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Algebra 1 View details

4. Rewriting and Solving Literal Equations

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Chapter 1

4.

Rewriting and Solving Literal Equations

This lesson delves into the realm of literal equations, emphasizing their significance in mathematical and real-world scenarios. Literal equations, primarily composed of variables, are pivotal in expressing relationships between different quantities. For instance, the area of a triangle or the volume of a cylinder can be represented using these equations. The lesson also highlights the process of isolating variables, a crucial step in solving these equations. Techniques such as the Properties of Equality and inverse operations play a vital role in this process. Real-life examples, like calculating the time taken for a truck to cover a certain distance or determining the height of a cylindrical object, further elucidate the practical applications of these equations. The lesson not only provides theoretical knowledge but also challenges learners with exercises to test their understanding.

Lesson Settings & Tools

	10 Theory slides
	13 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Rewriting and Solving Literal Equations

Slide of 10

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 $c .$

Triangle ABC

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

$LHS - 3 = RHS - 3$

y + 3 - 3 = 18 - 3

Simplify left-hand side

y + 3 - 3 = 18 - 3

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

$y = 15$

15 + 3 = ? 18

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

$LHS - 11 = RHS - 11$

t + 11 - 11 = 20 - 11

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

$LHS - 7 = RHS - 7$

7 - t - 7 = 16 - 7

CommutativePropAdd

Commutative Property of Addition

7 - 7 - t = 16 - 7

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

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

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

▼

Solve for

t

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

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

MoveNegDenomToFrac

Put minus sign in front of fraction

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

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

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

$LHS - 7 = RHS - 7$

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

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

$LHS \cdot 3 = RHS \cdot 3$

\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

$LHS + 40 = RHS + 40$

5 + 40 = 5 t - 40 + 40

Add terms

45 = 5 t

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

45 = 5 t

$LHS / 5 = RHS / 5$

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

Calculate quotient

9 = \frac{5 t}{5}

MovePartNumRight

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

9 = \frac{5}{5} t

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

9 = 1 t

Identity Property of Multiplication

9 = t

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

b

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

c

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

$C = 9667$

K = 9667 + 273

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

$LHS - 273 = RHS - 273$

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

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

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

$LHS + 32 = RHS + 32$

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

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

$K = 3500$

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

▼

Evaluate right-hand side

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

Calculate quotient

F = 5808.6 + 32

Add terms

F = 5840.6

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

$K = 5000$

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

▼

Evaluate right-hand side

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

Calculate quotient

F = 8508.6 + 32

Add terms

F = 8540.6

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.

Blue candle

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

$LHS / π r^{2} = RHS / π r^{2}$

\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}}

Simplify quotient

\frac{V}{π r ^{2}} = h

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}}

$V = 170$ , $r = 3$

h = \frac{1 7 0}{π ( 3 ^{2} )}

▼

Evaluate right-hand side

Calculate power

h = \frac{1 7 0}{π ( 9 )}

Use a calculator

h = 6.012520 \dots

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

$LHS / v = RHS / v$

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

CancelCommonFac

Cancel out common factors

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

Simplify quotient

\frac{d}{v} = t

Rearrange equation

t = \frac{d}{v}

Now, substitute

d = 150

and

v = 2.5 .

t = \frac{d}{v}

$d = 150$ , $v = 2.5$

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

Calculate quotient

t = 60

Rewriting and Solving Literal Equations

Exercises

A formula is given by $S = B (1 + p) .$

Isolate

B .

Isolate

p .

To isolate B, we will divide both sides of the formula by (1+p).

We will start by dividing both sides of the given formula by B. Then, we will use the inverse operations to isolate p.

The ideal gas law is often written in an empirical form.

P V = n R T

In this equation,

P

is the pressure,

V

is the volume of a gas,

n

is the amount of substance,

R

is the general gas constant, and

T

is the temperature — measured in Kelvin.

Which expression is obtained when the formula is solved for

R ?

A . \frac{n T}{P V} B . \frac{n P V}{T} C . \frac{P V}{n T}

Which expression is obtained when the formula is solved for

V ?

A . \frac{n R T}{P} B . P n R T C . \frac{P}{n R T}

Isolating R means that it should stand alone on one side of the equation. We can write the law in that form by dividing both sides by nT.

Therefore, R is equal to PVnT. This corresponds to C.

This time we will divide the formula by P to isolate V.

The volume V can be expressed as nRTP. This corresponds to A.

The formula $V = π r^{2} h$ is used to calculate the volume of a cylinder with radius $r$ and height $h .$

Solve the formula for

h .

Calculate the height of a cylindrical Coca-Cola jar with a radius of

3.1

centimeters and a volume of

330

cubic centimeters. Round to the answer to the nearest integer.

We need to solve the formula for h. Since h is multiplied by both π and r^2, we will divide the equation by π r^2.

The formula for h thus becomes h = Vπ r^2.

We will use the formula from the previous part and substitute the given values.

The Coca-Cola jar has a height about 11 centimeters.

Solve the following formulas for $b .$

a + b = 17

b - g = t

7 b = p

\frac{b}{y} = x

In the given equation, a is being added to b. Since the inverse operation of addition is subtraction, we can isolate the variable b by applying the Subtraction Property of Equality.

In this case, we have b and another variable g on the left-hand side of the equation. We can get rid of g by adding g to both sides.

Therefore, b is equal to the sum of t and g.

On the left side, b is multiplied by 7. To get rid of 7, we need to divide the equation by 7.

Now we have a y in the denominator of the fraction by. To isolate b, we multiply both sides of the equation by y.

An athletic facility is building an indoor track. The track is composed of three parts, a rectangle and two semicircles.

athletics track

Write a formula for the perimeter of the indoor track.

Solve the formula for

x .

The perimeter of the track is

562

meters and

r

is

45

meters. Find the value of

x ?

Round the value to the nearest meter.

The perimeter of the given figure is made up of two sides of a rectangle and two semicircles. The two sides each have a length of x, and the two semicircles make up a circle with a radius r. To find their length, let's substitute the radius into the formula for the circumference of a circle. C = 2 π r Now that we have the circumference of the circle, we can write the perimeter P of the entire figure, which is a sum of all the sides. P= x + x+ 2 π r Let's simplify the expression by adding like terms. P=x + x + 2 π r ⇕ P=2x+2 π r

In order to solve the formula for x, we need to isolate x on one of the sides.

The perimeter is given as 562 meters and the radius of the half circles is given as 45 meters. We can find x by substituting these values into the formula that we found in Part B.

When rounded to the nearest integer, the straight path x is about 140 meters long.

A truck travels on a highway at a rate of

108

kilometers per hour. How long does it take the truck to travel

162

kilometers?

Let's begin with recalling the formula for the distance traveled. d=r t In this formula d represents the distance, r is the rate, and t stands for time. In our exercise we are given that the truck travels 108 kilometers per hour and we want to determine the time the truck needs to travel 25 miles. Let's first isolate t. d=r t ⇒ t = d/r We can now substitute 108 for r and 162 for d into the formula.

The truck needs 1.5 hours to travel 162 kilometers. To convert it into minutes, we will multiply the time by the conversion factor 60 min1 h.

The truck travels 162 kilometers in 90 minutes.

Rewriting and Solving Literal Equations

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