Answer
$x = 12 \text{ or } x = -1$
Work Step by Step
The least common denominator, or LCD, is $3x^2$.
Convert each fraction to an equivalent one by multiplying its numerator with whatever factor is missing between its denominator and the LCD:
$$\dfrac{11(x)}{3x^2} - \dfrac{1(x^2)}{3x^2} = \dfrac{4(3)}{3x^2}$$
Multiply to simplify:
$$\dfrac{11x}{3x^2} - \dfrac{x^2}{3x^2} = \dfrac{12}{3x^2}$$
Subtract the fractions:
$$\dfrac{11x - x^2}{3x^2} = \dfrac{12}{3x^2}$$
Multiply each side of the equation by $3x^2$ to eliminate the fractions:
$$11x - x^2 = 12$$
Move all terms to the left side of the equation:
$$\begin{align*}
11x - x^2 - 12 &= 0\\
-x^2+11x-12&=0
\end{align*}$$
Divide each side of the equation by $-1$:
$$x^2 - 11x + 12 = 0$$
Factor the trinomial by looking for factors of $12$ whose sum is $-11$:
$$(x - 12)(x + 1) = 0$$
Use the Zero-Product Property by equating each factor to $0$, then solve each equation for $x$:
First factor:
$$\begin{align*}
x - 12 &= 0\\
x &= 12
\end{align*}$$
Second factor:
$$\begin{align*}
x + 1 &= 0\\
x&=-1
\end{align*}$$
To check the solution, plug in the values we just found for $x$ into the original equation:
For $x=12$:
$$\begin{align*}
\dfrac{11}{3(12)} - \dfrac{1}{3} &= \dfrac{-4}{12^2}\\
\\\dfrac{11}{36} - \dfrac{1}{3} &= \dfrac{-4}{144}\\
\end{align*}$$
Convert to equivalent fractions with $144$ as the LCD, then simplify:
$$\begin{align*}
\dfrac{44}{144} - \dfrac{48}{144} &= \dfrac{-4}{144}\\
\\\dfrac{-4}{144} &= \dfrac{-4}{144}\\
\\-\dfrac{1}{36} &= -\dfrac{1}{36}
\end{align*}$$
Both sides are equal to one another; therefore, this solution is correct.
For $x=-1$:
$$\begin{align*}
\dfrac{11}{3(-1)} - \dfrac{1}{3} &= \dfrac{-4}{(-1)^2}\\
\\-\dfrac{11}{3} - \dfrac{1}{3} &= -\frac{4}{1}\\
\end{align*}$$
Convert to equivalent fractions using the LCD. $2$, then simplify:
$$\begin{align*}
-\dfrac{11}{3} - \dfrac{1}{3} &= -\dfrac{12}{3}\\
\\-\dfrac{12}{3} &= -\dfrac{12}{3}\\
-4 &= -4\end{align*}$$
Both sides are equal to one another; therefore, this solution is correct.
