###### Example3.1.2

A water tank can hold \(140\) gallons of water, but it has only \(5\) gallons of water. A tap was turned on, pouring \(15\) gallons of water into the tank every minute. After how many minutes will the tank be full?

Let's find a pattern first.

Minutes since Tap | Amount of Water in |

Was Turned on | the Tank (in Gallons) |

\(0\) | \(5\) |

\(1\) | \(15\cdot1+5=20\) |

\(2\) | \(15\cdot2+5=35\) |

\(3\) | \(15\cdot3+5=50\) |

\(4\) | \(15\cdot4+5=65\) |

\(\vdots\) | \(\vdots\) |

\(m\) | \(15m+5\) |

We can see the tap can pour \(15m\) gallons of water into the tank in \(m\) minutes. The tank had \(5\) gallons of water in the beginning, so the amount of water in the tank can be modeled by \(15m+5\text{,}\) where \(m\) is the number of minutes since the tap was turned on. To find when the tank will be full (with \(140\) gallons of water), we can write the equation

\begin{equation*} 15m+5=140 \end{equation*}First, we need to isolate the variable term, \(15m\text{,}\) in the equation. In other words, we need to remove \(5\) from the left side of the equals sign. We can do this by subtracting \(5\) from both sides of the equation. Once the variable term is isolated, we can eliminate the coefficient and solve for \(m\text{.}\)

The full process appears as:

\begin{align*} 15m+5\amp=140\\ 15m+5\subtractright{5}\amp=140\subtractright{5}\\ 15m\amp=135\\ \divideunder{15m}{15}\amp=\divideunder{135}{15}\\ m\amp=9 \end{align*}Next, we need to substitute \(m\) with \(9\) in the equation \(15m+5=140\) to check the solution:

\begin{align*} 15m+5\amp=140\\ 15(\substitute{9})+5\amp\stackrel{?}{=}140\\ 135+5\amp\stackrel{?}{=}140\\ 140\amp\stackrel{\checkmark}{=}140 \end{align*}The solution \(9\) is checked.

In summary, the tank will be full after \(9\) minutes.