To start this section, we will use a scenario we have seen before in Example 13.1.4:
Scenario: Two towns are connected by a \(12\)-mile-long river, which flows from Town A to Town B at the velocity of \(2\) miles per hour. A boat will travel at a constant velocity, \(v\) miles per hour, from Town A to Town B, and then back to Town A. Due to the river's flow, the boat's actual speed is \(v+2\) miles per hour from Town A to Town B, and \(v-2\) miles per hour from Town B back to Town A. If the boat driver plans to spend \(8\) hours for the whole trip, how fast should he/she drive the boat?
The time it takes the boat to travel from Town A to Town B is \(\frac{12}{v+2}\text{,}\) and \(\frac{12}{v-2}\) from Town B to Town A. The function to model the whole trip's time is
Instead of using the function's graph, we will directly solve this equation. Please review the skill of Fraction Buster we learned in <<Unresolved xref, reference "subsection-fraction-buster"; check spelling or use "provisional" attribute>>. The same skill applies to rational functions. To remove fractions in this equation, we will multiply both sides of the equation by the common denominator \((v-2)(v+2)\text{,}\) and we have:
\begin{align*}
\frac{12}{v-2}+\frac{12}{v+2}\amp=8\\
\multiplyleft{(v+2)(v-2)}(\frac{12}{v-2}+\frac{12}{v+2})\amp=\multiplyleft{(v+2)(v-2)}8\\
(v+2)(v-2)\cdot\frac{12}{v-2}+(v+2)(v-2)\cdot\frac{12}{v+2}\amp=(v+2)(v-2)\cdot8\\
12(v+2)+12(v-2)\amp=8(v^2-4)\\
12v+24+12v-24\amp=8v^2-32\\
24v\amp=8v^2-32\\
0\amp=8v^2-24v-32\\
0\amp=8(v^2-3v-4)\\
0\amp=8(v-4)(v+1)\\
v-4=0\amp\text{ or }v+1=0\\
v=4\amp\text{ or }v=-1
\end{align*}
We could lose domain conditions when we remove fractional rational expressions, and thus we must check. In the original equation, we can see the domain conditions are \(v\ne2\) and \(v\ne-2\text{,}\) because those two values would cause a fraction to be undefined. Our solutions \(v=4\) and \(v=-1\) do not violate the domain conditions, so they are solutions to the equation.
In this context, the boat's speed shouldn't be negative, so we only take the solution \(v=4\text{.}\)
Solution: If the boat drives at \(4\) miles per hour, the whole trip would take \(8\) hours. This result matches the solution in Example 13.1.4.
Let's look at another application problems.
Example13.5.2
It takes Henry \(3\) hours to paint a room, and it takes Dennis \(6\) hours to paint the same room. If they work together, how long will it take them to paint the room?
It takes Henry \(3\) hours to paint the room. This implies Henry paints \(\frac{1}{3}\) of the room each hour. Similarly, Dennis paints \(\frac{1}{6}\) of the room each hour. If they work together, they paint \(\frac{1}{3}+\frac{1}{6}\) of the room.
Assume it takes \(x\) hours to paint the room if Henry and Dennis work together. This implies they paint \(\frac{1}{x}\) of the room when working together. Now we can write this equation:
In this equation, the domain conditions is \(x\ne0\text{.}\)
To remove all fractions, we multiply both sides of the equation by the common denominator of \(3\text{,}\) \(6\) and \(x\text{,}\) which is \(6x\text{:}\)
In this equation, we can see \(y\ne-1\) and \(y\ne0\text{.}\) The common denominator is \(y(y+1)\text{.}\) We will multiply both sides of the equation by \(y(y+1)\text{:}\)
In this equation, we can see \(z\ne4\text{.}\) The common denominator is \(z-4\text{.}\) We will multiply both sides of the equation by \(z-4\text{:}\)
\begin{align*}
z+\frac{1}{z-4}\amp=\frac{z-3}{z-4}\\
\multiplyleft{(z-4)}(z+\frac{1}{z-4})\amp=\multiplyleft{(z-4)}\frac{z-3}{z-4}\\
(z-4)\cdot z+(z-4)\cdot\frac{1}{z-4}\amp=z-3\\
z^2-4z+1\amp=z-3\\
z^2-5z+4\amp=0\\
(z-1)(z-4)\amp=0\\
z-1=0\amp\text{ or }z-4=0\\
z=1\amp\text{ or }z=4
\end{align*}
Since \(z=4\) violates the domain condition \(z\ne4\text{,}\) \(z=4\) is not a solution. The only solution of the equation is \(z=1\text{.}\)
Example13.5.5
Solve for \(p\) in \(\frac{3}{p-2}+\frac{5}{p+2}=\frac{12}{p^2-4}\)
Now we can see the common denominator is \((p+2)(p-2)\text{,}\) and the domain conditions are \(p\ne2\) and \(p\ne-2\text{.}\) We will multiply both sides of the equation by \((p+2)(p-2)\text{:}\)
Since \(p=2\) violates the domain condition \(p\ne2\text{,}\) this equation has no solution.
Example13.5.6
In physics, when two resistances, \(R_1\) and \(R_2\text{,}\) are connected in a parallel way, the combined resistance, \(R\text{,}\) can be calculated by the formula
In this equation, the domain condition is \(R\ne0\text{.}\) The common denominator is \(R R_1 R_2\text{.}\) We will multiply both sides of the equation by \(R R_1 R_2\text{:}\)
