2. Solving One-Step Equations with Rational Coefficients
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Chapter 6
2.
Solving One-Step Equations with Rational Coefficients
This lesson explores the methods for solving one-step equations where the coefficients are rational numbers. It emphasizes the importance of using the Multiplication Property of equality and the Division Property of Equality to find the solution. These techniques are not just theoretical; they have practical applications too. For example, they can be used in financial calculations, engineering problems, and various other fields that require precise mathematical reasoning. The goal is to make learners proficient in solving such equations, whether they are high school students preparing for advanced algebra or professionals needing these skills for their work.
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Lesson Settings & Tools
| | 12 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving One-Step Equations with Rational Coefficients
Slide of 12
Real-life situations modeled by an equation include a coefficient of the variable which can be a rational number. Both sides of the equation are multiplied or divided by a number to isolate the variable. This lesson begins its focus on solving equations involving multiplication and division. Then, the focus shifts to how to write these equations when modeling real-life problems.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
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A Vinyl Record Collection
Diego is going through some old stuff in storage. He discovers an entire rack of vinyls! There are 200 in total and they belong to his dad. Diego asks his father how long this collection took to gather.
a Write an equation in terms of m that represents the situation.
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b Solve the equation to find the time it took Diego's father to gather his collection of records.
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Multiplication Property of Equality
Multiplication and division are inverse operations. They can be used to solve equations by the following properties of equality.
Rule
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Multiplication Property of Equality
Given an equation, multiplying each side of the equation by the same number yields an equivalent equation. Let a, b, and c be real numbers.
If a = b, then a * c = b * c.
The Multiplication Property of Equality is an axiom, so it does not need a proof. This property is one of the Properties of Equality that can be used when solving equations. Consider the following example. x÷4&=2 x÷4 * 4&=2 * 4 x&=8
Here, by multiplying both sides of the equation by 4, the variable x was isolated and the solution of the equation was found.Discussion
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Division Property of Equality
Dividing each side of an equation by the same nonzero number yields an equivalent equation. Let a, b, and c be real numbers.
If a = b and c≠ 0, then a ÷ c = b ÷ c.
The Division Property of Equality is an axiom, so it does not need a proof to be accepted as true. This property is one of the Properties of Equality that can be used when solving equations. 5x&=10 5x ÷ 5&=10 ÷ 5 x&=2
As can be observed, by dividing both sides of the equation by 5, the variable x was isolated and the solution of the equation was found.Example
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Vinyl Collection Stored in Boxes
a Diego's father stored his vinyl collection in five boxes. Each box contains the same number of records and, in total, there are 200 vinyl records in the collection. This situation can be represented by an equation.
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b Diego's grandfather also found some records.
Diego wants to put an equal number of those records grandpa found into each of the five boxes. He notices that he would have to put seven records into each box to do that. The following equation represents the situation.
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Hint
a Use the Division Property of Equality.
b Use the Multiplication Property of Equality.
Solution
a When solving equations in one variable, the variable is isolated on one side of the equation. This can be done by using inverse operations because inverse operations undo each other. Consider the given equation.
5b=200
DivEqn
.LHS /5.=.RHS /5.
5b/5 = 200/5
CrossCommonFac
Cross out common factors
5b/5 = 200/5
SimpQuot
Simplify quotient
b = 200/5
CalcQuot
Calculate quotient
b = 40
b Recall that the variable is isolated on one side when solving equations in one variable. Inverse operations play a role in isolating the variable because they
undoeach other. Consider the given equation.
r/5 = 7
MultEqn
LHS * 5=RHS* 5
r/5 * 5 = 7* 5
FracMultDenomToNumber
a/5* 5 = a
r = 7* 5
Multiply
Multiply
r = 35
Discussion
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Solving Equations With Rational Coefficients
The Multiplication Property of Equality can be used instead of the Division Property of Equality anytime when solving an equation with a coefficient that is a rational number. Consider an equation in the form abx = c. a/bx = c The coefficient ab is a fraction. For that reason, the equation can be solved by multiplying both sides by the reciprocal of ab. As an example, consider the following equation. 3/5 x = 6 The coefficient next to the variable is 35. Interchange the numerator and denominator to find its reciprocal. The reciprocal of 35 is 53, so the equation can be solved by multiplying both sides by 53.
3/5x = 6
MultEqn
LHS * 5/3=RHS* 5/3
3/5x * 5/3 = 6 * 5/3
▼
Simplify
CommutativePropMult
Commutative Property of Multiplication
x * 3/5 * 5/3 = 6 * 5/3
MultFracByInverse
a/b* b/a=1
x * 1 = 6 * 5/3
MultByOne
a * 1=a
x = 6* 5/3
MoveLeftFacToNum
a*b/c= a* b/c
x = 6* 5/3
Multiply
Multiply
x = 30/3
CalcQuot
Calculate quotient
x = 10
Example
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Vinyl Records, Rotations, and Song Length
Diego finds himself wondering about the speed at which a record rotates. His father tells Diego that the record rotates 190 times while playing a song on the album. Diego later reads online that a record rotates 59 of a full rotation every second it plays.
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Hint
Identify the coefficient of the variable. Multiply both sides of the equation by the reciprocal of the coefficient.
Solution
Consider the given equation.
5/9 t = 190
Here, the variable is t and its coefficient is the fraction 59. Since it is a fraction, isolate the variable by multiplying both sides of the equation by the reciprocal of that fraction. Recall that the reciprocal of a fraction is found by interchanging the numerator and denominator. The reciprocal of 59 is 95. Now, multiply the equation by 95.
The solution to the equation is t = 342, which means that the song lasts 342 seconds.
5/9t = 190
MultEqn
LHS * 9/5=RHS* 9/5
5/9t*9/5= 190*9/5
▼
Simplify
CommutativePropMult
Commutative Property of Multiplication
t* 5/9*9/5= 190*9/5
MultFracByInverse
a/b* b/a=1
t* 1= 190*9/5
MultByOne
a * 1=a
t = 190*9/5
MoveLeftFacToNum
a*b/c= a* b/c
t = 190*9/5
Multiply
Multiply
t = 1710/5
CalcQuot
Calculate quotient
t = 342
Pop Quiz
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Solving Equations
Solve the equations using the Multiplication Property of Equality or the Division Property of Equality. If necessary, round answers to two decimal places.
Discussion
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Writing Equations
Many real-life situations can be algebraically modeled by equations. These equations can involve a variable that represents an unknown quantity. Consider modeling the following situation.
|
Diego categorizes his father's record collection by genre. He finds out that there are 8 different genres, each with the same number of records. The collection contains 200 records. How many records are there in each genre? |
8x = 200
DivEqn
.LHS /8.=.RHS /8.
8x/8 = 200/8
CrossCommonFac
Cross out common factors
8x/8 = 200/8
SimpQuot
Simplify quotient
x = 200/8
CalcQuot
Calculate quotient
x=25
Example
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Earning to Fix the Turntable
Diego's father's old turntable is broken. Diego is so eager to listen to the records that he decides to make some money to buy the replacement parts. The parts that Diego needs to buy costs $96. Diego's neighbor offers him $8 per walk to walk her dog.
a Write an equation for the number x of times Diego has to walk his neighbor's dog to earn enough money to buy the parts he needs.
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b Solve the equation.
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c Which of the following is a true statement?
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Hint
a Assign the variable to the unknown quantity.
b Use the Properties of Equality to solve the equation written in Part A.
c What does the variable represent?
Solution
a The number of times Diego has to walk the dog is represented by the variable x. Diego gets $ 8 per walk, so the total amount he earns is equal to 8 x. He wants to save $96, so the amount earned should be equal to $ 96. By combining all this information, an equation can be written as follows.
8 x = 96 This equation models the given situation.
b Solving an equation in one variable means isolating the variable on one side.
8x = 96
DivEqn
.LHS /8.=.RHS /8.
8x/8 = 96/8
CrossCommonFac
Cross out common factors
8x/8 = 96/8
SimpQuot
Simplify quotient
x = 96/8
CalcQuot
Calculate quotient
x = 12
c The variable x in the equation from Part A represents the number of times Diego needs to walk his neighbor's dog to earn enough money for the parts he needs to repair the broken turntable. The calculations show that x = 12. This means that Diego needs to walk the dog exactly 12 times to earn enough.
x = 12 ⇓ Diego needs to walk the dog12times.
Example
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Job Well Done
The dog Diego walked loved him so much — just look at the two of them! 
His neighbor, the dog owner, was so impressed that affter 12 walks, she decided to pay Diego handsomely. Diego received $10 for each walk.
a Write an equation for the amount t of money Diego got in total.
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b Solve the equation.
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c Which of the following is a true statement?
{"type":"choice","form":{"alts":["Diego got less than $120.","Diego got more than $120.","Diego got exactly $120."],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":2}
Hint
a Assign the variable to the unknown quantity.
b Use the Properties of Equality to solve the equation written in Part A.
c What does the variable represent?
Solution
a The total amount of money Diego got for walking the dog is represented by the variable t. Diego walked the dog 12 times, so the amount he earned for one walk is t 12. The neighbor paid Diego $ 10 per walk, so the amount earned for one walk should be equal to 10. Now, an equation can be written.
t/12 = 10 Note that different equations, equivalent to this one, can also be used to model this situation. An example of such an equation is t = 12 * 10.
b Isolate the variable t on one side to solve the equation.
t/12 = 10
MultEqn
LHS * 12=RHS* 12
t/12 * 12 = 10 * 12
FracMultDenomToNumber
a/12* 12 = a
t = 10 * 12
Multiply
Multiply
t = 120
c The variable t in the equation from Part A represents the total amount of money Diego earned walking his neighbor's dog. The calculations show that t = 120. This means that Diego earned exactly $120. t = 120 ⇒ Diego earned$120.
It is safe to say that Diego earned enough to fix the turntable. Play that jam! 
Closure
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How Long Did It Take To Compile the Collection?
The challenge presented at the beginning of the lesson can be solved by writing an equation that models the situation and then solving the equation. It stated that Diego's father collected 200 vinyl records and bought 10 records every month.
a Write an equation in terms of m that represents the situation.
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b Solve the equation to find the time it took Diego's father to gather his collection of records.
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Hint
a Assign a variable to the number of months.
b Use the Division Property of Equality.
Solution
a When modeling a real-life situation with an equation, the first step is to assign a variable to the unknown quantity. In this case, the unknown quantity is the number of months it took Diego's father to gather his collection, so let it be m.
Number of Months: m Diego's father bought 10 records every month. Then, the number of records Diego's father collected in m months is 10 m. There are 200 records in the collection, so 10 m must be equal to 200. 10 m = 200 Diego figured out the equation!
b The m-variable must be isolated to solve the equation. In this case, m is multiplied by 10. Divide both sides of the equation by 10 to isolate m. Recall that dividing both sides by the same number does not change the solution to the equation because of the Division Property of Equality.
10m = 200
DivEqn
.LHS /10.=.RHS /10.
10m/10 = 200/10
FracMultDenomToNumber
a/10* 10 = a
m = 200/10
CalcQuot
Calculate quotient
m=20
Extra
Equations With Decimal CoefficientsOn a final note, consider an equation where the coefficient is a decimal.
1.6x = 12
Equations where the coefficient is a decimal can be solved the same way as equations where the coefficient is an integer or a fraction. In this case, this means dividing both sides of the equation by the coefficient 1.6.
Similarly, when solving equations where the variable is divided by a decimal, or a decimal is added or subtracted from the variable, we use the same methods we would when solving an equation with integers or fractions.
1.6x = 12
DivEqn
.LHS /1.6.=.RHS /1.6.
1.6x/1.6 = 12/1.6
FracMultDenomToNumber
a/1.6* 1.6 = a
x = 12/1.6
CalcQuot
Calculate quotient
x = 7.5
Solving One-Step Equations with Rational Coefficients
Exercise 2.1
Mark went swimming at his local swimming pool. He and his friends decided to see who can swim the longest distance.
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We want to write an equation that represents the situation. We first represent the unknown quantity with a variable. Here, the unknown quantity is the length of the pool. Let's call it p. We know that Mark swam 8 lengths of the pool. Then, 8 p is equal to the total distance Mark swam. 8 p = total distance Mark swam We know that the total distance he swam is 200 meters. This means that the right-hand side of the equation is 200. 8 p= 200 This equation can help us find the length of the pool. Note that we can write different equations, but they will all be equivalent to the equation we wrote. These equivalent equations can also be obtained by applying the Properties of Equality to the above equation.
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Maya and her family went on a three-day sailing trip. They sailed 20 miles each day.
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We want to write an equation to model this situation. Note that the total distance sailed by Maya and her family divided by the number of days the trip lasted is equal to the distance sailed in one day. Total distance sailed/Number of days = Distance sailed in one day Let's represent the total distance sailed with the variable t. Then, we can substitute the given information and the variable into the verbal model to write the equation. Total distance sailed/Number of days = Distance sailed in one day ⇓ t/3 = 20 This equation can help us find the total distance Maya and her family sailed during their trip. Note that we can write different equations, but they will all be equivalent to the equation we wrote. These equivalent equations can also be obtained by applying the Properties of Equality to the above equation.
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Consider the following equation.
5b = 20
Choose the situation that can be modeled by the given equation.
A. & Ramsha buys four books, each for & the same price. She pays$ 5. [0.4em]
B. & Ramsha buys20books, each for & the same price. She pays$5. [0.4em]
C. & Ramsha buys four books, each for & the same price. She pays$20. [0.4em]
D. & Ramsha buys five books, each for & the same price. She pays$20.
{"type":"choice","form":{"alts":["A","B","C","D"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":3}
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We are given the following equation. 5b = 20 Here, the variable b is multiplied by 5 and equals 20. Recall that the variable represents some unknown quantity in an equation. We can notice that, in each of the given situations, the unknown quantity is the price of one book. We can assign it to the variable in the equation. Price of One Book: b The variable b is multiplied by 5 so the left-hand side of the equation 5 b can represent the price of 5 books. The equation tells us that this is equal to 20, so the total price of the 5 books Ramsha buys is $20. 5 b = 20 [0.3em] Ramsha buys 5books for $beach and pays $20. This situation is described in option D.
When we look at the given options, we can notice that they have the same structure. In each option, Ramsha buys a certain number of books at the same price and pays some money. Here, we can say that the price of a book in each case is unknown. Let's make a diagram that can be used to model each situation.
We can now write an equation modeling each situation. Let's start by Option A. Ramsha buys four books, each for the same price. She pays$ 5. We can use the variable b to represent the price of a book because it is the variable used in the given equation.
We found the equation. This is different than the given equation. We will follow the same reasoning to find the equations of the remaining options.
| Option | Number of Books | Price | Amount of Money Paid | Equation |
|---|---|---|---|---|
| A | 4 | b | 5 | 4 b = 5 |
| B | 20 | b | 5 | 20 b = 5 |
| C | 4 | b | 20 | 4 b = 20 |
| D | 5 | b | 20 | 5 b = 20 |
As we can see, the situation in Option D is represented by the given equation.
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Consider the following equation.
p/4 = 10
Choose the situation that can be modeled by the given equation.
A. & Tadeo collects postcards. He has10albums, & each containing40postcards. [0.4em]
B. & Tadeo collects postcards. He has4albums, & each containing10postcards. [0.4em]
C. & Tadeo collects postcards. He has40albums, & each containing4postcards. [0.4em]
D. & Tadeo collects postcards. He has4albums, & each containing40postcards.
{"type":"choice","form":{"alts":["A","B","C","D"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}
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We are given an equation. p/4 = 10 Here, the variable is p and divided by 4. The quotient is equal to 10. We know that the variable represents some unknown quantity in an equation. We can see that, in each of the given situations, the unknown quantity is the total number of postcards. We can use p to represent it. Number of Postcards: p The variable p is divided by 4, so the left-hand side of the equation can represent the number of postcards in a single album when p postcards are divided between 4 albums. The equation tells us that this is equal to 10, so there are 10 postcards in a single album. p/4 = 10 [0.5em] Tadeo has ppostcards split between 4albums. Each album contains 10postcards. This situation is described in option B.
When we look at the given options, we can notice that they have the same structure. In each option, Tadeo has a number of albums, each containing some postcards. Here, we can say that the total number of postcards is unknown. Let's make a diagram that can be used to model each situation.
We can now write an equation modeling each situation. Let's start with Option A. Tadeo collects postcards. He has 10albums, each containing 40postcards. We can use the variable p to represent the total number of postcards because it is the variable used in the given equation.
We found the equation. This is different than the given equation. We will follow the same reasoning to find the equations of the remaining options.
| Option | Total Number of Postcards | Number of Albums | Number of Postcards | Equation |
|---|---|---|---|---|
| A | p | 10 | 40 | p/10 = 40 |
| B | p | 4 | 10 | p/4 = 10 |
| C | p | 40 | 4 | p/40 = 4 |
| D | p | 4 | 40 | p/4 = 40 |
As we can see, the situation in Option B is represented by the given equation.
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Solving One-Step Equations with Rational Coefficients
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