in front!

$\quad c \quad - \quad (c-3) \quad = \quad ...$

And in the third term, the entire $c(c-3)$ in the numerator cancels out the $c(c-3)$ in the denominator, leaving us with just $3$.

$\quad c \quad - \quad (c-3) \quad = \quad 3$

See? By multiplying by the LCD, we've transformed a complex-looking equation with fractions into a much simpler linear equation. This is a game-changer, guys! It’s like clearing away all the clutter to see the clear path ahead. This step is fundamental to solving rational equations and often makes the subsequent steps much more manageable. The key is to be systematic and ensure that you multiply every single term on both sides of the equation by the LCD. Missing even one term can lead you down the wrong path and result in an incorrect solution. So, always double-check that you've applied the multiplier to all parts of the equation.

Solving the Simplified Equation

Now we're left with the simplified equation: $c - (c-3) = 3$. The next step is to simplify this equation further by distributing the negative sign and combining like terms. Remember that when you distribute a negative sign to a term in parentheses, you change the sign of each term inside the parentheses. So, $-(c-3)$ becomes $-c + 3$.

Our equation now looks like this:

$\quad c - c + 3 = 3$

Let's combine the 'c' terms: $c - c$ equals $0$. So, the equation simplifies even further to:

$\quad 0 + 3 = 3$

Which gives us:

$\quad 3 = 3$

This is a very interesting result, isn't it? We've arrived at a statement that is always true, regardless of the value of 'c'. In mathematics, when you simplify an equation and end up with a statement like $3=3$ (or $0=0$, or any other true statement), it means that the original equation is an identity. An identity is an equation that holds true for all possible values of the variable for which the equation is defined. This means that any real number could potentially be a solution, with one crucial exception.

Reconsidering the Excluded Values

Remember what we talked about at the very beginning? We identified that $c=0$ and $c=3$ are excluded values because they would cause division by zero in the original equation. Even though our simplified equation $3=3$ suggests that any 'c' is a solution, we must still adhere to these initial restrictions. Therefore, while the equation is true for all real numbers, we must exclude the values that make the original denominators zero. This is why it's so critical to identify excluded values first. They act as the gatekeepers for our final answer.

So, the solution set includes all real numbers except for $c=0$ and $c=3$. This means that if you plug in any other real number for 'c' into the original equation, it will hold true. For example, if $c=1$: $\frac{1}{1-3} - \frac{1}{1} = \frac{1}{-2} - 1 = -0.5 - 1 = -1.5$. And on the other side: $\frac{3}{1(1-3)} = \frac{3}{1(-2)} = \frac{3}{-2} = -1.5$. It checks out! If $c=-1$: $\frac{1}{-1-3} - \frac{1}{-1} = \frac{1}{-4} - (-1) = -0.25 + 1 = 0.75$. And on the other side: $\frac{3}{-1(-1-3)} = \frac{3}{-1(-4)} = \frac{3}{4} = 0.75$. It checks out again! This confirms our findings.

Conclusion: The Final Answer

So, to wrap things up, guys, when we solved the equation $\frac{1}{c-3}-\frac{1}{c}=\frac{3}{c(c-3)}$, we first identified that $c$ cannot be $0$ or $3$ because these values would lead to division by zero in the original equation. After clearing the denominators by multiplying by the LCD, $c(c-3)$, we arrived at a simplified equation $3=3$. This true statement indicates that the original equation is an identity, meaning it is true for all values of $c$ for which it is defined. Combining this understanding with our initial restrictions, the solution is all real numbers except for $c=0$ and $c=3$. This is a common type of problem in algebra, and understanding the concept of excluded values is key to getting it right every time. Keep practicing, and you'll master these kinds of equations in no time! That’s all for today, keep those brains buzzing!