Multi-step equations review

CCSS Math: 8.EE.C.7b
To solve an equation we find the value of the variable that makes the equation true. For more complicated, fancier equations, this process can take several steps.
When solving an equation, our goal is to find the value of the variable that makes the equation true.

Example 1: Two-step equation

Solve for $x$.
$3x+7=13$
We need to manipulate the equation to get $x$ by itself.
\begin{aligned} 3x+7&=13 \\\\ 3x+7\redD{-7}&=13\redD{-7} \\\\ 3x&=6 \\\\ \dfrac{3x}{\redD{3}}&=\dfrac{6}{\redD{3}} \\\\ x&=2 \end{aligned}
We call this a two-step equation because it took two steps to solve. The first step was to subtract $7$ from both sides, and the second step was to divide both sides by $3$. Want an explanation of why we do the same thing to both sides of the equation? Check out this video.
We check the solution by plugging $\redD2$ back into the original equation:
\begin{aligned} 3x+7&=13 \\\\ 3\cdot \redD 2 + 7 &\stackrel?= 13 \\\\ 6+7 &\stackrel?= 13 \\\\ 13 &= 13 ~~~~~~~\text{Yes!} \end{aligned}

Example 2: Variables on both sides

Solve for $a$.
$5 + 14a = 9a - 5$
We need to manipulate the equation to get $a$ by itself.
\begin{aligned} 5 + 14a &= 9a - 5 \\\\ 5 + 14a \blueD{- 9a} &= 9a - 5 \blueD{- 9a} \\\\ 5 + 5a &= -5 \\\\ 5 + 5a \blueD{-5} &= -5 \blueD{- 5}\\\\ 5a &= -10\\\\ \dfrac{5a}{\blueD5} &= \dfrac{-10}{\blueD5} \\\\ a &= \blueD{-2} \end{aligned}
$a = \blueD{-2}$
Check our work:
\begin{aligned} 5 + 14a &= 9a - 5 \\\\ 5 + 14(\blueD{-2}) &\stackrel?= 9(\blueD{-2}) - 5 \\\\ 5 + (-28) &\stackrel?= -18 - 5 \\\\ -23 &= -23 ~~~~~~~\text{Yes!} \end{aligned}

Example 3: Distributive property

Solve for $e$.
$7(2e-1)-11=6+6e$
We need to manipulate the equation to get $e$ by itself.
\begin{aligned} 7(2e-1)-11 &= 6+6e \\\\ 14e-7 -11&= 6+6e\\\\ 14e-18 &= 6+6e\\\\ 14e-18\purpleD{-6e} &= 6+6e\purpleD{-6e} \\\\ 8e-18&=6\\\\ 8e-18\purpleD{+18} &=6 \purpleD{+18} \\\\ 8e &=24\\\\ \dfrac{8e}{\purpleD{8}}&= \dfrac{24}{\purpleD{8}}\\\\ e &= \purpleD{3} \end{aligned}
$e= \purpleD{ 3 }$
\begin{aligned} 7(2e-1)-11 &= 6+6e \\\\ 7(2(\purpleD{3})-1) -11&\stackrel?= 6+6(\purpleD{3}) \\\\ 7(6-1)-11 &\stackrel?= 6+18 \\\\ 7(5)-11&\stackrel?=24 \\\\ 35-11&\stackrel?=24 \\\\ 24 &=24 ~~~~~~~\text{Yes!} \end{aligned}
Solve for $b$.
$4b+5=1+5b$
$b=$