Solving The Equation: $\frac{7}{x+1}-\frac{1}{2}=\frac{8}{3x+3}$

Hey guys! Today, we're diving into solving a pretty interesting equation: $\frac{7}{x+1}-\frac{1}{2}=\frac{8}{3x+3}$. Equations like these might seem intimidating at first, but don't worry! We'll break it down step by step so it's super clear and easy to understand. So, grab your pencils, and let's get started!

Understanding the Equation

Before we jump into the solution, let's take a closer look at the equation itself: $\frac{7}{x+1}-\frac{1}{2}=\frac{8}{3x+3}$. Notice that we have fractions, and the variable 'x' is in the denominator. This means we're dealing with a rational equation. Solving rational equations involves a few key steps, primarily focusing on eliminating the fractions to simplify the problem. The denominators we're working with are $(x+1)$, $2$, and $(3x+3)$. Spotting common factors and understanding how these denominators relate to each other is crucial for efficiently finding a common denominator, which is our first main goal.

In this particular equation, observe that $3x + 3$ can be factored. This is a common trick in algebra, and recognizing it early can save you a lot of trouble later on. Factoring $3x + 3$ gives us $3(x + 1)$. Now, looking at the denominators $(x + 1)$, $2$, and $3(x + 1)$, you might already see a common element emerging. Identifying these patterns is like having a secret weapon in algebra! It helps us plan our attack and choose the most effective method to solve the equation. Remember, the goal is to manipulate the equation in a way that gets rid of the fractions while still preserving the equality. This involves some clever algebraic techniques, but don't worry, we'll go through them all in detail. So, let's move on to the next step and start clearing those fractions!

Finding the Least Common Denominator (LCD)

Okay, so the first big step in tackling this equation is finding the Least Common Denominator (LCD). Think of the LCD as the magic key that unlocks the solution. It's the smallest multiple that all the denominators in our equation can divide into evenly. In our equation, $\frac{7}{x+1}-\frac{1}{2}=\frac{8}{3x+3}$, the denominators are $(x+1)$, $2$, and $(3x+3)$.

As we noticed earlier, $3x + 3$ can be factored into $3(x + 1)$. This is a crucial observation because it simplifies finding the LCD. Now we have the denominators $(x+1)$, $2$, and $3(x+1)$. To find the LCD, we need to consider each unique factor and the highest power it appears in any of the denominators. We've got the factors $(x+1)$, $2$, and $3$. So, the LCD will include each of these. The factor $(x+1)$ appears once in the first denominator and once in the third (after factoring), so we just include it as $(x+1)$. We also have the factors $2$ and $3$. Therefore, the LCD is the product of these factors: $2 * 3 * (x+1)$ which simplifies to $6(x+1)$.

So, our LCD is $6(x + 1)$. This is super important because we're going to use it to clear the fractions from the equation. Clearing the fractions makes the equation much easier to work with and solve. Now that we've found our LCD, we're ready to move on to the next step: multiplying both sides of the equation by the LCD. This is where the magic really happens, so let's see how it works!

Multiplying Both Sides by the LCD

Alright, we've found our LCD, which is $6(x+1)$. Now comes the fun part: multiplying both sides of the equation by this LCD. Remember, whatever we do to one side of the equation, we have to do to the other side to keep things balanced. It's like a seesaw – if you add weight to one side, you need to add the same weight to the other side to keep it level.

So, we take our equation, $\frac{7}{x+1}-\frac{1}{2}=\frac{8}{3x+3}$, and multiply both the left side and the right side by $6(x+1)$. This looks like: $6(x+1) * [\frac{7}{x+1}-\frac{1}{2}] = 6(x+1) * \frac{8}{3x+3}$. Now, we need to distribute $6(x+1)$ to each term on both sides. Let's break this down step by step.

On the left side, we have two terms: $\frac{7}{x+1}$ and $-\frac{1}{2}$. When we multiply $6(x+1)$ by $\frac{7}{x+1}$, the $(x+1)$ terms cancel out, leaving us with $6 * 7$, which is $42$. Then, we multiply $6(x+1)$ by $-\frac{1}{2}$. Here, $6$ divided by $2$ gives us $3$, so we have $-3(x+1)$. Distributing the $-3$ gives us $-3x - 3$. So, the left side of the equation becomes $42 - 3x - 3$.

On the right side, we multiply $6(x+1)$ by $\frac{8}{3x+3}$. Remember that $3x+3$ is the same as $3(x+1)$. So, we're multiplying $6(x+1)$ by $\frac{8}{3(x+1)}$. The $(x+1)$ terms cancel out, and $6$ divided by $3$ gives us $2$. So, we have $2 * 8$, which is $16$. The right side of the equation simplifies to $16$.

After multiplying by the LCD and simplifying, our equation now looks like this: $42 - 3x - 3 = 16$. Notice how the fractions are gone! We've transformed the equation into a much simpler form that we can easily solve. Next, we'll simplify and solve for 'x'.

Simplifying and Solving for x

Awesome, we've cleared the fractions, and our equation now looks like $42 - 3x - 3 = 16$. The next step is to simplify this equation by combining like terms. On the left side, we have the constants $42$ and $-3$. Combining these gives us $42 - 3 = 39$. So, the left side simplifies to $39 - 3x$. Now our equation is $39 - 3x = 16$.

Our goal now is to isolate 'x' on one side of the equation. To do this, we'll first subtract $39$ from both sides. This gives us $-3x = 16 - 39$, which simplifies to $-3x = -23$. We're almost there!

To finally solve for 'x', we need to get rid of the $-3$ that's multiplying it. We do this by dividing both sides of the equation by $-3$. So, we have $x = \frac{-23}{-3}$. A negative divided by a negative is a positive, so we get $x = \frac{23}{3}$.

So, we've found a potential solution: $x = \frac{23}{3}$. But wait, we're not quite done yet! When dealing with rational equations, it's super important to check our solution to make sure it doesn't make any of the original denominators equal to zero. If a solution does that, it's called an extraneous solution, and we have to discard it. So, let's move on to the final step: checking our solution.

Checking for Extraneous Solutions

Okay, we've arrived at a potential solution, $x = \frac{23}{3}$. But before we celebrate, we need to make sure this solution actually works in our original equation. Remember, extraneous solutions can pop up when we're dealing with rational equations, and they're like party crashers – we don't want them messing things up!

To check for extraneous solutions, we need to plug $x = \frac{23}{3}$ back into the original equation: $\frac{7}{x+1}-\frac{1}{2}=\frac{8}{3x+3}$. Let's substitute and see what happens. First, let's consider the denominators. We need to make sure that $x + 1$ and $3x + 3$ are not equal to zero when $x = \frac{23}{3}$.

If we plug $x = \frac{23}{3}$ into $x + 1$, we get $\frac{23}{3} + 1$, which is $\frac{23}{3} + \frac{3}{3} = \frac{26}{3}$. This is definitely not zero, so we're good so far. Now, let's plug it into $3x + 3$. We get $3(\frac{23}{3}) + 3$, which simplifies to $23 + 3 = 26$. Again, this is not zero. So, our solution doesn't make any of the denominators zero, which is a great sign!

Now, let's plug $x = \frac{23}{3}$ into the full equation to make sure both sides are equal. We have $\frac{7}{\frac{23}{3}+1} - \frac{1}{2} = \frac{8}{3(\frac{23}{3})+3}$. We already know that $\frac{23}{3} + 1 = \frac{26}{3}$ and $3(\frac{23}{3}) + 3 = 26$. So, the equation becomes $\frac{7}{\frac{26}{3}} - \frac{1}{2} = \frac{8}{26}$.

Let's simplify the left side. Dividing by a fraction is the same as multiplying by its reciprocal, so $\frac{7}{\frac{26}{3}}$ becomes $7 * \frac{3}{26} = \frac{21}{26}$. Now the left side is $\frac{21}{26} - \frac{1}{2}$. To subtract these, we need a common denominator, which is $26$. So, we rewrite $\frac{1}{2}$ as $\frac{13}{26}$. Now we have $\frac{21}{26} - \frac{13}{26} = \frac{8}{26}$.

On the right side, we have $\frac{8}{26}$. So, the equation is $\frac{8}{26} = \frac{8}{26}$. Hooray! Both sides are equal, which means our solution, $x = \frac{23}{3}$, is valid. We've successfully solved the equation and checked our answer. Great job, guys!

Conclusion

So, there you have it! We've successfully solved the equation $\frac{7}{x+1}-\frac{1}{2}=\frac{8}{3x+3}$. We walked through each step, from finding the LCD to multiplying both sides, simplifying, solving for 'x', and finally, checking for extraneous solutions. Remember, solving rational equations involves a few key steps, and it's all about being careful and methodical.

The solution we found was $x = \frac{23}{3}$, and after checking, we confirmed that it is indeed a valid solution. Equations like these might seem tough at first, but with practice and a clear understanding of the steps involved, you can tackle them with confidence. Keep practicing, and you'll become an equation-solving pro in no time! Keep up the great work, guys! You've got this!