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When solving real-life problems, it is usually possible to write equations in more than one variable. Sometimes solving the equation for the variable of interest is needed. In this lesson, equations involving two or more variables will be solved for a particular variable.

Catch-Up and Review

Here is some recommended reading before getting started with this lesson.

Properties of Equality
Inverse Operations
Solving One-Step Equations in One Variable
Solving Multi-Step Equations in One Variable

Challenge

Rewriting the Formula for the Perimeter of a Triangle

Consider with side lengths and

a Write an equation for the perimeter

of the triangle.

b If

and

are known, write an equation by solving the formula for

Challenge

Finding the Amount of Time Spent Jogging

Maya plans to jog meters at a constant speed of meters per second.

The distance traveled

is found using the following equation.

In this equation,

is the speed and

is the time.

a Determine how long it takes Maya to jog this distance by trying some values for

until a true statement is obtained.

b Find a different way to solve the question.

Discussion

Equivalent Equations

Two equations are called equivalent equations if they have the same solution. Equations are often solved by applying the Properties of Equality. Each time a property is applied, an equivalent equation is produced. Consider the following equation.

To solve this equation,

must be subtracted from both sides.

Simplify left-hand side

Subtract terms

Here, two equations that are equivalent to the given equation were created as a result of applying the Subtraction Property of Equality. They are equivalent because

is the solution to all these equations.

The obtained solution can be checked by substituting

in the original equation. If a true statement results, the solution is correct.

Add terms

\(18=18\ {\color{#009600}{\bm{\Large{\checkmark }}}}\)

A true statement was obtained. Therefore,

is indeed a solution to the original equation.

Example

Determining Equivalent Equations

Dominika is finishing up her homework on Friday so she can enjoy the weekend. She is told that three of the following equations are equivalent equations.

Help Dominika identify the equation that is not equivalent to the other three.

Hint

Equivalent equations have the same solution(s). Use the Properties of Equality and inverse operations to solve for the variable

Solution

Each equation will be solved for and then the solutions will be compared.

Equation I

In the first equation,

is being added to

Since the inverse operation of addition is subtraction, applying the Subtraction Property of Equality would simplify the equation.

Subtract terms

The solution to the first equation is

Equation II

To isolate the variable, the Subtraction Property of Equality will be applied first.

CommutativePropAdd

Commutative Property of Addition

Subtract terms

To remove the negative sign from the variable

each side of the equation will be divided by

In other words, the Division Property of Equality will be applied.

▼

Solve for

MoveNegDenomToFrac

Put minus sign in front of fraction

The solution to Equation II is

Equation III

To solve Equation III for

should be subtracted from both sides of the equation.

Subtract terms

To remove the fraction, the equation needs to be multiplied by

To do so, apply the Multiplication Property of Equality.

FracMultDenomToNumber

Multiply

The solution to Equation III is

which is the same as the solution to Equation I. Therefore, the first and third equations are equivalent.

Equation IV

In this equation,

is being subtracted from

Since the inverse operation of subtraction is addition, applying the Addition Property of Equality will simplify the equation.

Add terms

By applying the Division Property of Equality, the equation can be further simplified.

Calculate quotient

MovePartNumRight

Identity Property of Multiplication

Rearrange equation

The last equation has the same solution as Equations I and III. This means that Equations I, III, and IV are equivalent equations.

Therefore, Equation II is not equivalent to any of those equations.

Discussion

Literal Equation

A literal equation is an equation that is comprised mostly or entirely of variables. The area of a triangle, for example, is expressed by the following literal equation.

Here,

represents the length of the base and

represents the height of the triangle. Therefore, formulas can be seen as literal equations. Using the Properties of Equality, a literal equation can be solved for a variable, which means isolating the variable on one side of the equation.

Example

Solving Literal Equations

While at the planetarium on Saturday, Dominika learns about Annie Jump Cannon, known as the census taker of the sky, who introduced the Harvard Spectral Classification. In its simplest form, this system classifies stars according to their surface temperatures.

Stellar Classification
Class	Temperature (K)	Apparent Color
O		blue
B		blue white
A		white
F		yellow white
G		yellow
K		light orange
M		orange red

a To convert temperatures in degrees Celsius

to temperatures in Kelvin

the following literal equation is used.

If Sirius has a surface temperature of

what color does it appear?

b The following literal equation shows the relationship between temperatures in Kelvin

and temperatures in degrees Fahrenheit

Solve the formula for

c Determine a range in degrees Fahrenheit for Pollux, which appears as a light orange point of light. If necessary, round the values to the nearest integer.

Answer

a White

b

c

Hint

a Substitute the value into the given equation.

b Use the Properties of Equality and inverse operations to isolate

c Substitute the minimum and maximum range values for a light orange star into the derived formula from Part B.

Solution

a To determine the apparent color of Sirius, its surface temperature should be converted into Kelvin. To do so,

will be substituted into the given formula.

Add terms

The surface temperature of Sirius is about

Since its temperature is between

and

it appears as a white star in the night sky.

b In the given formula,

is the variable of interest. To isolate

inverse operations will be used.

Rearrange equation

This formula can now be used to convert Kelvin to degrees Fahrenheit.

c It is known that Pollux appears as a light orange point of light. Therefore, its surface temperature is between

and

Kelvin. These values should be substituted into the derived formula from Part B. Start with minimum value

▼

Evaluate right-hand side

Subtract term

MoveRightFacToNum

Calculate quotient

Add terms

Round to nearest integer

Therefore,

Kelvin is approximately equivalent to

Fahrenheit. Next, the maximum value

will be substituted.

▼

Evaluate right-hand side

Subtract term

MoveRightFacToNum

Calculate quotient

Add terms

Round to nearest integer

This means that

Kelvin is approximately

Fahrenheit. Therefore, the range for a light orange star in degrees Fahrenheit is as follows.

Pop Quiz

Practice Solving Literal Equations

Use the Properties of Equality to solve the given literal equation for the indicated variable. Some equations may require dividing both sides of the equation by a variable. Since division by zero is not defined, for these equations assume that the variable is not equal to

Example

Rewriting a Formula for Volume

On Sunday Dominika spends time on her hobby, which is candle making. She uses cubic centimeters of wax to produce a cylindrical candle of radius centimeters.

The volume of a cylinder with radius

and height

is given by the following formula.

a Solve the formula for

b What is the height of the candle? Round the answer to the nearest integer.

Hint

a Use the Division Property of Equality.

b Substitute the given values into the derived formula.

Solution

a On the right-hand side of the formula,

is multiplied by

To isolate

the inverse operation of multiplication should be applied to both sides of the equation. This means applying the Division Property of Equality.

CancelCommonFac

Cancel out common factors

Simplify quotient

Rearrange equation

b It is known that

cubic centimeters of wax were used, and that the radius of the cylindrical candle is

centimeters. Therefore,

and

will be substituted into the derived formula.

,

▼

Evaluate right-hand side

Calculate power

Use a calculator

Round to nearest integer

The height of the candle is about

centimeters.

Closure

Finding the Amount of Time Spent Jogging

A literal equation shows the relationship of two or more variables. In some cases, a literal equation needs to be solved for a specific variable. Coming back to the challenge presented at the beginning of the lesson, the time Maya spent jogging can also be found by solving the given formula for

It is known that Maya will run

meters at a speed of

meters per second.

a Determine how long it takes Maya to jog this distance by trying some values for some

values until a true statement is obtained.

b Find a different way to solve the question.

Answer

a Table:

Time (s)	Distance Traveled (m)

It takes Maya seconds to run meters.

b See solution.

Hint

a Use multiples of

for

to make a table.

b Solve the formula for

and substitute the given values.

Solution

a It is known that Maya's speed is

meters per second. By the formula, the speed multiplied by the time equals the distance traveled. For different

values, the distance traveled is shown in the table.

Time (s)	Speed Time	Distance Traveled (m)

Maya will run meters in seconds at a speed of meters per second.

b The same result can be found by solving the formula for

and then substituting in the given values.

▼

Solve for

CancelCommonFac

Cancel out common factors

Simplify quotient

Rearrange equation

Now, substitute

and

,

Calculate quotient

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