Solve 3/(x-2) = 6/(x-5) For X

Dealing with algebraic equations that involve fractions can seem a bit daunting at first, but with a systematic approach, they become quite manageable. Our focus today is to solve the equation $\frac{3}{x-2}=\frac{6}{x-5}$. This type of equation, where the variable appears in the denominator, is known as a rational equation. The core strategy for solving these is to eliminate the denominators, thereby transforming the rational equation into a simpler polynomial equation. We need to be mindful of any values of $x$ that would make a denominator zero, as these are extraneous solutions and must be excluded from our final answer. In this specific equation, the denominators are $x-2$ and $x-5$. Therefore, $x$ cannot be equal to $2$ or $5$, because these values would lead to division by zero. Keep this in mind as we proceed; we'll check our final answers against these restrictions.

To start tackling $\frac{3}{x-2}=\frac{6}{x-5}$, our primary goal is to get rid of those pesky denominators. A very effective method for this is to multiply both sides of the equation by the least common denominator (LCD). The denominators here are $x-2$ and $x-5$. Since they don't share any common factors (other than 1), their LCD is simply their product: $(x-2)(x-5)$. Let's multiply both sides of the equation by this LCD:

$$(x-2)(x-5) \times \frac{3}{x-2} = (x-2)(x-5) \times \frac{6}{x-5}$$

Now, we can simplify. On the left side, the $(x-2)$ terms cancel out, leaving us with $3(x-5)$. On the right side, the $(x-5)$ terms cancel out, leaving us with $6(x-2)$. Our equation now looks much friendlier:

$$3(x-5) = 6(x-2)$$

This is a linear equation, and we can solve it using standard algebraic techniques. Let's distribute the constants on both sides:

$$3x - 15 = 6x - 12$$

Our next step is to gather all the terms containing $x$ on one side of the equation and all the constant terms on the other. It's often convenient to move the $x$ terms to the side where the coefficient of $x$ will be positive. In this case, $6x$ is greater than $3x$, so let's move the $3x$ term to the right side by subtracting $3x$ from both sides:

$$-15 = 6x - 3x - 12$$

$$-15 = 3x - 12$$

Now, let's move the constant term $-12$ to the left side by adding $12$ to both sides:

$$-15 + 12 = 3x$$

$$-3 = 3x$$

Finally, to isolate $x$, we divide both sides by $3$:

$$\frac{-3}{3} = \frac{3x}{3}$$

$$-1 = x$$

So, we have found a potential solution: $x = -1$. Remember our initial restriction? We noted that $x$ cannot be $2$ or $5$. Our solution $x = -1$ does not violate these restrictions. Therefore, $x = -1$ is the valid solution to the equation $\frac{3}{x-2}=\frac{6}{x-5}$.

Let's double-check our solution by substituting $x=-1$ back into the original equation. This is a crucial step in solving rational equations to ensure we haven't introduced any extraneous solutions and that our arithmetic is correct.

Left Side:

$$\frac{3}{x-2} = \frac{3}{(-1)-2} = \frac{3}{-3} = -1$$

Right Side:

$$\frac{6}{x-5} = \frac{6}{(-1)-5} = \frac{6}{-6} = -1$$

Since the left side equals the right side (both are $-1$), our solution $x = -1$ is indeed correct. This systematic approach—identifying restrictions, clearing denominators, solving the resulting equation, and checking the solution—is key to mastering rational equations.

In summary, to solve the equation $\frac{3}{x-2}=\frac{6}{x-5}$, we first identified that $x \neq 2$ and $x \neq 5$. We then multiplied both sides by the LCD, $(x-2)(x-5)$, to obtain $3(x-5) = 6(x-2)$. Distributing and simplifying led to $3x - 15 = 6x - 12$. Rearranging terms to solve for $x$ gave us $-3 = 3x$, which simplifies to $x = -1$. Since $x=-1$ does not violate our initial restrictions, it is the valid solution. The question asks for the value(s) that is (are) the solution(s). In this case, there is only one solution, which is $x=-1$. Looking at the options provided:

A. $x=-1,1$ only B. $x=-1$ only C. $x=2,5$ only D. $x=5$ only

Our derived solution, $x=-1$, matches option B. It's important to note why the other options are incorrect. Option A includes $x=1$, which is not our solution. Option C includes $x=2$ and $x=5$, which are precisely the values that make the denominators zero and are therefore extraneous solutions – they are never solutions to the original rational equation. Option D includes only $x=5$, which is also an extraneous solution.

This problem highlights the importance of understanding the properties of rational expressions and the process of solving rational equations, including the critical step of identifying and excluding extraneous solutions. By carefully applying these methods, you can confidently tackle similar problems in mathematics. For further exploration into solving algebraic equations and understanding extraneous solutions, you might find resources from Khan Academy very helpful.