We now turn back to solving word problems. In this topic, the first goal is to build an equation that has rational expressions in it, and then use the techniques from the last topic to find appropriate solutions.
Literal equations are formulas that have multiple variables, the area of a trapezoid for example: $$A=\frac{1}{2}h(b_1+b_2)$$ where the \(h\) is the height, and the \(b_1,\;b_2\) are the lengths of the two parallel bases. Normally, we plug in information to this formula and calculate the area of the trapezoid shape. What if we knew the area, and needed to find one of the base lengths, like \(b_2\)? Solving for a different variable to create a new formula is the goal of literal equations.
In physics, the focal length, \(f\), of a camera lens is given by the formula: $$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$$ where \(p\) is the distance from the object to the lens and \(q\) is the distance from the lens to the image. Rewrite the formula by solving for \(p\).
| Get a CD: | \(\implies\) CD = \(fpq\) |
| Use shortcut: | \(\implies pq=fq+fp\) |
| Solve for \(p\): | \(\begin{array}{l} \implies pq-fp=fq \\ \implies p\left(q-f\right)=fq \\ \implies p=\large{\frac{fq}{q-f}}\;\normalsize{\color{green}{\checkmark}} \end{array}\) |
The next examples use the formula that \(d=rt\). A good technique to be successful here is to create a table from the information given in the problem to help create the equation.
At O'Hare International Airport in Chicago, Cheryl and Bill are walking at the same rate to the gate to catch their flight to Denver. Bill steps onto the moving sidewalk and continues to walk while Cheryl uses the stationary sidewalk. If the sidewalk moves at 1 meter per second and Bill saves 50 seconds covering the 300 meter distance, what is their walking speed?
Let \(x\) denote the walking speed (rate) for Cheryl. Since Bill is on the sidewalk moving at 1 mps, his rate is \(x+1\). Using \(d=rt\), solve for time by dividing by the rate: \(t=\frac{d}{r}\). Fill in the table with the information from the problem:
| Distance | Rate | Time | |
|---|---|---|---|
| Cheryl | \(300\) | \(x\) | \(\frac{300}{x}\) |
| Bill | \(300\) | \(x+1\) | \(\frac{300}{x+1}\) |
To build an equation, the time for Bill is 50 seconds faster than the time for Cheryl: $$\begin{array}{ccccc} \color{blue}{\underbrace{\text{Bill's time}}} & \text{is} & \color{green}{\underbrace{\text{Cheryl's time}}} & & \color{purple}{\underbrace{\text{less 50 secs}}} \\ \color{blue}{\large{\frac{300}{x+1}}} & \normalsize{=} & \color{green}{\large{\frac{300}{x}}} & \normalsize{-} & \normalsize{\color{purple}{50}} \end{array}$$ Now solve for \(x\):
| Get a CD: | \(\implies x(x+1)\) |
| Shortcut: | \(\begin{array}{l} \implies300\color{green}{x}=300\color{green}{(x+1)}-50\color{green}{x(x+1)} \\ \implies 300x=300x+300-50x^2-50x \\ \implies 50x^2+50x-300=0 \end{array}\) |
| Factor to solve: | \(\begin{array}{l} \implies50\left(x^2+x-6\right)=0 \\ \implies 50(x+3)(x-2)=0 \\ \implies x+3=0, \; x-2=0 \\ \implies x=-3, \; x=2\;\color{green}{\checkmark} \end{array}\) |
Since a rate cannot be negative, the only solution is \(x=2\), their walking speed is 2 mps.
A work problem depends on the rate \(r\) at which the work is being done. The time \(t\) for that amount \(A\) of work to be done is related by the equation: \(A=rt\) (Very similar to the distance, rate & time formula!). Our problems a job to be completed, so \(A=1\). This give an equation for the rate of work one person does on any job:
$$r=\frac{1}{t}$$Stan needs 45 minutes to clean all the dishes. Deb can do them in 30 minutes. How long would it take to clean all the dishes if they work together?
Let \(x\) denote the time they work together. The table summarizes the information from the problem:
| Rate | Time Working Together |
Part of Job Completed |
|
|---|---|---|---|
| Stan | \(\large{\frac{1}{45}}\) | \(x\) | \(\large{\frac{x}{45}}\) |
| Deb | \(\large{\frac{1}{30}}\) | \(x\) | \(\large{\frac{x}{30}}\) |
To build an equation, take the fractional part for each person and add them together to complete the job: $$\begin{array}{ccccc} \color{orange}{\underbrace{\text{Stan's part}}} & \text{plus} & \color{brown}{\underbrace{\text{Deb's part}}} & \text{makes} & \color{red}{\underbrace{\text{job complete}}} \\ \color{orange}{\large{\frac{x}{45}}} & \normalsize{+} & \color{brown}{\large{\frac{x}{30}}} & = & \color{red}{1} \end{array}$$ Solve for \(x\): $$\text{CD}=90\implies 90\left(\frac{x}{45}+\frac{x}{30}=1\right)$$ $$2x+3x=90\implies 5x=90 \implies x=18\;\color{green}{\checkmark}$$ It will take them 18 minutes to clean all the dishes if they work together.