First, we multiply both sides of the equation by $3$ to get rid of the fraction:

\begin{eqnarray} \require {cancel} \eqalign{ \dfrac 4 3 x &= 8 \\[5pt] 3 \cdot \dfrac 4 3 x &= 8 \cdot 3 \\[5pt] \cancel{3} \cdot \dfrac{4}{\cancel{3}} x &= 8 \cdot 3 \\[5pt] 4x &= 24 } \end{eqnarray}

Then, we divide both sides of the equation by $4$ to isolate $x\mathbin{:}$

\begin{eqnarray} \require {cancel} \eqalign{ 4x &= 24 \\[5pt] \dfrac{4x}{4} &= \dfrac{24}{4} \\[5pt] \dfrac{\cancel{4}x}{\cancel{4}} &= \dfrac{24}{4} \\[5pt] x &= 6 } \end{eqnarray}

First, we multiply both sides of the equation by $5$ to get rid of the fraction:

\begin{eqnarray} \require {cancel} \eqalign{ \dfrac 2 5 x &= 4 \\[5pt] 5 \cdot \dfrac2 5 x &= 5 \cdot4 \\[5pt] \cancel{5} \cdot \dfrac{2}{\cancel{5}} x &= 5 \cdot4 \\[5pt] 2x &= 20 } \end{eqnarray}

Then, we divide both sides of the equation by $2$ to isolate $x\mathbin{:}$

\begin{eqnarray} \require {cancel} \eqalign{ 2x &= 20 \\[5pt] \dfrac{2x}{2} &= \dfrac{20}{2} \\[5pt] \dfrac{\cancel{2}x}{\cancel{2}} &= \dfrac{20}{2} \\[5pt] x &= 10 } \end{eqnarray}

We multiply both sides of the equation by the reciprocal of $-\dfrac{12}{5},$ which is \[\left( -\dfrac{12}{5} \right)^{-1} = -\dfrac{5}{12}.\] This gives: \begin{eqnarray} \require {cancel} \eqalign{ -\dfrac{12}{5}v &= 7 \\[5pt] -\dfrac{5}{12} \cdot \left(-\dfrac{12}{5}v\right) &= -\dfrac{5}{12}\cdot 7 \\[5pt] 1 v &= -\dfrac{35}{12} \\[5pt] v &= -\dfrac{35}{12} } \end{eqnarray}

First, we express the fraction as a coefficient:

$$\dfrac{7}{17}w=-7$$

We multiply both sides of the equation by the reciprocal of $\dfrac{7}{17},$ which is \[\left( \dfrac{7}{17} \right)^{-1} = \dfrac{17}{7}.\] This gives: \begin{align*} \dfrac{7}{17}w &= -7 \\[5pt] \dfrac{17}{7}\cdot \dfrac{7}{17}w &= \dfrac{17}{7}\cdot (-7) \\[5pt] 1w &= -\dfrac{17}{7}\cdot 7 \\[5pt] w &= -\dfrac{17}{ \cancel{7} }\cdot \cancel{7} \\[5pt] w &= -17 \end{align*}

First, we isolate the fractional term $\dfrac{5z}{3}$ by subtracting $\dfrac 1 2$ from both sides:

\begin{eqnarray} \eqalign{ \dfrac {5z} 3 + \dfrac 1 2 &= 1 \\[5pt] \dfrac {5z} 3 + \dfrac 1 2 - \dfrac 1 2 &= 1 - \dfrac 1 2 \\[5pt] \dfrac {5z} 3 &= \dfrac 1 2 } \end{eqnarray}

The fractional coefficient is $\dfrac{5}{3},$ so we multiply by its reciprocal $\left( \dfrac{5}{3} \right)^{-1} = \dfrac{3}{5}.$ This gives:

\begin{eqnarray} \require{cancel} \eqalign{ \dfrac{3}{5} \cdot \dfrac {5}{3} z &= \dfrac 3 5 \cdot \dfrac 1 2 \\[5pt] 1 z&= \dfrac 3 {10} \\[5pt] z &= \dfrac 3 {10} } \end{eqnarray}

First, we isolate the fractional term $\dfrac{7y}{2}$ by adding $3$ to both sides:

\begin{eqnarray} \require{cancel} \eqalign{ \dfrac {7y} 2 - 3 &= 11 \\[5pt] \dfrac {7y} 2 - 3 + 3 &= 11 + 3 \\[5pt] \dfrac {7y} 2 &= 14 } \end{eqnarray}

The fractional coefficient is $\dfrac{7}{2}$, so we multiply by its reciprocal $\left( \dfrac{7}{2} \right)^{-1} = \dfrac{2}{7}.$ This gives:

\begin{eqnarray} \require{cancel} \eqalign{ \dfrac {7}{2} y &= 14 \\[5pt] \dfrac{2}{7} \cdot \dfrac {7} {2} y &= \dfrac{2}{7} \cdot 14\\[5pt] 1 y &= \dfrac{2}{ \cancel{7} } \cdot \cancel{7} \cdot 2 \\[5pt] y &= 4 } \end{eqnarray}