We need to isolate $x.$ To do this, we divide both sides of the equation by $3k\mathbin{:}$

\begin{align*} \require{cancel} 3kx &= 5 \\[5pt] \dfrac{3kx}{3k} &= \dfrac{5}{3k} \\[5pt] \dfrac{\cancel{3k}x}{\cancel{3k}} &= \dfrac{5}{3k} \\[5pt] x &= \dfrac{5}{3k} \end{align*}

We need to isolate $n.$ To do this, we first add $3$ to both sides of the equation:

\begin{align*} 2kn - 3 &= 5 \\[5pt] 2kn - 3 + 3 &= 5 + 3 \\[5pt] 2kn &= 8 \end{align*}

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

\begin{align*} \require{cancel} 2kn &= 8 \\[5pt] \dfrac{2kn}{2k} &= \dfrac{8}{2k} \\[5pt] \dfrac{\cancel{2k}n}{\cancel{2k}} &= \dfrac{8}{2k} \\[5pt] n &= \dfrac{8}{2k} \end{align*}

We can further simplify the right-hand side as follows:

\begin{align*} n &= \dfrac{2\cdot 4}{2k} \\[5pt] n &= \dfrac{\cancel{2}\cdot 4}{\cancel{2}k} \\[5pt] n &=\dfrac{4}{k} \end{align*}

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

\begin{align*} 2lx &= \dfrac{6}{5} \\[5pt] 5 \cdot 2lx &= 5 \cdot \dfrac{6}{5} \\[5pt] 10lx &= 6\\[5pt] \end{align*}

Now, we need to isolate $x.$ To do this, we divide both sides of the equation by $10l{:}$

\begin{align*} \require{cancel} 10lx &= 6\\[5pt] \dfrac{10lx}{10l} &= \dfrac{6}{10l}\\[5pt] \dfrac{\cancel{10l}x}{\cancel{10l}} &= \dfrac{6}{10l}\\[5pt] %x &= \dfrac{6}{10l}\\[5pt] x &= \dfrac{6}{10l} \end{align*}

We can further simplify the right-hand side as follows:

\begin{align*} x &= \dfrac{2\cdot 3}{2\cdot 5l} \\[5pt] x &= \dfrac{\cancel{2}\cdot 3}{\cancel{2}\cdot 5l} \\[5pt] x &= \dfrac{3}{5l} \end{align*}

First, we multiply both sides of the equation by $2$ to get rid of the fraction:

\begin{align*} \dfrac{1}{2}kn-5 &= -2 \\[5pt] 2 \cdot \left( \dfrac{1}{2}kn-5 \right) &= 2 \cdot (-2) \\[5pt] kn-10 &= -4 \end{align*}

Now, we need to isolate $n.$ To do this, we first add $10$ to both sides of the equation:

\begin{align*} kn-10 &= -4 \\[5pt] kn-10+10 &= -4 +10\\[5pt] kn &= 6 \end{align*}

Then, we divide both sides of the equation by $k\mathbin{:}$

\begin{align*} \require{cancel} kn &= 6 \\[5pt] \dfrac{kn}{k} &= \dfrac{6}{k} \\[5pt] \dfrac{\cancel{k}n}{\cancel{k}} &= \dfrac{6}{k} \\[5pt] n &= \dfrac{6}{k} \end{align*}

We need to isolate $x.$ To do this, we first cross-multiply:

\begin{align*} \dfrac{8nx}{3} &= \dfrac{6}{5} \\[5pt] 5 \cdot 8nx &= 3 \cdot 6 \\[5pt] 40nx &= 18 \\[5pt] \end{align*}

Then, we divide both sides of the equation by $40n\mathbin{:}$

\begin{align*} \require{cancel} 40nx &= 18 \\[5pt] \dfrac{40nx}{40n} &= \dfrac{18}{40n} \\[5pt] \dfrac{\cancel{40n}x}{\cancel{40n}} &= \dfrac{18}{40n} \\[5pt] x &= \dfrac{18}{40n} \\[5pt] \end{align*}

We can further simplify the right-hand side as follows:

\begin{align*} x &= \dfrac{2\cdot 9}{2\cdot 20n} \\[5pt] x &= \dfrac{\cancel{2}\cdot 9}{\cancel{2}\cdot 20n} \\[5pt] x &= \dfrac{9}{20n} \\[5pt] \end{align*}

We need to isolate $w.$ To do this, we first cross-multiply:

\begin{align*} -\dfrac{5zw}{4} &= \dfrac{10}{3} \\[5pt] \dfrac{-5zw}{4} & =\dfrac{10}{3} \\[5pt] 3 \cdot (-5zw) &= 4 \cdot 10 \\[5pt] -15zw &= 40 \\[5pt] \end{align*}

Then, we divide both sides of the equation by $-15z\mathbin{:}$

\begin{align*} \require{cancel} -15zw &= 40 \\[5pt] \dfrac{-15zw}{-15z} &= \dfrac{40}{-15z} \\[5pt] \dfrac{\cancel{-15z}w}{\cancel{-15z}} &= \dfrac{40}{-15z} \\[5pt] w &= -\dfrac{40}{15z}\\[5pt] \end{align*}

We can further simplify the right-hand side as follows:

\begin{align*} w &= -\dfrac{5\cdot 8}{5\cdot 3z}\\[5pt] w &= -\dfrac{\cancel{5}\cdot 8}{\cancel{5}\cdot 3z}\\[5pt] w&= -\dfrac{8}{3z} \end{align*}