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the following rational equation has denominators that contain variables…

Question

the following rational equation has denominators that contain variables. for this equation, a. write the value or values of the variable restrictions on the variable. b. keeping the restrictions in mind, solve the equation.\\(\frac{6x}{x + 1}=9-\frac{6}{x + 1}\\)\
a. what are the value or values of the variable that makes the denominators zero?\\(x = -1\\)\\(\text{(simplify your answer. use a comma to separate answers as needed.)}\\)\
b. solve the equation. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\
\\(\bigcirc\\) a. the solution set is ( ).\\(\text{(type an integer or a simplified fraction. use a comma to separate answers as needed.)}\\)\
\\(\bigcirc\\) b. the solution set is \\(\\{x|x\\) is a real number\\(\\}\\).\
\\(\bigcirc\\) c. the solution set is \\(\varnothing\\).

Explanation:

Response
Part b: Solving the Rational Equation

Step 1: Eliminate the Denominator

Since the denominators are the same ($x + 1$) and $x
eq - 1$ (from part a), we can multiply both sides of the equation $\frac{6x}{x + 1}=9-\frac{6}{x + 1}$ by $x + 1$ to eliminate the denominators.
$$(x + 1)\times\frac{6x}{x + 1}=(x + 1)\times9-(x + 1)\times\frac{6}{x + 1}$$
Simplifying each term, we get:
$$6x = 9(x + 1)-6$$

Step 2: Expand and Simplify the Right - Hand Side

Expand $9(x + 1)$ using the distributive property $a(b + c)=ab+ac$. Here, $a = 9$, $b=x$ and $c = 1$, so $9(x + 1)=9x+9$.
The equation becomes:
$$6x=9x + 9-6$$
Simplify the right - hand side: $9x+9 - 6=9x + 3$. So, $6x=9x + 3$.

Step 3: Solve for x

Subtract $9x$ from both sides of the equation:
$$6x-9x=9x + 3-9x$$
$$- 3x=3$$
Divide both sides by $-3$:
$$x=\frac{3}{-3}=-1$$

Step 4: Check the Restriction

But from part a, we know that $x=-1$ makes the denominator of the original rational equation zero. So, $x = - 1$ is not a valid solution.

Answer:

C. The solution set is $\varnothing$.