We need to isolate the variable $h.$ To do this, we treat $k$ as though it were a constant and apply the addition and multiplication principles as usual.

In this case, we simply add $\dfrac 3 2 k$ to both sides of the equation:

\begin{align*} h-\dfrac{3}{2}k &= 2 \\[5pt] h-\dfrac{3}{2}k+\dfrac{3}{2}k&= 2+\dfrac{3}{2}k\\[5pt] h - 0 &= 2+\dfrac{3}{2}k \\[5pt] h &= 2+\dfrac{3}{2}k \end{align*}

We need to isolate the variable $x.$ To do this, we treat $k$ as though it were a constant and apply the addition and multiplication principles as usual.

In this case, we simply subtract $k$ from both sides of the equation:

\begin{align*} x+k &= 5 \\[3pt] x+k-k &= 5-k \\[3pt] x + 0 &= 5-k \\[3pt] x &= 5-k \end{align*}

We need to isolate the variable $x.$ To do this, we treat $y$ as though it were a constant and apply the addition and multiplication principles as usual.

In this case, we simply multiply both sides of the equation by $5\mathbin{:}$

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

We need to isolate the variable $x.$ To do this, we treat $y$ as though it were a constant and apply the addition and multiplication principles as usual.

First, we add $4y$ to both sides of the equation:

\begin{align*} 3x-4y &=5y\\[3pt] 3x-4y+4y &=5y+4y\\[3pt] 3x &=5y+4y\\[3pt] 3x &=9y \end{align*}

Then, we divide both sides of the equation by $3\mathbin{:}$

\begin{align*} \require{cancel} 3x&=9y\\[5pt] \dfrac{3x}{3}&=\dfrac{9y}{3}\\[5pt] \dfrac{\cancel{3}x}{\cancel{3}}&=\dfrac{9y}{3}\\[5pt] x&=\dfrac{9y}{3}\\[5pt] x&=3y \end{align*}

We need to isolate the variable $s.$ To do this, we treat $t$ as though it were a constant and apply the addition and multiplication principles as usual.

First, we subtract $\dfrac{t}{2}$ from both sides of the equation:

\begin{align*} \dfrac{s}{4}+\dfrac{t}{2} &=1\\[3pt] \dfrac{s}{4}+\dfrac{t}{2}-\dfrac{t}{2} &=1-\dfrac{t}{2}\\[3pt] \dfrac{s}{4}+0 &=1-\dfrac{t}{2}\\[3pt] \dfrac{s}{4} &=1-\dfrac{t}{2} \end{align*}

Then, we multiply both sides of the equation by $4\mathbin{:}$

\begin{align*} \require{cancel} \dfrac{s}{4} &= 1-\dfrac{t}{2}\\[5pt] 4 \cdot \dfrac{s}{4} &= 4 \cdot \left(1-\dfrac{t}{2}\right) \\[5pt] \cancel{4}\cdot \dfrac{s}{\cancel{4}} &= 4 \cdot \left(1-\dfrac{t}{2}\right) \\[5pt] s &= 4 \cdot \left(1-\dfrac{t}{2}\right) \\[5pt] s &= 4\cdot 1 - 4 \cdot \dfrac{t}{2} \\[5pt] s &= 4 - 2t \\[5pt] \end{align*}

We need to isolate the variable $a.$ To do this, we treat $b$ as though it were a constant and apply the addition and multiplication principles as usual.

First, we subtract $3a$ from both sides of the equation:

\[ \eqalign{ 3a + b &= 12a \\[5pt] 3a + b -3a&= 12a - 3a \\[5pt] b &= 9a } \]

Then, we divide both sides of the equation by $9\mathbin{:}$

\begin{align*} \require{cancel} b &= 9a \\[5pt] \dfrac{b}{9} &=\dfrac{a}{9} \\[5pt] \dfrac{b}{9} &=\dfrac{\cancel{9}a}{\cancel{9}} \\[5pt] \dfrac{b}{9} &=a \\[5pt] \end{align*}

Thus, $a=\dfrac b 9.$

We need to isolate the variable $n.$ To do this, we treat $m$ as though it were a constant and apply the addition and multiplication principles as usual.

First, we add $6n$ to both sides of the equation:

\begin{align*} 4m - 6n &= 3n - 7 \\[5pt] 4m - 6n + 6n &= 3n - 7 + 6n \\[5pt] 4m &= 9n - 7 \\[5pt] \end{align*}

Then, we add $7$ to both sides of the equation:

\begin{align*} 4m &= 9n - 7 \\[5pt] 4m + 7 &= 9n - 7 + 7 \\[5pt] 4m + 7 &= 9n \end{align*}

Finally, we divide both sides of the equation by $9\mathbin{:}$

\begin{align*} 4m + 7 &= 9n \\[5pt] \dfrac{4m + 7}{9} &= \dfrac{9n}{9}\\[5pt] \dfrac{4m + 7}{9} &= \dfrac{\cancel{9}n}{\cancel{9}}\\[5pt] \dfrac{4m + 7}{9} &= n \end{align*}

Therefore, $n = \dfrac{4m + 7}{9}.$

We need to isolate the variable $x.$ First, we cross-multiply:

\begin{align*} \dfrac{4x-3y}{2} &=\dfrac{1-y}{3} \\[5pt] 3\cdot (4x-3y) &= 2\cdot (1-y) \\[5pt] 12x - 9y &= 2-2y \\[5pt] \end{align*}

Next, we add $9y$ to both sides of the equation:

\begin{align*} 12x - 9y &= 2-2y \\[5pt] 12x - 9y + 9y &= 2-2y + 9y \\[5pt] 12x = 2+7y \end{align*}

Finally, we divide both sides of the equation by $12\mathbin{:}$

\begin{align*} \require{cancel} 12x = 2+7y \\[5pt] \dfrac{12x}{12} = \dfrac{2+7y}{12} \\[5pt] \dfrac{\cancel{12}x}{\cancel{12}} = \dfrac{2+7y}{12} \\[5pt] x = \dfrac{2+7y}{12} \\[5pt] \end{align*}

We need to isolate the variable $f.$ First, we cross-multiply:

\begin{align*} \dfrac{f - 2}{5} &= \dfrac{5g - 2f}{6} \\[5pt] 6\cdot (f - 2) &= 5\cdot(5g - 2f) \\[5pt] 6f - 12 &= 25g - 10f \end{align*}

Next, we add $10f $ to both sides of the equation:

\begin{align*} 6f - 12 &= 25g - 10f \\[5pt] 6f - 12 + 10f &= 25g - 10f + 10f \\[5pt] 6f + 10f - 12 &= 25g - 10f + 10f \\[5pt] 16f - 12 &= 25g \end{align*}

Next, we add $12 $ to both sides of the equation:

\begin{align*} 16f - 12 &= 25g \\[5pt] 16f - 12 + 12 &= 25g + 12 \\[5pt] 16f &= 25g + 12 \end{align*}

Finally, we divide both sides of the equation by $16\mathbin{:}$

\begin{align*} \require{cancel} 16f &= 25g + 12 \\[5pt] \dfrac{16f}{16} &= \dfrac{25g + 12}{16} \\[5pt] \dfrac{\cancel{16}f}{\cancel{16}} &= \dfrac{25g + 12}{16} \\[5pt] f &= \dfrac{25g + 12}{16} \end{align*}