3.3 Solving Literal Equations and Formulas
Introduction
If you were to stop a random person on the street and ask them to give you a fact from math, there's a good chance they'd give you an equation with lots of letters in it that they memorized in school, like \(a^2+b^2=c^2\).
Equations like this, that have more than one variable, are called literal equations. Literal equations that expresses well-defined rules in specific contexts, like geometry, physics, or finance, are called formulas.
From adjusting recipes and paying taxes to predicting planetary orbits and biochemical reactions, formulas are versatile tools used in a wide range of fields. In this section, we'll explore some of these practical formulas and learn how to work with and solve literal equations.
Solving Literal Equations
In this section, we'll take the strategies and rules we've already learned about solving equations and transfer them over to solving formulas and other literal equations.
Formulas are often written with one variable isolated to make the calculations straightforward. For instance, the classic formula for calculating distance traveled \(d\) using rate \(r\) and time \(t\) is
\[ d = rt \]This formula quickly gives the distance for any rate and time.
But what if we needed to find \(t\)? If we knew that \(d=100\) and \(r=50\), for instance, then solving \(100=50t\) once would be manageable.
\begin{align} d &= r t && \small \color{#5fa2ce} {\text{Original formula}} \\ 100 &= 50t && \small \color{#5fa2ce} {\text{Substitute} d=100 \text{ and }r=50} \\ \frac{100}{50} &= \frac{50t}{50} && \small \color{#5fa2ce} {\text{Divide to isolate }t} \\ 2 &= t && \small \color{#5fa2ce} {\text{Simplify}} \end{align}But if we needed to find \(t\) for several different values it would be more convenient if we had a formula for that specific purpose. To create a formula like that we can take the distance formula and solve for \(t\) without putting in any values for \(d\) and \(r\). We follow the same steps as in the specific case above, dividing to isolate \(t\) and simplifying.
\begin{align} d &= r t && \small \color{#5fa2ce} {\text{Original formula}} \\ \frac{d}{r} &= \frac{rt}{r} && \small \color{#5fa2ce} {\text{Divide to isolate }t} \\ \frac{d}{r} &= t && \small \color{#5fa2ce} {\text{Simplify}} \\ t &= \frac{d}{r} && \small \color{#5fa2ce} {\text{Final answer}} \end{align}As we can see from this example, solving formulas with generic values transforms specific cases into universal rules. Unlike single-variable equations where the answer is a number, solutions for formulas usually don't simplify to a single value. Isolating a variable in a formula creates a new formula!
And since we already know how to use inverse operations to solve, the primary challenge with formulas lies in maintaining clarity and organization.
Solve the literal equation \(ax + b = c\) for \(x\)
Solution \begin{align} ax + b &= c && \small \color{#5fa2ce} {\text{Original equation}} \\ ax &= c - b && \small \color{#5fa2ce} {\text{Subtract }b} \\ x &= \frac{c-b}{a} && \small \color{#5fa2ce} {\text{Divide by }a} \end{align}
This solution cannot be simplified further since \(a\), \(b\), and \(c\) are distinct variables that cannot be combined.
Practical Examples
The formula \(I = Prt\) gives the amount of simple interest produced by a prinicipal balance \(P\) earning an interest rate of \(r\) for \(t\) years. Solve the formula for \(P\)
Solution \begin{align} I &= Prt && \small \color{#5fa2ce} {\text{Original formula}} \\ \frac{I}{rt} &= P && \small \color{#5fa2ce} {\text{Divide by }rt} \\ P &= \frac{I}{rt} && \small \color{#5fa2ce} {\text{New formula}} \end{align}
The perimeter of a rectangle is given by the formula \(P = 2L + 2W\) where \(L\) is the length and \(W\) is the width. Solve the formula for the width \(W\).
Solution \begin{align} P &= 2L + 2W && \small \color{#5fa2ce} {\text{Original formula}} \\ P - 2L &= 2W && \small \color{#5fa2ce} {\text{Subtract }2L} \\ \frac{P-2L}{2} &= W && \small \color{#5fa2ce} {\text{Divide by }2} \\ W &= \frac{P-2L}{2} && \small \color{#5fa2ce} {\text{New formula}} \end{align}
Newtons's Second Law of Motion states that force \(F\) is equal to the product of mass \(m\) and acceleration \(a\), i.e., \(F = ma\). Solve this formula for \(m\).
Solution \begin{align} F &= ma && \small \color{#5fa2ce} {\text{Original formula}} \\ \frac{F}{a} &= m && \small \color{#5fa2ce} {\text{Divide by }a} \\ m &= \frac{F}{a} && \small \color{#5fa2ce} {\text{New formula}} \end{align}
The formula for calculating the area \(A\) of a triangle is \(A = \frac{1}{2}bh\), where \(b\) is the base length and \(h\) is the height. Solve this formula for \(h\).
Solution \begin{align} A &= \frac{1}{2}bh && \small \color{#5fa2ce} {\text{Original formula}} \\ 2A &= bh && \small \color{#5fa2ce} {\text{Multiply by }2} \\ \frac{2A}{b} &= h && \small \color{#5fa2ce} {\text{Divide by }b} \\ h &= \frac{2A}{b} && \small \color{#5fa2ce} {\text{New formula}} \end{align}
The formula for converting temperature from degrees Celsius \(C\) to degrees Fahrenheit \(F\) is \(F = \frac{9}{5}C + 32\). Solve this formula for \(C\).
Solution \begin{align} F &= \frac{9}{5}C + 32 && \small \color{#5fa2ce} {\text{Original formula}} \\ F - 32 &= \frac{9}{5}C && \small \color{#5fa2ce} {\text{Subtract }32} \\ \frac{5}{9}(F - 32) &= C && \small \color{#5fa2ce} {\text{Divide by }\frac{9}{5}\text{ or multiply by }\frac{5}{9}} \\ C &= \frac{5}{9}(F - 32) && \small \color{#5fa2ce} {\text{New formula}} \end{align}
The Pythagorean Theorem states that in a right triangle, the square of the length of the hypotenuse \(c\) is equal to the sum of the squares of the other two sides \(a\) and \(b\), i.e., \(c^2 = a^2 + b^2\). Solve this formula for \(a\).
Solution \begin{align} c^2 &= a^2 + b^2 && \small \color{#5fa2ce} {\text{Original formula}} \\ c^2 - b^2 &= a^2 && \small \color{#5fa2ce} {\text{Subtract }b^2} \\ \sqrt{c^2 - b^2} &= a && \small \color{#5fa2ce} {\text{Take the square root}} \\ a &= \sqrt{c^2 - b^2} && \small \color{#5fa2ce} {\text{New formula}} \end{align}
Normally when using a square root to solve we would take the \(\pm\) root. In this case, we are solving for the side of a triangle which must be positive.
The formula for calculating the volume \(V\) of a cylinder is \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height. Solve this formula for \(r\).
Solution \begin{align} V &= \pi r^2 h && \small \color{#5fa2ce} {\text{Original formula}} \\ \frac{V}{\pi h} &= r^2 && \small \color{#5fa2ce} {\text{Divide by }\pi h} \\ \sqrt{\frac{V}{\pi h}} &= r && \small \color{#5fa2ce} {\text{Take the square root}} \\ r &= \sqrt{\frac{V}{\pi h}} && \small \color{#5fa2ce} {\text{New formula}} \end{align}
Conclusion
Later in Chapter 5 we will explore graphing equations containing variables \( x\) and \( y\). Having equations solved for \( y\) often makes them easier to graph by hand or input into a graphing calculator.