\begin{align} 6x + 3 &= x + 8 &&\text{Given. Subtract \( 3 \) from both sides.} \tag 1 \\[2ex] 6x &= x + 5 &&\text{Subtract \( 1x \) from both sides.} \tag 2 \\[2ex] 5x &= 5 &&\text{Divide both sides by \( 5 \).} \tag 3 \\[2ex] x &= 1 &&\text{Solution.} \tag 4 \\[2ex] \end{align}

\begin{align} \dfrac{3p}{8} + \dfrac{7p}{16} - \dfrac{3}{4} &= \dfrac{1}{4} + \dfrac{p}{16} + \dfrac{1}{2} &&\text{Multiply all terms by 16.} \tag 1 \\[2ex] {\color{red}9p} - {\color{green}12} &= {\color{green}12} + {\color{red}p} &&\text{Subtract 1p from both sides.} \tag 2 \\[2ex] {\color{red}8p} - {\color{green}12} &= {\color{green}12} &&\text{Add 12 to both sides.} \tag 3 \\[2ex] {\color{red}8p} &= {\color{green}24} &&\text{Divide both sides by 8.} \tag 4 \\[2ex] p &= 3 &&\text{Solution.} \tag 5 \\[2ex] \end{align}

Find the number of minutes.

1. Let \( m \) = number of minutes
2. Company A: \( 2 + 0.027m \)
3. Company B: \( 0.035m \)
4. Set the expressions equal to each other and solve for $m$.

\begin{align} 2 + 0.027m &= 0.035m &&\text{Given. Multiply all terms by 16.} \tag 1 \\[2ex] 2 &= 0.008m &&\text{Divide both sides by 0.008.} \tag 2 \\[2ex] 250 &= m &&\text{Solution.} \tag 3 \\[2ex] \end{align}

Find the monthly charge at which both companies have the same fee.

1. Let $c$ = the cost at which the charges from both companies are the same.
2. Multiply the Number of Minutes (250) by the Unit Rate for Company B ($0.035).

\begin{align} c &= 250(0.035) &&\text{Given. Multiply the left side.} \tag 1 \\[2ex] c &= $8.75 &&\text{Solution.} \tag 2 \\[2ex] \end{align}

Solutions

1. 250 minutes
2. $8.75 per month