First, note that $x=0$ cannot be a solution. This is because if $x=0,$ then the rational term would be $\dfrac{12}{0},$ which is undefined because it has a zero denominator.

To solve the equation, we first multiply both sides by the denominator $x$ to get rid of the fraction:

\begin{align*} \require{cancel} \dfrac {12} {x} &= -4\\[5pt] x \cdot \dfrac {12} {x} &= x \cdot (-4)\\[5pt] \cancel{x} \cdot \dfrac {12} {\cancel{x}} &= -4x \\[5pt] 12 &= -4x \end{align*}

Then, we apply the multiplication principle:

\begin{align*} \require{cancel} 12 &= -4x \\[5pt] \dfrac{12}{-4} &= \dfrac{-4x}{-4} \\[5pt] -3 &= x \end{align*}

Since $x = -3$ is different from $x=0$, we can accept it as a solution. Thus, the solution is $x=-3.$

First, note that $x=0$ cannot be a solution. This is because if $x=0,$ then the rational term would be $\dfrac{4}{5\cdot 0},$ which is undefined because it has a zero denominator.

To solve the equation, we first multiply both sides by $(5x)$ to get rid of the fraction:

\begin{align*} \require{cancel} \dfrac{4}{5x} &= 2 \\[5pt] 5x \cdot \dfrac{4}{5x} &= 2 \cdot 5x \\[5pt] \cancel{5x} \cdot \dfrac{4}{\cancel{5x}} &= 10x \\[5pt] 4 &= 10x \end{align*}

Then, we apply the multiplication principle, as usual:

\begin{align*} 4 &= 10x \\[5pt] \dfrac{4}{10} &= \dfrac{10x}{10} \\[5pt] \dfrac{2}{5} &= x. \end{align*}

Since $x = \dfrac{2}{5}$ is different from $x=0$, we can accept it as a solution. Thus, the solution is $x = \dfrac{2}{5}.$

First, note that $x=4$ cannot be a solution. This is because if $x=4,$ then the rational term would be $\dfrac{3}{0},$ which is undefined because it has a zero denominator.

To solve the equation, we first multiply both sides by the denominator $(x-4)$ to get rid of the fraction:

\begin{align*} \require{cancel} \dfrac {3} {x-4} &= 2\\[5pt] (x-4) \cdot \dfrac {3} {x-4} &= (x-4) \cdot 2\\[5pt] \cancel{(x-4)} \cdot \dfrac {3} {\cancel{x-4}} &= 2x-8 \\[5pt] 3 &= 2x-8 \end{align*}

Then, we apply the addition and multiplication principles, as usual:

\begin{align*} \require{cancel} 3 &= 2x-8 \\[5pt] 3+8 &= 2x-8+8 \\[5pt] 11 &= 2x \\[5pt] \dfrac{11}{2} &= \dfrac{2x}{2} \\[5pt] \dfrac{11}{2} &= x \end{align*}

Since $x = \dfrac{11}{2}$ is different from $x=4$, we can accept it as a solution. Thus, the solution is $x=\dfrac{11}{2}.$

First, note that $x=5$ cannot be a solution since division by $0$ is undefined.

To solve the equation, we first multiply both sides by the denominator $(x-5)$ to get rid of the fraction:

\begin{align*} \require{cancel} \dfrac{x}{x-5} &= 6\\[5pt] (x-5) \cdot \dfrac{x}{x-5} &= (x-5) \cdot 6\\[5pt] \cancel{(x-5)} \cdot \dfrac{x}{\cancel{x-5}} &= (x-5) \cdot 6\\[5pt] x &= 6x - 30 \end{align*}

Then, we apply the addition and multiplication principles, as usual:

\begin{align*} \require{cancel} x &= 6x - 30 \\[5pt] x - 6x &= 6x - 30 - 6x \\[5pt] - 5x &= -30\\[5pt] \dfrac{- 5x}{-5} &= \dfrac{-30}{-5}\\[5pt] x &= 6 \end{align*}

Since $x = 6$ is different from $x=5$, we can accept it as a solution. Thus, the solution is $x = 6.$

First, note that $x = -1$ cannot be a solution since division by $0$ is undefined.

To solve the equation, we first multiply both sides by the denominator $(x+1)$ to get rid of the fraction:

\begin{align*} \require{cancel} 2 & = \dfrac{5x + 1}{x+1} \\[5pt] (x+1) \cdot 2 & = (x+1) \cdot \dfrac{5x + 1}{x+1}\\[5pt] (x+1) \cdot 2 & = \cancel{(x+1)} \cdot \dfrac{5x + 1}{\cancel{x+1}}\\[5pt] 2x + 2 & = 5x + 1 \end{align*}

Then, we apply the addition and multiplication principles, as usual:

\begin{align*} 2x + 2 & = 5x + 1 \\[5pt] 2x + 2 - 1 & = 5x + 1 - 1\\[5pt] 2x + 1 & = 5x\\[5pt] 2x + 1 - 2x & = 5x - 2x \\[5pt] 1 & = 3x \\[5pt] \dfrac{1}{3} & = \dfrac{3x}{3} \\[5pt] \dfrac{1}{3} & = x \end{align*}

Since $x = \dfrac{1}{3}$ is different from $x=-1$, we can accept it as a solution. Thus, the solution is $x = \dfrac{1}{3}.$

First, note that $x = \dfrac{3}{5}$ is an excluded solution since division by $0$ is undefined.

To solve the equation, we first multiply both sides by the denominator $(5x-3)$ to get rid of the fraction:

\begin{align*} \require{cancel} \dfrac{x}{5x - 3} & = 9\\[5pt] \dfrac{x(5x-3)}{5x-3} & = 9(5x-3)\\[5pt] \dfrac{x\cancel{(5x-3)}}{\cancel{5x-3}} & = 9(5x-3)\\[5pt] x & = 45x - 27 \end{align*}

Then, we apply the addition and multiplication principles, as usual:

\begin{align*} x & = 45x - 27\\[5pt] x - 45 x & = 45x - 27 - 45 x\\[5pt] -44x & = -27\\[5pt] \dfrac{-44x}{-44} & = \dfrac{-27}{-44}\\[5pt] x & = \dfrac{27}{44} \end{align*}

Since $x = \dfrac{27}{44}$ is different from $x= \dfrac{3}{5}$, we can accept it as a solution. Thus, the solution is $x = \dfrac{27}{44}.$

First, note that $x = 1$ cannot be a solution since division by $0$ is undefined.

To solve the equation, we first multiply both sides by the denominator $(3x-3)$ to get rid of the fraction:

\begin{align*} \require{cancel} \dfrac{19x - 7}{3x - 3} &= 3\\[5pt] (3x - 3) \cdot \dfrac{19x - 7}{3x - 3} &= (3x - 3) \cdot 3\\[5pt] \cancel{(3x-3)} \cdot \dfrac{19x - 7}{\cancel{3x - 3}} &= (3x - 3) \cdot 3\\[5pt] 19x - 7 &= 9x - 9 \end{align*}

Then, we apply the addition and multiplication principles, as usual:

\begin{align*} \require{cancel} 19x - 7 &= 9x - 9\\[5pt] 19x - 7 -9x &= 9x - 9 -9x \\[5pt] 10x -7& = -9\\[5pt] 10x -7+7& = -9+7\\[5pt] 10x & = -2\\[5pt] x & = -\dfrac{2}{10}\\[5pt] x & = -\dfrac{1}{5}. % x &= 2. \end{align*}

Since $x = -\dfrac{1}{5}$ is different from $x=1$, we can accept it as a solution. Thus, the solution is $x = -\dfrac{1}{5}.$