First, we apply the addition principle, subtracting $4$ from both sides:

\begin{eqnarray} \eqalign{ 5x+4&=19 \\[3pt] 5x+4-4 &= 19-4\\[3pt] 5x + 0 &= 15 \\[3pt] 5x &=15 } \end{eqnarray}

Second, we apply the multiplication principle, dividing both sides by $5\mathbin{:}$

\begin{eqnarray} \require{cancel} \eqalign{ 5x &=15 \\[5pt] \dfrac {5x} {5} &= \dfrac {15} {5} \\[5pt] \dfrac {\cancel{5}x} {\cancel{5}} &= 3 \\[5pt] x& = 3 } \end{eqnarray}

First, we rewrite the left-hand side by placing the term containing the variable first: \begin{align*} -7 - z &= 10\\[5pt] -z - 7&= 10 \end{align*}

Next, we apply the addition principle, adding $7$ to both sides: \begin{align*} -z - 7 &= 10\\[5pt] -z - 7 + 7&= 10+7\\[5pt] -z + 0 &= 17\\[5pt] -z &= 17 \end{align*}

Then, we apply the multiplication principle. We multiply by $-1$ to get rid of the negative sign. \begin{align*} \require{cancel} -z &= 17 \\[3pt] (-1) \cdot \left( -z \right) &= (-1) \cdot 17 \\[5pt] z &= \bbox[3pt,Gainsboro]{\color{blue}-17} \end{align*}

First, we rewrite the left-hand side by placing the term containing the variable first: \begin{align} 7-3k&=16\\[3pt] -3k+7&=16 \end{align}

Next, we apply the addition principle, subtracting $7$ from both sides:

\begin{align} -3k+7&=16\\[3pt] -3k+7-7&=16-7\\[3pt] -3k+0&=9\\[3pt] -3k&=9 \end{align}

Next, we apply the multiplication principle, dividing both sides by $-3\mathbin{:}$

\begin{align} \require{cancel} - 3k &= 9 \\[3pt] \dfrac{ -3 k}{-3} &= \dfrac{ 9}{-3} \\[3pt] k &= -3 \end{align}

To save time, we will keep the variable on the right-hand side as we solve. First, we apply the addition principle, subtracting $8$ from both sides: \begin{align*} 20 &=6k+8 \\[5pt] 20-8 &=6k+8-8 \\[5pt] 12 &=6k+0 \\[5pt] 12 &=6k \end{align*}

Second, we apply the multiplication principle, dividing both sides by $6\mathbin{:}$ \begin{align*} \require{cancel} 12 &=6k \\[5pt] \dfrac{12}{6} &= \dfrac{6k}{6} \\[5pt] 2 &= \dfrac{\cancel{6} k}{\cancel{6} } \\[5pt] 2 &= k \end{align*}

Since $2=k,$ we conclude that the solution is $k=\bbox[3pt,Gainsboro]{\color{blue}2}.$

To save time, we will keep the variable on the right-hand side as we solve. First, we rewrite the right-hand side by placing the term containing the variable first:

\begin{align*} 30 & = -6+4s \\[3pt] 30 & = 4s - 6 \end{align*}

Next, we apply the addition principle, adding $6$ to both sides:

\begin{align*} 30 & = 4s - 6\\[3pt] 30 + 6 & = 4s - 6 + 6 \\[3pt] 36 & = 4s + 0 \\[3pt] 36 & = 4s \end{align*}

Next, we apply the multiplication principle, dividing both sides by $4\mathbin{:}$

\begin{align*} \require{cancel} 36 & = 4s \\[5pt] \dfrac{36}{4} &= \dfrac{4s}{4} \\[5pt] 9 & = \dfrac{\cancel{4} s}{\cancel{4} } \\[5pt] 9 & = s \end{align*}

Since $9 = s,$ we conclude that the solution is $s = 9.$

To save time, we will keep the variable on the right-hand side as we solve. First, we rewrite the right-hand side by placing the term containing the variable first: \begin{align*} 15 &= 8 - x \\[3pt] 15 &= -x + 8 \end{align*}

Next, we apply the addition principle, subtracting $8$ from both sides: \begin{align*} 15 &= -x + 8 \\[3pt] 15 - 8 &= -x + 8 - 8 \\[3pt] 7 &= -x + 0 \\[3pt] 7 &= -x \end{align*}

Next, we apply the multiplication principle. We multiply by $-1$ to get rid of the negative sign. \begin{align*} \require{cancel} 7 &= -x \\[5pt] (-1) \cdot 7 &= (-1) \cdot \left( -x \right) \\[3pt] -7 &= x \end{align*}

Since $-7=x,$ we conclude that the solution is $x=\bbox[3pt,Gainsboro]{\color{blue}-7}.$

To save time, we will keep the variable on the right-hand side as we solve. First, we rewrite the right-hand side by placing the term containing the variable first: \begin{align*} 18 &=3-5x \\[3pt] 18 &=-5x+3 \end{align*}

Next, we apply the addition principle, subtracting $3$ from both sides: \begin{align*} 18 &=-5x+3 \\[3pt] 18-3 &=-5x+3-3 \\[3pt] 15 &=-5x+0 \\[3pt] 15 &=-5x \end{align*}

Next, we apply the multiplication principle, dividing both sides by $-5\mathbin{:}$ \begin{align*} \require{cancel} 15 &=-5x \\[5pt] \dfrac{15}{-5} &= \dfrac{-5x}{-5} \\[5pt] -3 &= \dfrac{\cancel{-5} x}{\cancel{-5} } \\[5pt] -3 &= x \end{align*}

Since $-3=x,$ we conclude that the solution is $x=\bbox[3pt,Gainsboro]{\color{blue}-3}.$

First, we apply the addition principle, subtracting $2$ from both sides:

\begin{eqnarray} \require{cancel} \eqalign{ \dfrac 1 6 x + 2&= 4 \\[5pt] \dfrac 1 6 x + 2-2&= 4-2 \\[5pt] \dfrac 1 6 x + 0&= 2 \\[5pt] \dfrac 1 6 x &= 2 } \end{eqnarray}

Second, we apply the multiplication principle, multiplying both sides by $6\mathbin{:}$

\begin{eqnarray} \require{cancel} \eqalign{ \dfrac 1 6 x &= 2 \\[5pt] \cancel{6} \cdot \dfrac{1}{\cancel{6}} x &= 6 \cdot 2 \\[5pt] x &= 12 } \end{eqnarray}

First, we rewrite the left-hand side by placing the term containing the variable first:

\begin{eqnarray} \require{cancel} \eqalign{ 12 - \dfrac b 4 &= 14 \\[5pt] - \dfrac b 4 + 12 &= 14 } \end{eqnarray}

Next, we apply the addition principle, subtracting $12$ from both sides:

\begin{eqnarray} \require{cancel} \eqalign{ - \dfrac b 4 + 12 &= 14 \\[5pt] - \dfrac b 4 +12-12 &= 14-12 \\[5pt] - \dfrac b 4 +0 &= 2 \\[5pt] - \dfrac b 4 &= 2 } \end{eqnarray}

Then, we apply the multiplication principle. We multiply by $-1$ to get rid of the negative sign, and again by $4$ to get rid of the fraction.

\begin{eqnarray} \require{cancel} \eqalign{ - \dfrac b 4 &= 2 \\[5pt] (-1) \cdot \left( - \dfrac b 4 \right) &= (-1) \cdot 2 \\[5pt] \dfrac b 4 &= -2 \\[5pt] \cancel{4} \cdot \dfrac{b}{\cancel{4}} &= 4 \cdot (-2) \\[5pt] b &= -8 } \end{eqnarray}

First, we apply the addition principle, subtracting $2$ from both sides of the equation:

\begin{eqnarray} \require{cancel} \eqalign{ \dfrac{x}{5} + 2 &= \dfrac{2}{3} \\[5pt] \dfrac{x}{5} + 2-2 &= \dfrac{2}{3}-2 \\[5pt] \dfrac{x}{5} + 0&= \dfrac{2}{3}-\dfrac{6}{3} \\[5pt] \dfrac{x}{5} &= -\dfrac{4}{3} } \end{eqnarray}

Then, we apply the multiplication principle, multiplying both sides of the equation by $5\mathbin{:}$

\begin{eqnarray} \require{cancel} \eqalign{ \dfrac{x}{5} &= -\dfrac{4}{3} \\[5pt] \cancel{5} \cdot \dfrac{x}{\cancel{5}} &= 5 \cdot \left( -\dfrac{4}{3} \right) \\[5pt] x &= -\dfrac{20}{3} } \end{eqnarray}

First, we apply the addition principle, adding $\dfrac{1}{5}$ to both sides:

\begin{eqnarray} \require{cancel} \eqalign{ \dfrac{x}{2} - \dfrac{1}{5} &= 1 \\[5pt] \dfrac{x}{2} - \dfrac{1}{5} + \dfrac{1}{5} &= 1 + \dfrac{1}{5} \\[5pt] \dfrac{x}{2} + 0&= \dfrac{5}{5}+\dfrac{1}{5} \\[5pt] \dfrac{x}{2} &= \dfrac{6}{5} } \end{eqnarray}

Then, we apply the multiplication principle, multiplying both sides by $2\mathbin{:}$

\begin{eqnarray} \require{cancel} \eqalign{ \dfrac{x}{2} &= \dfrac{6}{5} \\[5pt] \cancel{2} \cdot \dfrac{x}{\cancel{2}} &= 2 \cdot \dfrac{6}{5} \\[5pt] x &= \dfrac{12}{5} } \end{eqnarray}