We must bring all variables to one side of the equation and all constants to the other side.

Notice that we have $3z$ on the left-hand side and $z$ on the right-hand side. Since the variable on the left-hand side has a larger coefficient, let's bring all the variables to this side.

The first step is to remove the $z$ from the right-hand side. We do this by subtracting ${\color{blue}z}$ from both sides:

\begin{align} 3z & = z + 8\\[5pt] 3z - {\color{blue}z} & = z + 8 - {\color{blue}z}\\[5pt] 2z & = 8 \end{align}

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

\begin{align} \require{cancel} 2z & = 8\\[5pt] \dfrac{2z}{2} & = \dfrac{8}{2}\\[5pt] \dfrac{\cancel{2}z}{\cancel{2}} & = 4\\[5pt] z & = 4 \end{align}

We must bring all variables to one side of the equation and all constants to the other side.

Notice that we have $p$ on the left-hand side and $5p$ on the right-hand side. Since the variable on the right-hand side has a larger coefficient, let's bring all the variables to this side.

The first step is to remove the $p$ from the left-hand side. We do this by subtracting ${\color{blue}p}$ from both sides:

\begin{align*} p+12 &= 5p \\[5pt] p+12 - {\color{blue}p} &= 5p-{\color{blue}p}\\[5pt] 12 &= 4p \end{align*}

Finally, we divide both sides of the equation by $4\mathbin{:}$

\begin{align*} \require{cancel} 12 &= 4p \\[5pt] \dfrac{12}{4} &= \dfrac {4p}{4} \\[5pt] 3 &= \dfrac {\cancel 4p}{\cancel 4} \\[5pt] 3 &= p\\[5pt] p &=3. \end{align*}

We must bring all variables to one side of the equation and all constants to the other side.

Notice that we have $-2y$ on the left-hand side and $5y$ on the right-hand side. Since the variable on the right-hand side has a larger coefficient, let's bring all the variables to this side.

The first step is to remove the $-2y$ from the left-hand side. We do this by adding ${\color{blue}2y}$ to both sides:

\begin{align} 28−2y & = 5y\\[5pt] 28 −2y + {\color{blue}2y} & = 5y + {\color{blue}2y}\\[5pt] 28 & = 7y\\[5pt] \end{align}

Finally, we divide both sides of the equation by $7\mathbin{:}$

\begin{align} \require{cancel} 28 & = 7y\\[5pt] \dfrac{28}{7} & = \dfrac{7y}{7}\\[5pt] 4 & = \dfrac{\cancel{7}y}{\cancel{7}}\\[5pt] 4 & = y \\[5pt] y & = 4 \end{align}

We must bring all variables to one side of the equation and all constants to the other side.

Notice that we have $5t$ on the left-hand side and $2t$ on the right-hand side. Since the variable on the left-hand side has a larger coefficient, let's bring all the variables to this side.

The first step is to remove the $2t$ from the right-hand side of the equation. We do this by subtracting ${\color{blue}2t}$ from both sides:

\begin{align} 5t-1&=5+2t \\[5pt] 5t - 1 - {\color{blue}2t}& = 5 + 2t - {\color{blue}2t}\\[3pt] 3t - 1 &= 5\\[5pt] \end{align}

Next, we remove the $-1$ from the left-hand side. We do this by adding ${\color{red}1}$ to both sides:

\begin{align} \require{cancel} 3t - 1 &= 5\\[5pt] 3t - 1 + {\color{red}1} &= 5 + {\color{red}1}\\[5pt] 3t & = 6\\[5pt] \end{align}

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

\begin{align} 3t & = 6\\[5pt] \dfrac{3t}{3} & = \dfrac{6}{3}\\[5pt] \dfrac{\cancel{3}t}{\cancel{3}} & = 2\\[5pt] t & = 2 \end{align}

We must bring all variables to one side of the equation and all constants to the other side.

Notice that we have $2p$ on the left-hand side and $-4p$ on the right-hand side. Since the variable on the left-hand side has a larger coefficient, let's bring all the variables to this side.

The first step is to remove the $-4p$ from the right-hand side of the equation. We do this by adding ${\color{blue}4p}$ to both sides:

\begin{align} 2p + 1 & = 4 - 4p \\[3pt] 1 + 2p + {\color{blue}4p} & = 4 - 4p + {\color{blue}4p} \\[3pt] 1 + 6p & = 4 \\[3pt] \end{align}

Next, we remove the $1$ from the left-hand side. We do this by subtracting ${\color{red}1}$ from both sides:

\begin{align} \require{cancel} 1 + 6p &= 4\\[5pt] 1 + 6p - {\color{red}1} &= 4 - {\color{red}1}\\[5pt] 6p & = 3\\[5pt] \end{align}

Finally, we divide both sides of the equation by $6\mathbin{:}$ \begin{align} \require{cancel} 6p & = 3\\[5pt] \dfrac{6p}{6} & = \dfrac{3}{6}\\[5pt] \dfrac{\cancel{6}p}{\cancel{6}} & = \dfrac{1}{2}\\[5pt] p & = \dfrac{1}{2} \end{align}

We must bring all variables to one side of the equation and all constants to the other side.

Notice that we have $-k$ on the left-hand side and $3k$ on the right-hand side. Since the variable on the right-hand side has a larger coefficient, let's bring all the variables to this side.

The first step is to remove the $-k$ from the left-hand side of the equation. We do this by adding ${\color{blue}k}$ to both sides:

\begin{align*} 2-k &= 3k + 9 \\[5pt] 2 - k + {\color{blue}k} &= 3k + 9 + {\color{blue}k}\\[5pt] 2 &= 4k + 9 \end{align*}

Next, we remove the $9$ from the right-hand side. We do this by subtracting ${\color{red}9}$ from both sides:

\begin{align*} \require{cancel} 2 &= 4k + 9 \\[5pt] 2 - {\color{red}9}&= 4k + 9 - {\color{red}9} \\[5pt] -7 &= 4k \end{align*}

Finally, we divide both sides of the equation by $4\mathbin{:}$

\begin{align*} - 7 &= 4k \\[5pt] \dfrac{-7}{4} &= \dfrac{4k}{4} \\[5pt] -\dfrac{7}{4} &= \dfrac{\cancel{4}k}{\cancel{4}} \\[5pt] -\dfrac74 &=k \\[5pt] k &= -\dfrac{7}{4} \end{align*}

First, we expand the parentheses on the left-hand-side:

\begin{align*} 3(a - 2) &= a \\[5pt] 3 \cdot a - 3 \cdot 2 &= a \\[5pt] 3a - 6 &= a \end{align*}

Now, we must now bring all variables to one side of the equation and all constants to the other side.

Notice that we have $3a$ on the left-hand side and $a$ on the right-hand side. Since the variable on the left-hand side has a larger coefficient, let's bring all the variables to this side.

The first step is to remove the $a$ from the right-hand side of the equation. We do this by subtracting ${\color{blue}a}$ from both sides:

\begin{align*} 3a - 6 &= a \\[5pt] 3a - 6 -{\color{blue}a} &= a-{\color{blue}a} \\[5pt] 2a - 6 &= 0 \end{align*}

Next, we remove the $-6$ from the left-hand side. We do this by adding ${\color{red}6}$ to both sides:

\begin{align*} 2a - 6 &= 0 \\[5pt] 2a - 6 + {\color{red}6} &= 0+{\color{red}6} \\[5pt] 2a &= 6 \\[5pt] \end{align*}

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

\begin{align*} \dfrac{2a}{2} &= \dfrac{6}{2} \\[5pt] \dfrac{\cancel{2} a}{\cancel{2}} &= 3 \\[5pt] a &= 3 \end{align*}

First, we expand the parentheses on the left-hand-side:

\begin{align} 2(2p + 1) & = - p\\[5pt] 2\cdot 2p + 2\cdot 1 & = - p\\[5pt] 4p + 2 & = -p \end{align}

Now, we must now bring all variables to one side of the equation and all constants to the other side.

Notice that we have $4p$ on the left-hand side and $-p$ on the right-hand side. Since the variable on the left-hand side has a larger coefficient, let's bring all the variables to this side.

The first step is to remove the $-p$ from the right-hand side of the equation. We do this by adding ${\color{blue}p}$ to both sides:

\begin{align} 4p + 2 & = -p\\[5pt] 4p + 2 + {\color{blue}p} & = -p + {\color{blue}p}\\[5pt] 5p + 2 & = 0 \end{align}

Next, we remove the $2$ from the left-hand side. We do this by subtracting ${\color{red}2}$ from both sides:

\begin{align*} 5p +2 &= 0 \\[5pt] 5p +2 - {\color{red}2} &= 0- {\color{red}2} \\[5pt] 5p &= -2 \\[5pt] \end{align*}

Finally, we divide both sides of the equation by $5\mathbin{:}$

\begin{align*} \dfrac{5p}{5} &= \dfrac{-2}{5} \\[5pt] \dfrac{\cancel{5} p}{\cancel{5}} &=-\dfrac{2}{5} \\[5pt] p &= -\dfrac{2}{5} \end{align*}

First, we expand the parentheses on both sides:

\begin{align} 5(x-3) & = 3(x + 7) + 5x\\[5pt] 5 \cdot x - 5 \cdot 3 & = 3 \cdot x + 3 \cdot 7 + 5x\\[5pt] 5x - 15 & = 3x + 21 + 5x\\[5pt] 5x - 15 & = 21 + 3x + 5x\\[5pt] 5x - 15 & = 21 + 8x \end{align}

Now, we must bring all variables to one side of the equation and all constants to the other side.

Notice that we have $5x$ on the left-hand side and $8x$ on the right-hand side. Since the variable on the right-hand side has a larger coefficient, let's bring all the variables to this side.

The first step is to remove the $5x$ from the left-hand side of the equation. We do this by subtracting ${\color{blue}5x}$ from both sides:

\begin{align} 5x - 15 & = 21 + 8x\\[5pt] 5x - 15 - {\color{blue}5x}& = 21 + 8x - {\color{blue}5x}\\[5pt] -15 &=21 + 3x \end{align}

Next, we remove the $21$ from the right-hand side. We do this by subtracting ${\color{red}21}$ from both sides:

\begin{align} \require{cancel} -15 &=21 + 3x \\[5pt] -15 - {\color{red}21} &=21 + 3x - {\color{red}21} \\[5pt] -36 &=3x \end{align}

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

\begin{align*} \require{cancel} -36 &=3x \\[5pt] \dfrac{-36}{3} &=\dfrac{3x}{3}\\[5pt] -12 &=\dfrac{\cancel{3}x}{\cancel{3}}\\[5pt] -12 &=x\\[5pt] x &=-12 \end{align*}