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.
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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.
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Challenge
Rewriting the Formula for the Perimeter of a Triangle
Consider △ ABC with side lengths a, b, and c.
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b If P, a, and b are known, write an equation by solving the formula for c.
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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* 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.
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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
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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 [-1em] rl I: & t+11 =20 [0.5em] II: & 7-t = 16 [0.5em] III: & t/3+7 = 10 [0.8em] IV: & 5 = 5t-40 Help Dominika identify the equation that is not equivalent to the other three.
{"type":"choice","form":{"alts":["Equation I","Equation II","Equation III","Equation IV"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
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.
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).
- t/- 1= 9/- 1
▼
Solve for t
DivNegNeg
- a/- b=a/b
t/1= 9/- 1
MoveNegDenomToFrac
Put minus sign in front of fraction
t/1= - 9/1
DivByOne
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.
To remove the fraction, the equation needs to be multiplied by 3. To do so, apply the Multiplication Property of Equality.
t/3=3
MultEqn
LHS * 3=RHS* 3
t/3 * 3=3 * 3
FracMultDenomToNumber
a/3* 3 = a
t=3* 3
Multiply
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 5t. Since the inverse operation of subtraction is addition, applying the Addition Property of Equality will simplify the equation.
By applying the Division Property of Equality, the equation can be further simplified.
45 = 5t
DivEqn
.LHS /5.=.RHS /5.
45/5 = 5t/5
CalcQuot
Calculate quotient
9 = 5t/5
MovePartNumRight
a* b/c=a/c* b
9 = 5/5t
QuotOne
a/a=1
9 = 1t
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 & Solution I, III, and IV & 9 Therefore, Equation II is not equivalent to any of those equations.
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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 = 1/2bh
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.
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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 | ≥ 30 000 | blue |
| B | 10 000 - 30 000 | blue white |
| A | 7500 - 10 000 | 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 9667^(∘)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= 59(F-32)+273 Solve the formula for F.
Answer
a White
b F=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.
The surface temperature of Sirius is about 9940 K. Since its temperature is between 7500 K and 10 000 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=5/9(F-32)+273
SubEqn
LHS-273=RHS-273
K -273=5/9(F-32)
MultEqn
LHS * 9/5=RHS* 9/5
9/5(K -273)=F-32
AddEqn
LHS+32=RHS+32
9/5(K -273) + 32 =F
RearrangeEqn
Rearrange equation
F=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=9/5(K -273) + 32
Substitute
K= 3500
F=9/5( 3500 -273) + 32
▼
Evaluate right-hand side
SubTerm
Subtract term
F=9/5(3227) + 32
MoveRightFacToNum
a/c* b = a* b/c
F=29 043/5 + 32
CalcQuot
Calculate quotient
F=5808.6+32
AddTerms
Add terms
F=5840.6
RoundInt
Round to nearest integer
F ≈ 5841
Therefore, 3500 Kelvin is approximately equivalent to 5841^(∘) Fahrenheit. Next, the maximum value 5000 will be substituted.
F=9/5(K -273) + 32
Substitute
K= 5000
F=9/5( 5000 -273) + 32
▼
Evaluate right-hand side
SubTerm
Subtract term
F=9/5(4727) + 32
MoveRightFacToNum
a/c* b = a* b/c
F=42 543/5 + 32
CalcQuot
Calculate quotient
F=8508.6+32
AddTerms
Add terms
F=8540.6
RoundInt
Round to nearest integer
F ≈ 8541
This means that 3500 Kelvin is approximately 8541^(∘) Fahrenheit. Therefore, the range for a light orange star in degrees Fahrenheit is as follows. Range(^(∘)F) 5840 - 8540
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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.
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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.
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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^2h
DivEqn
.LHS /π r^2.=.RHS /π r^2.
V/π r^2=π r^2h/π r^2
CancelCommonFac
Cancel out common factors
V/π r^2=π r^2h/π r^2
SimpQuot
Simplify quotient
V/π r^2=h
RearrangeEqn
Rearrange equation
h = 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.
The height of the candle is about 6 centimeters.
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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 * 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 * 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* t
▼
Solve for t
DivEqn
.LHS /v.=.RHS /v.
d/v = v * t/v
CancelCommonFac
Cancel out common factors
d/v = v * t/v
SimpQuot
Simplify quotient
d/v = t
RearrangeEqn
Rearrange equation
t = d/v
Now, substitute d = 150 and v=2.5.
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Rewriting and Solving Literal Equations
Exercise 2.1
{"type":"choice","form":{"alts":["A","B","C"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
{"type":"choice","form":{"alts":["A","B","C"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
{"type":"choice","form":{"alts":["A","B","C","D"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":3}
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We see that m is multiplied by g and h. We can isolate the variable m by first dividing the equation by g and then by h.
Therefore, the answer is B.
We first multiply the equation by 2 to get rid of the fraction on the right hand side. Then, we divide both sides by v^2.
This corresponds to A.
We equate the expressions of the different energies and solve for v.
We can exclude the negative solution because in this context the sign of the expression only indicates the direction. v=sqrt(2gh) The answer is D.
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E
Hint & Answer
S
Solution
There is a relationship between the speed of light c, the electric constant ε_0, and the magnetic constant μ_0. c=1/sqrt(μ_0ε_0) Which of the following equations is formed when the equation is solved for μ_0? A.& μ_0= 1cε_0 && B. μ_0=c^2ε_0 C.& μ_0= 1c^2ε_0 && D. μ_0=cε_0
{"type":"choice","form":{"alts":["A","B","C","D"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":2}
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We will start by multiplying both sides by sqrt()μ_0 ε_0. Then we will raise both sides by 2 to get rid of the radical sign.
This equation matches with C.
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Hint & Answer
S
Solution
The square shown has side lengths a and diagonal length d.
Write a formula for a that contains d. Simplify the answer.
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Notice that two sides of the square and the diagonal forms a right triangle. Since the whole figure is a square, the legs have length a. The diagonal is the hypotenuse of the right triangle.
For any right triangle, we can use the Pythagorean Theorem to relate its side lengths. Let's use it to see relationship between a and d. a^2+a^2=d^2 Now we will solve this equation for a.
Because a is a length, it is always positive. Therefore, we can ignore the negative root. a=d/sqrt(2)
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Solution
The gravitational force between any two objects in the universe is calculated by the following formula. F=Gm_1 m_2/r^2 Here, m_1 and m_2 are the masses of objects, G is the universal gravitational constant and r is the distance between the centers of the objects. Which of the following is equal to r? A.& Gm_1m_2/F && B. sqrt(F/Gm_1m_2) [1em] C.& sqrt(Gm_1m_2/F) && D. Gm_1/m_2F
{"type":"choice","form":{"alts":["A","B","C","D"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":2}
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We need to find an expression equivalent to r. To do so, we will use the Properties of Equality. Let's first multiply both sides by r^2, then isolate r.
Since distances are always positive, the negative solution is irrelevant in this context and thus we can remove it. r=sqrt(Gm_1m_2/F) The expression on the right-hand side corresponds to option C.
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Hint & Answer
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Solution
Rewriting and Solving Literal Equations
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