Discover How to Solve Literal Equations: Key Techniques

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.

Challenge

Rewriting the Formula for the Perimeter of a Triangle

Consider $△ A B C$ with side lengths $a,$ $b,$ and $c .$

a Write an equation for the perimeter

P

of the triangle.

b If

P,

a,

and

b

are known, write an equation by solving the formula for

c .

Challenge

Finding the Amount of Time Spent Jogging

Maya plans to jog $150$ meters at a constant speed of $2.5$ meters per second.

The distance traveled

d

is found using the following equation.

d = v \cdot t

In this equation,

v

is the speed and

t

is the time.

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

t

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.

y + 3 = 18

To solve this equation,

3

must be subtracted from both sides.

y + 3 = 18

SubEqn

$LHS - 3 = RHS - 3$

y + 3 - 3 = 18 - 3

Simplify left-hand side

y + 3 - 3 = 18 - 3

SubTerms

Subtract terms

y = 15

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

15

is the solution to all these equations.

I . II . III . y + 3 = 18 y + 3 - 3 = 18 - 3 y = 15

The obtained solution can be checked by substituting

y = 15

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

y + 3 = 18

Substitute

$y = 15$

15 + 3 = ? 18

AddTerms

Add terms

18 = 18 ✓

A true statement was obtained. Therefore,

y = 15

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.

Equations I : II : III : IV : t + 11 = 20 7 - t = 16 \frac{t}{3} + 7 = 10 5 = 5 t - 40

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 $t .$

Solution

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

Equation I

In the first equation,

11

is being added to

t .

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

t + 11 = 20

SubEqn

$LHS - 11 = RHS - 11$

t + 11 - 11 = 20 - 11

SubTerms

Subtract terms

t = 9

The solution to the first equation is

t = 9 .

Equation II

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

7 - t = 16

SubEqn

$LHS - 7 = RHS - 7$

7 - t - 7 = 16 - 7

CommutativePropAdd

Commutative Property of Addition

7 - 7 - t = 16 - 7

SubTerms

Subtract terms

- t = 9

To remove the negative sign from the variable

t,

each side of the equation will be divided by

- 1 .

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

- t = 9

DivEqn

$LHS / (- 1) = RHS / (- 1)$

\frac{- t}{- 1} = \frac{9}{- 1}

▼

Solve for

t

DivNegNeg

$\frac{- a}{- b} = \frac{a}{b}$

\frac{t}{1} = \frac{9}{- 1}

MoveNegDenomToFrac

Put minus sign in front of fraction

\frac{t}{1} = - \frac{9}{1}

DivByOne

$\frac{a}{1} = a$

t = - 9

The solution to Equation II is

t = - 9 .

Equation III

To solve Equation III for

t,

7

should be subtracted from both sides of the equation.

\frac{t}{3} + 7 = 10

SubEqn

$LHS - 7 = RHS - 7$

\frac{t}{3} + 7 - 7 = 10 - 7

SubTerms

Subtract terms

\frac{t}{3} = 3

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

3 .

To do so, apply the Multiplication Property of Equality.

\frac{t}{3} = 3

MultEqn

$LHS \cdot 3 = RHS \cdot 3$

\frac{t}{3} \cdot 3 = 3 \cdot 3

FracMultDenomToNumber

$\frac{a}{3} \cdot 3 = a$

t = 3 \cdot 3

Multiply

t = 9

The solution to Equation III is

t = 9,

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

Equation IV

In this equation,

40

is being subtracted from

5 t .

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

5 = 5 t - 40

AddEqn

$LHS + 40 = RHS + 40$

5 + 40 = 5 t - 40 + 40

AddTerms

Add terms

45 = 5 t

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

45 = 5 t

DivEqn

$LHS / 5 = RHS / 5$

\frac{4 5}{5} = \frac{5 t}{5}

CalcQuot

Calculate quotient

9 = \frac{5 t}{5}

MovePartNumRight

$\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

9 = \frac{5}{5} t

QuotOne

$\frac{a}{a} = 1$

9 = 1 t

IdPropMult

Identity Property of Multiplication

9 = t

RearrangeEqn

Rearrange equation

t = 9

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

Equaivalent Equations I, III, and IV Solution 9

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.

A = \frac{1}{2} b h

Here,

b

represents the length of the base and

h

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	$\geq 30000$	blue
B	$10000 - 30000$	blue white
A	$7500 - 10000$	white
F	$6000 - 7500$	yellow white
G	$5000 - 6000$	yellow
K	$3500 - 5000$	light orange
M	$2000 - 3500$	orange red

a To convert temperatures in degrees Celsius

C

to temperatures in Kelvin

K,

the following literal equation is used.

K = C + 273

If Sirius has a surface temperature of

966 7^{\circ} C,

what color does it appear?

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

K

and temperatures in degrees Fahrenheit

F .

K = \frac{5}{9} (F - 32) + 273

Solve the formula for

F .

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

F = \frac{9}{5} (K - 273) + 32

5840 - 8540

Hint

a Substitute the value into the given equation.

b Use the Properties of Equality and inverse operations to isolate

F .

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,

9667

will be substituted into the given formula.

K = C + 273

Substitute

$C = 9667$

K = 9667 + 273

AddTerms

Add terms

K = 9940

The surface temperature of Sirius is about

9940 K .

Since its temperature is between

7500 K

and

10000 K,

it appears as a white star in the night sky.

b In the given formula,

F

is the variable of interest. To isolate

F,

inverse operations will be used.

K = \frac{5}{9} (F - 32) + 273

SubEqn

$LHS - 273 = RHS - 273$

K - 273 = \frac{5}{9} (F - 32)

MultEqn

$LHS \cdot \frac{9}{5} = RHS \cdot \frac{9}{5}$

\frac{9}{5} (K - 273) = F - 32

AddEqn

$LHS + 32 = RHS + 32$

\frac{9}{5} (K - 273) + 32 = F

RearrangeEqn

Rearrange equation

F = \frac{9}{5} (K - 273) + 32

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

3500

and

5000

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

3500 .

F = \frac{9}{5} (K - 273) + 32

Substitute

$K = 3500$

F = \frac{9}{5} (3500 - 273) + 32

▼

Evaluate right-hand side

SubTerm

Subtract term

F = \frac{9}{5} (3227) + 32

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

F = \frac{2 9 0 4 3}{5} + 32

CalcQuot

Calculate quotient

F = 5808.6 + 32

AddTerms

Add terms

F = 5840.6

RoundInt

Round to nearest integer

F \approx 5841

Therefore,

3500

Kelvin is approximately equivalent to

584 1^{\circ}

Fahrenheit. Next, the maximum value

5000

will be substituted.

F = \frac{9}{5} (K - 273) + 32

Substitute

$K = 5000$

F = \frac{9}{5} (5000 - 273) + 32

▼

Evaluate right-hand side

SubTerm

Subtract term

F = \frac{9}{5} (4727) + 32

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

F = \frac{4 2 5 4 3}{5} + 32

CalcQuot

Calculate quotient

F = 8508.6 + 32

AddTerms

Add terms

F = 8540.6

RoundInt

Round to nearest integer

F \approx 8541

This means that

3500

Kelvin is approximately

854 1^{\circ}

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

\underline{Range (^{\circ} F)} 5840 - 8540

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 $0 .$

Example

Rewriting a Formula for Volume

On Sunday Dominika spends time on her hobby, which is candle making. She uses $170$ cubic centimeters of wax to produce a cylindrical candle of radius $3$ centimeters.

The volume of a cylinder with radius

r

and height

h

is given by the following formula.

V = π r^{2} h

a Solve the formula for

h .

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,

h

is multiplied by

π r^{2} .

V = π r^{2} h

To isolate

h,

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

V = π r^{2} h

DivEqn

$LHS / π r^{2} = RHS / π r^{2}$

\frac{V}{π r ^{2}} = \frac{π r ^{2} h}{π r ^{2}}

CancelCommonFac

Cancel out common factors

\frac{V}{π r ^{2}} = \frac{π r ^{2} h}{π r ^{2}}

SimpQuot

Simplify quotient

\frac{V}{π r ^{2}} = h

RearrangeEqn

Rearrange equation

h = \frac{V}{π r ^{2}}

b It is known that

170

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

3

centimeters. Therefore,

V = 170

and

r = 3

will be substituted into the derived formula.

h = \frac{V}{π r ^{2}}

SubstituteII

$V = 170$ , $r = 3$

h = \frac{1 7 0}{π ( 3 ^{2} )}

▼

Evaluate right-hand side

CalcPow

Calculate power

h = \frac{1 7 0}{π ( 9 )}

UseCalc

Use a calculator

h = 6.012520 \dots

RoundInt

Round to nearest integer

h \approx 6

The height of the candle is about

6

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

t .

d = v \cdot t

It is known that Maya will run

150

meters at a speed of

2.5

meters per second.

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

t -

values until a true statement is obtained.

b Find a different way to solve the question.

Answer

a Table:

Time (s)	Distance Traveled (m)
$10$	$25$
$20$	$50$
$30$	$75$
$40$	$100$
$50$	$125$
$60$	$150$

It takes Maya $60$ seconds to run $150$ meters.

b See solution.

Hint

a Use multiples of

10

for

t

to make a table.

b Solve the formula for

t

and substitute the given values.

Solution

a It is known that Maya's speed is

2.5

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

t -

values, the distance traveled is shown in the table.

Time (s)	Speed $\times$ Time	Distance Traveled (m)
$10$	$2.5 (10) = 25$	$25$
$20$	$2.5 (20) = 50$	$50$
$30$	$2.5 (30) = 75$	$75$
$40$	$2.5 (40) = 100$	$100$
$50$	$2.5 (50) = 125$	$125$
$60$	$2.5 (60) = 150$	$150$

Maya will run $150$ meters in $60$ seconds at a speed of $2.5$ meters per second.

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

t

and then substituting in the given values.

d = v \cdot t

▼

Solve for

t

DivEqn

$LHS / v = RHS / v$

\frac{d}{v} = \frac{v \cdot t}{v}

CancelCommonFac

Cancel out common factors

\frac{d}{v} = \frac{v \cdot t}{v}

SimpQuot

Simplify quotient

\frac{d}{v} = t

RearrangeEqn

Rearrange equation

t = \frac{d}{v}

Now, substitute

d = 150

and

v = 2.5 .

t = \frac{d}{v}

SubstituteII

$d = 150$ , $v = 2.5$

t = \frac{1 5 0}{2 . 5}

CalcQuot

Calculate quotient

t = 60

Rewriting and Solving Literal Equations

Catch-Up and Review

Rewriting the Formula for the Perimeter of a Triangle

Finding the Amount of Time Spent Jogging

Equivalent Equations

Determining Equivalent Equations

Hint

Solution

Equation I

Equation II

Equation III

Equation IV

Literal Equation

Solving Literal Equations

Answer

Hint

Solution

Practice Solving Literal Equations

Rewriting a Formula for Volume

Hint

Solution

Finding the Amount of Time Spent Jogging

Answer

Hint

Solution

Rewriting and Solving Literal Equations

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