First, note that $x=0$ can't be a solution since division by $0$ is undefined.

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

\begin{align*} \dfrac 4 x &= \dfrac 2 5 \\[5pt] 4 \cdot 5 &= x \cdot 2 \\[5pt] 20 &= 2x \end{align*}

Then, we apply the multiplication principle: \begin{align*} 20 &= 2x\\[5pt] \dfrac{20}{2} &= \dfrac{2x}{2}\\[5pt] \dfrac {20} 2 & =x \\[5pt] 10 &= x \end{align*}

Since $x = 10$ is different from $x = 0$, we can accept it as a solution. Thus, the solution is $x = 10.$

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

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

\begin{align} \dfrac{1}{8} &= \dfrac{3}{4a} \\[5pt] 1 \cdot 4a &= 8 \cdot 3\\[5pt] 4a &= 24 \end{align}

Then, we apply the multiplication principle:

\begin{align} 4a &= 24 \\[5pt] \dfrac{4a}{4} &= \dfrac{24}{4}\\[5pt] a &= 6 \end{align}

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

First, note that $a=0$ cannot be a solution 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:

\[ \eqalign{ \dfrac{4}{3a}&=-\dfrac{2}{9}\\[5pt] \dfrac{4}{3a}&=\dfrac{-2}{9}\\[5pt] 4 \cdot 9 &= 3a \cdot (-2) \\[5pt] 36 &= -6a } \]

Then, we apply the multiplication principle:

\[ \eqalign{ 36 &= -6a \\[5pt] \dfrac{36}{-6} &= a \\[5pt] -6 &= a } \]

Since $a = -6$ is different from $a=0,$ we can accept it as a solution. Therefore, the solution is $a=-6.$

First, note that $t=0$ cannot be a solution 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{4}{3} &= \dfrac{2}{6t}\\[5pt] \dfrac{-4}{3} &= \dfrac{2}{6t}\\[5pt] (-4) \cdot 6t &= 3 \cdot 2 \\[5pt] -24t &= 6 \end{align*}

Then, we apply the multiplication principle: \begin{align*} -24t &= 6 \\[5pt] t &= \dfrac{6}{-24} \\[5pt] t &= -\dfrac{1}{4} \end{align*}

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

First, note that the denominator $3x-8$ cannot be zero, since division by $0$ is undefined. Therefore we cannot have $3x-8=0,$ which means $x=\dfrac{8}{3}$ cannot be a solution.

To solve the equation, we first apply cross-multiplication: \begin{align*} \dfrac{10}{3x-8} &=\dfrac{5}{2} \\[5pt] 10 \cdot 2 &= (3x-8) \cdot 5 \\[5pt] 20 &= 15x - 40 \end{align*}

Then, we apply the addition and multiplication principles, as usual: \begin{align*} 20 &= 15x - 40 \\[5pt] 20 + 40 &= 15x - 40 + 40 \\[5pt] 60 &= 15x \\[5pt] \dfrac{60}{15} &= \dfrac{15x}{15} \\[5pt] 4 &= x \end{align*}

Since $x = 4$ is different from $x=\dfrac{8}{3}$, we can accept it as a solution. Therefore, the solution is $x = \bbox[4pt,border: 1px solid lightgray]{4}.$

First, note that the denominator $3z-9$ cannot be zero, since division by $0$ is undefined. Therefore we cannot have $3z-9=0,$ which means $z=3$ cannot be a solution.

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

\[ \eqalign{ \dfrac{z}{3z-9} &= \dfrac{2}{3} \\[5pt] z \cdot 3 &= (3z-9) \cdot 2 \\[5pt] 3z &= 6z - 18 } \]

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

\[ \eqalign{ 3z &= 6z - 18 \\[5pt] 3z - 6z &= 6z - 18 - 6z \\[5pt] -3z &= -18 \\[5pt] \dfrac{-3z}{-3} &= \dfrac{-18}{-3} \\[5pt] z &= 6 } \]

Since $z = 6$ is different from $z=3$, we can accept it as a solution. Thus, the solution is $z=6.$

First, note that the denominator $2s+6$ cannot be zero, since division by $0$ is undefined. Therefore we cannot have $2s+6=0,$ which means $s=-3$ cannot be a solution.

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

\begin{align*} \dfrac{2s+4}{2s+6} &= \dfrac{6}{5} \\[5pt] (2s+4) \cdot 5 &= (2s+6)\cdot 6 \\[5pt] 10s + 20 &= 12s + 36 \end{align*}

Then, we apply the addition and multiplication principles, as usual: \begin{align*} 10s + 20 &= 12s + 36\\[5pt] 10s + 20 - 10s &= 12s + 36 - 10s \\[5pt] 20 &= 2s + 36 \\[5pt] 20 - 36 &= 2s + 36 - 36 \\[5pt] -16 &= 2s \\[5pt] \dfrac{-16}{2} &=\dfrac{2s}{2} \\[5pt] -8 &= s \end{align*}

Since $s = - 8$ is different from $s=-3$, we can accept it as a solution. Thus, the solution is $s = - 8.$