(8.5.1) – Solve a rational formula for a specified variable

Rational formulas can be useful tools for representing real-life situations and for finding answers to real problems. Equations representing direct, inverse, and joint variation are examples of rational formulas that can model many real-life situations. As you will see, if you can find a formula, you can usually make sense of a situation.

When solving problems using rational formulas, it is often helpful to first solve the formula for the specified variable. For example, work problems ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, [latex]W=rt[/latex]. The amount of work done [latex]W[/latex] is the product of the rate of work [latex]r[/latex] and the time spent working [latex]t[/latex]. Using algebra, you can write the work formula 3 ways:

[latex]W=rt[/latex]

Find the time [latex]t[/latex]:

[latex]\displaystyle t=\frac{W}{r}[/latex] (divide both sides by [latex]r[/latex])

Find the rate [latex]r[/latex]:

[latex]\displaystyle r=\frac{W}{t}[/latex](divide both sides by [latex]t[/latex])

Now let’s look at an example using the formula for the volume of a cylinder.

In the following video we give another example of solving for a variable in a formula, or as they are also called, a literal equation.

(8.5.2) – Solving applications of rational equations

Average cost problems

Suppose we know that the cost of making a product is dependent on the number of items, [latex]x[/latex], produced. This is given by the equation [latex]C\left(x\right)=15,000+500x[/latex]. Note, the cost function consists of a fixed cost ([latex]\$15,000[/latex]) and a variable cost [latex](500x)[/latex]. The fixed cost doesn’t change when more items are produced, whereas the variable cost increases as more cars are produced.  If we want to know the average cost for producing [latex]x[/latex] items, we would divide the cost function by the number of items, [latex]x[/latex].

The average cost function, which yields the average cost per item for [latex]x[/latex] items produced, is

[latex]\displaystyle \overline{C}\left(x\right)=\frac{15,000 +500x}{x}[/latex]

Many other application problems require finding an average value in a similar way, giving us variables in the denominator.

Work problems

Rational equations can be used to solve a variety of problems that involve rates, times and work. Using rational expressions and equations can help you answer questions about how to combine workers or machines to complete a job on schedule.

A Good Day’s Work

A “work problem” is an example of a real life situation that can be modeled and solved using a rational equation. Work problems often ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, [latex]W=rt[/latex]. (Notice that the work formula is very similar to the relationship between distance, rate, and time, or [latex]d=rt[/latex].) The amount of work done [latex]W[/latex] is the product of the rate of work [latex]r[/latex] and the time spent working [latex]t[/latex]. The work formula has 3 versions.

[latex]\large \begin{array}{l}W=rt\\\\\,\,\,\,\,t=\frac{W}{r}\\\\\,\,\,\,\,r=\frac{W}{t}\end{array}[/latex]

Some work problems include multiple machines or people working on a project together for the same amount of time but at different rates. In that case, you can add their individual work rates together to get a total work rate. Let’s look at an example.

Other work problems go the other way. You can calculate how long it will take one person to do a job alone when you know how long it takes people working together to complete the job.

In the video that follows, we show another example of finding one person’s work rate given a combined work rate.

As shown above, many work problems can be represented by the equation [latex]\displaystyle \frac{t}{a}+\frac{t}{b}=1[/latex], where [latex]t[/latex] is the time to do the job together, [latex]a[/latex] is the time it takes person A to do the job, and [latex]b[/latex] is the time it takes person B to do the job. The 1 refers to the total work done—in this case, the work was to paint 1 house.

The key idea here is to figure out each worker’s individual rate of work. Then, once those rates are identified, add them together, multiply by the time [latex]t[/latex], set it equal to the amount of work done, and solve the rational equation. Here is a summary:

We present another example of two people painting at different rates in the following video.

Motion problems

We have solved uniform motion problems using the formula [latex]D = rt[/latex] in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

[latex]\begin{array}{|c|c|c|c|} \hline & \,\text{rate}\,&\,\text{time}\,&\,\text{distance}\\ \hline \,\text{First}\,&\,&\,&\\ \hline \,\text{Second}\,&\,&\,&\\ \hline \end{array} [/latex]

The formula [latex]D=rt[/latex] assumes we know [latex]r[/latex] and  [latex]t[/latex] and use them to find  [latex]D[/latex]. If we know  [latex]D[/latex] and [latex]r[/latex] and need to find  [latex]t[/latex], we would solve the equation for  [latex]t[/latex] and get the formula [latex]\displaystyle t=\frac{D}{r}[/latex].

Concentration of a mixture problems

Mixtures are made of ratios of different substances that may include chemicals, foods, water, or gases. There are many different situations where mixtures may occur both in nature and as a means to produce a desired product or outcome.  For example, chemical spills, manufacturing and even biochemical reactions involve mixtures.  The thing that can make mixtures interesting mathematically is when components of the mixture are added at different rates and concentrations. In our last example we will define an equation that models the concentration  – or ratio of sugar to water – in a large mixing tank over time. You are asked whether the final concentration of sugar is greater than the concentration at the beginning.

In the following video, we show another example of how to use rational functions to model mixing.