Rational Equation: $\frac{1}{s-6}-3=\frac{6}{s-6}$ Solved

Hey guys! Today, we're diving deep into the nitty-gritty of solving a rational equation. We've got this beast: $\frac{1}{s-6}-3=\frac{6}{s-6}$. Now, I know some of you might see fractions and think, "Oh no, not again!" But trust me, with a few straightforward steps, this equation will be as clear as day. Rational equations, at their core, are just algebraic equations that involve fractions where the variable you're solving for appears in the denominator. The biggest trick with these is to always be mindful of the denominator. Remember, you can never, ever divide by zero. So, before we even start manipulating the equation, we need to identify any values of the variable that would make a denominator zero. In our case, the denominator is $(s-6)$. If $s-6=0$, then $s=6$. This means that $s=6$ is an extraneous solution, and if we get $s=6$ as a potential answer at the end, we have to throw it out. Keep that little nugget in your back pocket as we work through this. Solving rational equations often involves clearing out those denominators to simplify the equation into a form we're more comfortable with, like a linear or quadratic equation. The most common way to do this is by multiplying both sides of the equation by the least common denominator (LCD). The LCD is the smallest expression that is a multiple of all the denominators in the equation. For our equation, $\frac{1}{s-6}-3=\frac{6}{s-6}$, there's only one unique denominator, which is $(s-6)$. So, our LCD is simply $(s-6)$. Once we multiply every term in the equation by the LCD, those pesky denominators will vanish, leaving us with a much simpler equation to solve. This technique is super powerful because it transforms a potentially complex problem into something much more manageable. We'll go through each step meticulously, so by the end of this, you'll feel confident tackling any rational equation that comes your way. So, buckle up, and let's get this equation solved!

Let's get down to business, shall we? We're staring at the equation $\frac{1}{s-6}-3=\frac{6}{s-6}$. As we discussed, our first priority is to identify any values of $s$ that would make our denominators zero. In this equation, the only denominator is $(s-6)$. Setting this to zero, we get $s-6=0$, which means $s=6$. So, $s=6$ is an excluded value. We absolutely cannot have $s=6$ as our final answer. Got it? Good. Now, to simplify this equation and get rid of those fractions, we're going to multiply every single term on both sides of the equation by the least common denominator (LCD). In this case, the LCD is pretty straightforward: it's just $(s-6)$. So, let's multiply:

$\qquad (s-6) \times \left(\frac{1}{s-6}\right) - (s-6) \times 3 = (s-6) \times \left(\frac{6}{s-6}\right)$

Look what happens! On the left side, the $(s-6)$ in the numerator cancels out the $(s-6)$ in the denominator of the first term:

$\qquad 1 - (s-6) \times 3 = (s-6) \times \left(\frac{6}{s-6}\right)$

And on the right side, the $(s-6)$ in the numerator cancels out the $(s-6)$ in the denominator of the second term:

$\qquad 1 - (s-6) \times 3 = 6$

See? No more fractions! This is exactly what we wanted. Now, the equation looks much friendlier. We just need to distribute the $-3$ on the left side:

$\qquad 1 - 3s + 18 = 6$

Combine the constant terms on the left side:

$\qquad 19 - 3s = 6$

Now, we want to isolate the term with $s$. Let's subtract 19 from both sides:

$\qquad -3s = 6 - 19$

$\qquad -3s = -13$

Finally, to solve for $s$, we divide both sides by $-3$:

$\qquad s = \frac{-13}{-3}$

$\qquad s = \frac{13}{3}$

And there you have it! Our potential solution is $s = \frac{13}{3}$.

Now, here's the crucial final step for any rational equation: checking for extraneous solutions. Remember way back when we identified that $s=6$ is an excluded value because it would make the denominator zero? We need to make sure our solution, $s = \frac{13}{3}$, is not equal to 6. Since $\frac{13}{3}$ is definitely not 6, our solution is valid! It's always a good practice to plug this value back into the original equation to double-check, but in this case, since it's not an excluded value, we can be pretty confident. Let's quickly verify it just for kicks:

Original equation: $\frac{1}{s-6}-3=\frac{6}{s-6}$

Substitute $s = \frac{13}{3}$:

Left side: $\frac{1}{\frac{13}{3}-6}-3 = \frac{1}{\frac{13}{3}-\frac{18}{3}}-3 = \frac{1}{\frac{-5}{3}}-3 = \frac{-3}{5}-3 = \frac{-3}{5}-\frac{15}{5} = \frac{-18}{5}$

Right side: $\frac{6}{\frac{13}{3}-6} = \frac{6}{\frac{13}{3}-\frac{18}{3}} = \frac{6}{\frac{-5}{3}} = 6 \times \frac{3}{-5} = \frac{18}{-5} = \frac{-18}{5}$

Since the left side equals the right side ($\frac{-18}{5} = \frac{-18}{5}$), our solution $s = \frac{13}{3}$ is correct. You guys nailed it! Solving rational equations involves a few key steps: identify excluded values, find the LCD, multiply to clear denominators, solve the resulting simpler equation, and always check your answer against the excluded values. Practice makes perfect, so keep working through these, and you'll be a rational equation pro in no time. If you ever find yourself stuck, just remember the process and double-check your calculations. Happy solving!