Topics
1. Equations and Inequalities
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United States of America
Grades 9-12

1.05 Literal equations

Lesson
Practice

Introduction

Rearranging an equation to isolate one variable is called solving for the variable. As we will see in this lesson, the rearranged equations are all equivalent. Sometimes a rearranged equation provides a more efficient method for finding values, especially when repeated calculations are needed.

Literal equations

Literal equation

An equation that involves two or more variables.

Formula

A type of literal equation that describes a relationship between real-world quantities.

Example:

A = l\cdot w

where A is area,

l is length, and

w is width.

There are many different formulas in science, mathematics, business, and other subjects that allow us to measure quantities such as area, volume, speed, etc. We can use the properties of equality to isolate any variable in a literal equation or formula.

The same variable might be used to represent different quantities across different formulas. For example, in the formula for the area of a rectangle, w is used to represent the width of the rectangle. However, in another context, w might be used to represent a weight or other value. To avoid any confusion, formulas will always state what the variables represent.

It is also possible that the same letter will be used multiple times in a formula but one capitalized and one lower case. These represent different quantities and should be treated as completely different variables.

For example, sometimes the formula for the area of a trapezoid is written as A=\dfrac{1}{2}h\left(B+b\right) where capital B represents the length of one base of the trapezoid and lowercase b represents the length of the other base. These bases typically have different lengths and so we cannot combine them as if they are the same variable. Other times the formula is written as A=\dfrac{1}{2}h\left(b_1+b_2\right) in which case the subscripts indicate that they should be treated as two different variables.

Exploration

Consider the formula for distance:

\displaystyle d=rt
\bm{d}
distance
\bm{r}
rate
\bm{t}
time

Use the distance formula to solve each of the following:

  • The speed at which a person travels if they drive 100 \text{ mi} in 75 \text{ min}
  • The distance traveled when walking 5 \text{ km} in 30 \text{ min}
  • The distance traveled by a car driving 65 \text{ mph} for 5 \text{ hr}
  • The time it will take to travel 1\,000 \text{ km} if you walk 80 \text{ m/min}
  • How fast a plane is traveling if it can fly 2\,789 \text{ mi} in 6 \text{ hr}
  • How long it will take to walk to the store 2 \text{ mi} away if you walk 176 \text{ ft/min}

  1. Which ones took the most effort to solve?

  2. How could we reduce the effort when making repeated calculations for the same variable?

Using the division property of equality, we can rearrange the equation relating distance, rate, and time to be r=\dfrac{d}{t} or t=\dfrac{d}{r}.

By rearranging a formula for a variable of interest, we can reduce the number of repeated calculations needed, depending on which variable is unknown.

Examples

Example 1

Solve for x in the following equation:

y = 5 \left(1+\dfrac{x}{k}\right)

Worked Solution
Create a strategy

We need to rearrange the equation to isolate x. We can use the properties of equality and inverse operations to solve literal equations for a variable, just as we would for linear equations.

Apply the idea
\displaystyle y\displaystyle =\displaystyle 5 \left(1+\dfrac{x}{k}\right)Given equation
\displaystyle \dfrac{y}{5}\displaystyle =\displaystyle 1+\dfrac{x}{k}Division property of equality
\displaystyle \dfrac{y}{5} -1\displaystyle =\displaystyle \dfrac{x}{k}Subtraction property of equality
\displaystyle k\left(\dfrac{y}{5} -1\right)\displaystyle =\displaystyle xMultiplication property of equality
\displaystyle x\displaystyle =\displaystyle k\left(\dfrac{y}{5} -1\right)Symmetric property of equality
Reflect and check

Remember, when rearranging an equation, we reverse the operations acting on the variable we want to isolate, in the reverse order of operations. Whatever is done to one side of the equation, must be done to the other to keep the equation balanced.

Example 2

Ohm's law states:

V=IR

where V is voltage, I is current, and R is resistance.

Write the formula for current.

Worked Solution
Create a strategy

The formula for current is Ohm's law with I isolated. We use inverse operations and properties of equality to get the solution.

Apply the idea
\displaystyle V\displaystyle =\displaystyle IRGiven equation
\displaystyle \dfrac{V}{R}\displaystyle =\displaystyle IDivision property of equality
\displaystyle I\displaystyle =\displaystyle \dfrac{V}{R}Symmetric property of equality

Example 3

The perimeter of a semicircle is given by the formula:

P = 2r+ \pi r

where r is the radius of the semicircle.

a

Find the radius of a semicircle with a perimeter of 24.85 \text{ cm}. Round your answer to the nearest tenth.

Worked Solution
Create a strategy

We need to substitute 24.85 for P, then solve for r.

Apply the idea
\displaystyle P\displaystyle =\displaystyle 2r+ \pi rGiven equation
\displaystyle 24.85\displaystyle =\displaystyle 2r+\pi rSubstitute P=24.85
\displaystyle 24.85\displaystyle =\displaystyle (2+\pi)rDistributive property
\displaystyle \dfrac{24.85}{2+\pi}\displaystyle =\displaystyle rDivision property of equality
\displaystyle 4.8\displaystyle \approx\displaystyle rApproximate using a calculator

The radius of the semicircle is about 4.8\text{ cm}.

Reflect and check

Notice the variable we are trying to solve for was in more than one term. If these terms cannot be combined, we may be required to factor part of the expression to isolate the variable.

b

Write an equation to solve for r using the properties of equality.

Worked Solution
Create a strategy

We need to rearrange the equation to isolate r. We can use the properties of equality and inverse operations to get the solution.

Apply the idea
\displaystyle P\displaystyle =\displaystyle 2r+ \pi rGiven equation
\displaystyle P\displaystyle =\displaystyle r \left(2+ \pi\right)Distributive property
\displaystyle \dfrac{P}{\left(2+ \pi\right)}\displaystyle =\displaystyle rDivision property of equality
\displaystyle r\displaystyle =\displaystyle \dfrac{P}{\left(2+ \pi\right)}Symmetric property of equality
Reflect and check

These are the same steps we used in part (a) when given a specific value for P.

c

Use the equation from part (b) to find the radius of a semicircle with a perimeter of 37.27\text{ in}. Round your answer to the nearest tenth.

Worked Solution
Create a strategy

Since the formula is already solved for r, we only need to plug 37.27\text{ in} for P and use our calculators to simplify.

Apply the idea
\displaystyle r\displaystyle =\displaystyle \dfrac{P}{\left(2+ \pi\right)}Formula solved for r
\displaystyle r\displaystyle =\displaystyle \dfrac{37.27}{\left(2+ \pi\right)}Substitute P=37.27
\displaystyle r\displaystyle \approx\displaystyle 7.2Approximate using a calculator

The radius is about 7.2\text{ in}.

Reflect and check

In part (a), we solved for the radius after plugging in the perimeter. In part (c), we solved for the radius first, then plugged in the perimeter. Both methods use the same properties of equality. The second method is most useful when we need to solve multiple problems to find the same unknown value.

Idea summary

In the same way we solve one-variable equations, we can use inverse operations and the properties of equality to isolate a variable in a literal equation.

Outcomes

A.CED.A.4

Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations.

A.REI.B.3

Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

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