We will use a linear equation of the form

\[ \text{(total value)} = \text{(starting value)} + \text{(rate of change)}n. \]

We fill in the formula as follows:

• The total value is $G,$ the amount of gas, in gallons, remaining in the tank.

• The starting value is $30,$ since the truck has $30$ gallons of gas in the tank at the start of the journey.

• The rate of change is $-0.2,$ since the amount of gas in the tank decreases at a rate of $0.2$ gallons per mile. Note that the rate of change is negative because the amount of gas decreases as the truck travels.

Substituting into the formula, we reach

\[ G = 30 - 0.2 n. \]

We will use a linear equation of the form \[ \text{(total value)} = \text{(starting value)} + \text{(rate of change)}n. \]

We fill in the formula as follows:

• The total value is $D,$ the total distance that the car has traveled.

• The starting value is $10,$ since the car had traveled $10$ miles before the purchase.

• The rate of change is $46,$ since Sam's distance increases at a rate of $46$ miles per hour.

Substituting into the formula, we reach

\begin{align*} D &= 10+46n. \end{align*}

We will use a linear equation of the form \[ \text{(total value)} = \text{(starting value)} + \text{(rate of change)}n. \]

We fill in the formula as follows:

• The total value is $P,$ the number of pages remaining.

• We do not know the number of pages in the book. We will call it $A.$

• The rate of change is $-15,$ since Mandy reads $15$ pages each day. Note that the rate of change is negative because the number of pages remaining decreases as Mandy reads.

Substituting into the formula, we reach

\begin{align*} P &= A - 15n. \end{align*}

To figure out the value of $A,$ we can use the fact that after $n=7$ days, Mandy has $P=13$ pages remaining. We substitute this information into our equation and solve for $A{:}$

\begin{align*} P &= A - 15n \\ (13) &= A - 15(7) \\ 13 &= A - 105 \\ 118 &= A \end{align*}

So, the formula is

\begin{align*} P &= 118 - 15n. \end{align*}

We will use a linear equation of the form \[ \text{(total value)} = \text{(starting value)} + \text{(rate of change)}n. \]

We fill in the formula as follows:

• The total value is $P,$ the number of pizzas in stock.

• We do not know the number of pizzas that were in stock before Jerry started working. We will call it $A.$

• The rate of change is $2,$ since Jerry makes $2$ additional pizzas each hour.

Substituting into the formula, we reach

\begin{align*} P &= A + 2n. \end{align*}

To figure out the value of $A,$ we can use the fact that after $n=5$ hours, there are $P=22$ pizzas in stock. We substitute this information into our equation and solve for $A\mathbin{:}$

\begin{align*} P &= A + 2n \\ (22) &= A + 2(5) \\ 22 &= A + 10 \\ 12 &= A \end{align*}

So, the formula is

\begin{align*} P &= 12 + 2n. \end{align*}

We want to solve for the number of hours $n$ it takes until Jacklyn is $0$ miles from her school.

We can do this by substituting into the equation $D = 3 - 6n$ and solving for $n,$ as follows:

\begin{align*} 0 &= 3 - 6n \\[3pt] 6n &= 3 \\[3pt] n &= \dfrac{1}{2} \end{align*}

Therefore, Jacklyn takes $n=\dfrac{1}{2}$ hour, or $30$ minutes, to walk to school.

We want to solve for the number of years $n$ it takes until Harry has $H=60$ hats.

We can do this by substituting into the equation $H=8+4n$ and solving for $n,$ as follows:

\begin{align*} 60 &= 8 + 4n \\[3pt] 52 &= 4n \\[5pt] n &= \dfrac{52}{4} \\[5pt] n &= 13 \end{align*}

Therefore, it takes Harry $n=13$ years to collect $60$ hats.