First, note that $x=-\dfrac{2}{3}$ and $x=0$ cannot be solutions since division by $0$ is undefined.

To solve the equation, we first apply cross-multiplication: \[ \eqalign{ \dfrac{2}{3x+2} & = \dfrac{1}{x}\\[5pt] 2 \cdot x & = (3x+2) \cdot 1 \\[5pt] 2x & = 3x+2 } \]

Then, we apply the addition and multiplication principles, as usual: \[ \eqalign{ 2x & = 3x+2\\[5pt] 2x - 3x& = 3x + 2 - 3x\\[5pt] -x & = 2\\[5pt] (-1)x & = 2\\[5pt] \dfrac{(-1)x}{(-1)} & =\dfrac{2}{(-1)}\\[5pt] x & = -2 } \]

Since $x = -2$ is different from $x=-\dfrac{2}{3}$ and $x=0,$ we can accept it as a solution. Thus, the solution is $x = -2.$

First, note that $y=0$ and $y = 7$ cannot be solutions since division by $0$ is undefined.

To solve the equation, we first apply cross-multiplication:

\begin{align*} \dfrac {9} {y-7} &= \dfrac {6} {y}\\[5pt] 9 \cdot y& = (y-7)\cdot 6 \\[5pt] 9y & = 6y-42 \end{align*}

Then, we apply the addition and multiplication principles, as usual: \begin{align*} 9y & = 6y-42\\[5pt] 9y - 6y & = 6y-42-6 y\\[5pt] 3y & = -42 \\ \dfrac {3y}{3} & = \dfrac {-42}{3} \\ y & = -14 \end{align*}

Since $y = -14$ is different from $y = 0$ and $y = 7$, we can accept it as a solution. Thus, the solution is $y = -14.$

First, note that $z=0$ and $z=-10$ cannot be solutions since division by $0$ is undefined.

To solve the equation, we first apply cross-multiplication: \begin{align*} \dfrac{3}{4z}&=\dfrac{2}{z+10}\\[5pt] 3 \cdot (z +10) &=(4z) \cdot 2\\[5pt] 3z + 30&= 8z \end{align*}

Then, we apply the addition and multiplication principles, as usual: \begin{align*} 3z +30&= 8z\\[5pt] 3z + 30 - 3z&= 8z - 3z \\[5pt] 30 &=5z \\[5pt] \dfrac{30}{5} &=\dfrac{5z}{5}\\[5pt] 6 &= z \end{align*}

Since $z= 6$ is different from $z=0$ and $z=-10$, we can accept it as a solution. Thus, the solution is $z = \bbox[4pt,border: 1px solid lightgray]{6}.$

First, note that $y=-\dfrac{3}{2}$ and $y=0$ cannot be solutions since division by $0$ is undefined.

To solve the equation, we move the negative sign onto the numerator of the fraction and then apply cross-multiplication: \begin{align*} -\dfrac{3}{2y+3}&=\dfrac{2}{y}\\[5pt] \dfrac{-3}{2y+3}&=\dfrac{2}{y}\\[5pt] (-3) \cdot y & = (2y+3) \cdot 2\\[5pt] -3y& =4y + 6 \end{align*}

Then, we apply the addition and multiplication principles, as usual: \begin{align*} -3y & = 4y + 6\\[5pt] -3y - 4y& = 4y + 6 - 4y\\[5pt] -7y & = 6\\[5pt] \dfrac{-7y}{-7} & = \dfrac{6}{-7}\\[5pt] y & = -\dfrac{6}{7} \end{align*}

Since $y = - \dfrac{6}{7}$ is different from $x=-\dfrac{3}{2}$ and $y=0,$ we can accept it as a solution. Thus, the solution is $y = \bbox[4pt,border: 1px solid lightgray]{-\dfrac{6}{7}}.$

First, note that $s=0$ and $s=-2$ cannot be solutions since division by $0$ is undefined.

To solve the equation, we move the negative sign onto the numerator of the fraction and then apply cross-multiplication:

\[ \eqalign{ \dfrac{4}{s}&=-\dfrac{6}{s+2}\\[5pt] \dfrac{4}{s}&=\dfrac{-6}{s+2}\\[5pt] 4 \cdot (s+2) &=s \cdot (-6) \\[5pt] 4s+8&=-6s } \]

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

\[ \eqalign{ 4s+8&=-6s\\[5pt] 4s+8-4s&=-6s-4s\\[5pt] 8&=-10s\\[5pt] \dfrac{-8}{10}&=\dfrac{-10s}{-10} \\[5pt] -\dfrac{4}{5} &= s } \]

Since $s = - \dfrac{4}{5}$ is different from $s=0$ and $s=-2$, we can accept it as a solution. Thus, the solution is $s = -\dfrac{4}{5}.$

First, note that $x=0$ and $x=-5$ cannot be solutions since division by $0$ is undefined.

To solve the equation, we move the negative sign onto the numerator of the fraction and then apply cross-multiplication:

\[ \eqalign{ -\dfrac{3}{x}&=\dfrac{2}{x+5}\\[5pt] \dfrac{-3}{x}&=\dfrac{2}{x+5}\\[5pt] (-3) \cdot (x+5) &=x \cdot 2 \\[5pt] -3x - 15 & =2x } \]

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

\[ \eqalign{ -3x-15&=2x\\[5pt] -3x -15 + 3x&=2x + 3x\\[5pt] -15&=5x\\[5pt] \dfrac{-15}{5}&=\dfrac{5x}{5} \\[5pt] -3 &= x } \]

Since $x = -3$ is different from $x=0$ and $x=-5$, we can accept it as a solution. Thus, the solution is $x =\bbox[4pt,border: 1px solid lightgray]{-3}.$

First, note that $x=-4 $ and $x=2$ cannot be solutions since division by $0$ is undefined.

To solve the equation, we first apply cross-multiplication: \begin{align*} \dfrac{1}{x-2} & = \dfrac{2}{x+4} \\[5pt] 1 \cdot (x+4) & = (x-2) \cdot 2\\[5pt] x+4 & = 2x-4 \end{align*}

Then, we apply the addition and multiplication principles, as usual: \begin{align*} x+4 & = 2x-4\\[5pt] x+4-x & = 2x-4-x \\[5pt] 4 & = x-4\\[5pt] 4+4 & = x-4+4\\[5pt] 8 & = x \end{align*}

Since $x = 8$ is different from $x=-4$ and $x=2,$ we can accept it as a solution. Therefore, the solution is $x=\bbox[4pt,border: 1px solid lightgray]{8}.$

First, note that $x=-3$ and $x=-14$ cannot be solutions since division by $0$ is undefined.

To solve the equation, we move the negative sign onto the numerator of the fraction and then apply cross-multiplication: \begin{align} -\dfrac{1}{2x+6} & = \dfrac{5}{x+14} \\[5pt] \dfrac{-1}{2x+6} & = \dfrac{5}{x+14} \\[5pt] (-1) \cdot (x+14) & = (2x+6) \cdot 5\\[5pt] -x -14 & = 10x + 30 \end{align}

Then, we apply the addition and multiplication principles, as usual: \begin{align} -x - 14& = 10x + 30 \\[5pt] -x -14 +x & = 10x + 30 +x \\[5pt] -14& = 11x + 30\\[5pt] -14 -30 & = 11x + 30 - 30\\[5pt] -44 & = 11x\\[5pt] \dfrac{-44}{11} & = \dfrac{11x}{11}\\[5pt] -4 & =x \end{align}

Since $x= -4$ is different from $x=-3$ and $x=-14,$ we can accept it as a solution. Thus, the solution is $x=\bbox[4pt,border: 1px solid lightgray]{-4}.$

First, note that $x=-4$ cannot be a solution since division by $0$ is undefined.

To solve the equation, we first apply cross-multiplication:

\[ \eqalign{ \dfrac{2}{3x+12} &= \dfrac{2}{x+4} \\[5pt] 2 \cdot (x+4) &= (3x+12) \cdot 2 \\[5pt] 2x + 8 &= 6x + 24 } \]

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

\[ \eqalign{ 2x + 8 &= 6x + 24 \\[5pt] 2x + 8-24 &= 6x + 24-24 \\[5pt] 2x-16 &= 6x \\[5pt] 2x-16-2x &= 6x-2x \\[5pt] -16 &= 4x \\[5pt] \dfrac{-16}{4} &= \dfrac{4x}{4} \\[5pt] -4 &= x } \]

Since $x=-4$ cannot be a solution, the equation has no solutions.