We need to isolate $w,$ so we will treat the other variables $x$ and $y$ as constants. We can use the addition and multiplication principles to solve for $w.$

First, we subtract $3x$ from both sides: \begin{align*} 2w + 3x &= y \\[5pt] 2w + 3x - 3x&= y -3x \\[5pt] 2w &= y - 3x \end{align*}

Then, we divide both sides by $2\mathbin{:}$ \begin{align*} 2w &= y - 3x \\[5pt] \dfrac{2w}{2} &= \dfrac{y-3x}{2} \\[5pt] w &= \bbox[4pt,border: 1px solid lightgray]{\dfrac{y-3x}{2}} \end{align*}

We need to isolate $c,$ so we will treat the other variables $a$ and $b$ as constants. We can use the addition and multiplication principles to solve for $c.$

First, we add $2c$ to both sides:

\[ \eqalign{ -6a&=9b-2c\\[3pt] -6a+2c&=9b-2c+2c\\[3pt] -6a+2c&=9b } \] Next, we add $6a$ to both sides: \[ \eqalign{ -6a+2c&=9b\\[3pt] -6a+2c+6a&=9b+6a\\[3pt] 2c&=9b+6a } \] Then, we divide both sides by $2\mathbin{:}$ \[ \eqalign{ 2c&=9b+6a\\[5pt] \dfrac{2c}{2}&=\dfrac{9b+6a}{2}\\[5pt] c&=\dfrac{9b+6a}{2} } \]

We need to isolate $l,$ so we will treat the other variables $p$ and $w$ as constants. We can use the addition and multiplication principles to solve for $l.$

First, we subtract $2w$ from both sides: \begin{align*} p &= 2w + 2l\\[3pt] p-2w &= 2l+2w -2w \\[3pt] p-2w &=2l \end{align*}

Then, we divide by $2$ to get $l$ alone: \begin{align*} p-2w & = 2l \\[5pt] \dfrac{p-2w}{2} & = \dfrac{2l}{2} \\[5pt] \dfrac{p-2w}{2} & = l \end{align*}

Thus, $l = \bbox[4pt,border: 1px solid lightgray]{\dfrac{p-2w}{2}}.$

We need to isolate $c,$ so we will treat the other variables $a$ and $b$ as constants. Since $c$ is being multiplied by $b,$ we need to divide both sides of the equation by $b\mathbin{:}$

\begin{align*} a &= bc \\[5pt] \dfrac{a}{b} &= \dfrac{bc}{b} \\[5pt] \dfrac{a}{b} &= c \end{align*}

Thus, $c = \bbox[4pt,border: 1px solid lightgray]{\dfrac{a}{b}}.$

Note: This result assumes that $b \neq 0.$

We need to isolate $z,$ so we will treat the other variables $w,x,$ and $y$ as constants. We can use the multiplication principle to solve for $z.$

First, we rearrange and then divide both sides by $4xy\mathbin{:}$ \begin{align*} w & = 4xyz \\[5pt] w & = (4xy)z \\[5pt] \dfrac{w}{4xy} & = \dfrac{(4xy)z}{4xy}\\[5pt] \dfrac{w}{4xy} & = z \end{align*}

Therefore, $z = \bbox[4pt,border: 1px solid lightgray]{\dfrac{w}{4xy}}.$

Note: This result assumes that $x \neq 0$ and $y \neq 0.$

We need to isolate $t,$ so we will treat the other variables $v$ and $g$ as constants.

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

\[ \eqalign{ v &= \dfrac{1}{2}gt\\[5pt] 2 \cdot v &= 2 \cdot \dfrac{1}{2}gt\\[5pt] 2v &= gt } \]

Since $t$ is being multiplied by $g,$ we need to divide both sides of the equation by $g\mathbin{:}$

\[ \eqalign{ 2v &= gt\\[5pt] \dfrac{2v}{g} &= \dfrac{gt}{g} \\[5pt] \dfrac{2v}{g} &= t } \]

Thus, $t = \dfrac{2v}{g}.$

Note: This result assumes that $g \neq 0.$

We need to isolate $c,$ so we will treat the other variables $a,b,$ and $d$ as constants. We can use the multiplication principle to solve for $c.$

Since $c$ is being multiplied by $5d,$ we need to divide both sides of the equation by $5d\mathbin{:}$

\[ \eqalign{ 5cd &= a+b \\[5pt] (5d)c &= a+b \\[5pt] \dfrac{(5d)c}{5d}&= \dfrac{a+b}{5d} \\[5pt] c &= \bbox[4pt,border: 1px solid lightgray]{\dfrac{a+b}{5d}} } \]

Note: This result assumes that $d \neq 0.$

We need to isolate $q,$ so we will treat the other variables $p$ and $r$ as constants. We can use the addition and multiplication principles to solve for $q.$

First, we add $2r$ to both sides of the equation: \begin{align*} \dfrac{pq}{4} - 2r &= 5p\\[5pt] \dfrac{pq}{4} - 2r+ 2r &= 5p + 2r \\[5pt] \dfrac{pq}{4} &= 5p + 2r \end{align*}

Then, we multiply by $4$ to get rid of the fraction: \begin{align*} \dfrac{pq}{4} &= 5p + 2r \\[5pt] 4 \cdot \dfrac{pq}{4} &= 4 \cdot (5p +2r) \\[5pt] pq &=4 (5p + 2r) \end{align*}

Finally, we divide by $p$ to get $q$ alone: \begin{align*} pq &= 4 (5p+2r) \\[5pt] \dfrac{pq}{p} &= \dfrac{4 (5p+2r)}{p} \\[5pt] q &= \bbox[4pt,border: 1px solid lightgray]{\dfrac{4(5p + 2r)}{p}} \end{align*}

Note: This result assumes that $p \neq 0.$

We need to isolate $h,$ so we will treat the other variables $s,$ $b,$ and $r$ as constants. We can use the addition and multiplication principles to solve for $h.$

First, we subtract $b$ from both sides:

\[ \eqalign{ s &= b+rh \\ s-b &= b+rh-b \\ s-b &= rh } \]

Then, we divide by $r$ to get $h$ alone:

\[ \eqalign{ s-b & = rh \\[5pt] \dfrac{s-b}{r} & = \dfrac{rh}{r}\\[5pt] \dfrac{s-b}{r} & = h } \]

Thus, $h = \bbox[4pt,border: 1px solid lightgray]{\dfrac{s-b}{r}}.$

Note: This result assumes that $r \neq 0.$