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.
Show less Show more expand_more| Student Learning Objectives: |
|---|
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| | 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
Equations can include variables multiplied or divided by rational numbers. To solve these equations, we multiply or divide both sides by the same number. This lesson explains how to solve these equations and use them to model real-life situations.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
A Vinyl Record Collection
Kriz has a collection of 200 vinyl records.
They bought 10 records each month while building the collection.
a
Write an equation in terms of m that represents the situation.
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b
Solve the equation to find how long it took Kriz to build their collection.
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Discussion
Multiplication Property of Equality
Multiplication and division are inverse operations. They can be used to solve equations using the following Properties of Equality.
Rule
Multiplication Property of Equality
Given an equation, multiplying each side of the equation by the same number results in 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 is accepted without 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
Multiplying both sides of the equation by 4 isolates the variable x and allows us to find the solution of the equation.
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Discussion
Division Property of Equality
Dividing each side of an equation by the same nonzero number results in 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 is accepted without proof. This property is one of the Properties of Equality that can be used when solving equations. 5x&=10 5x ÷ 5&=10 ÷ 5 x&=2
Dividing both sides of the equation by 5 isolates the variable x and allows us to find the solution of the equation.
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Example
Bakery Orders
a
A baker packs cookies into 5 boxes. Each box contains the same number of cookies. There are a total of 180 cookies. This situation can be represented by an equation.
5b = 180 Here, b represents the number of cookies in one box. Solve the equation to find the number of cookies b in each box.
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b
The baker is also packing a cupcake order. The baker wants to pack the cupcakes into 5 boxes as well. Each box will contain 6 cupcakes. The following equation represents this situation.
c/5 = 6 Solve this equation to find the total number of cupcakes c in the order.
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Hint
a
Use the Division Property of Equality.
b
Use the Multiplication Property of Equality.
Solution
a
Let's consider the given equation.
5b = 180 We can solve it by isolating the variable b on one side of the equation using inverse operations. Since b is multiplied by 5, let's divide each side by 5 using the Division Property of Equality.
5b=180
DivEqn
.LHS /5.=.RHS /5.
5b/5 = 180/5
CrossCommonFac
Cross out common factors
5b/5 = 180/5
SimpQuot
Simplify quotient
b = 180/5
CalcQuot
Calculate quotient
b = 36
Therefore, since b is the number of cookies in each box, each box contains 36 cookies. Let's also check our answer by substituting 36 for b in the equation and simplifying.
We got Substituting 36 for b into the equation results in a true statement, so b = 36 is the correct solution.
b
Let's consider the given equation.
c/5 = 6 Here, the variable c is divided by 5. To isolate it, we will multiply each side of the equation by 5 using the Multiplication Property of Equality.
c/5 = 6
MultEqn
LHS * 5=RHS* 5
c/5 * 5 = 6* 5
FracMultDenomToNumber
a/5* 5 = a
c = 6* 5
Multiply
Multiply
c = 30
Therefore, since c represents the number cupcakes in the order, there are 30 cupcakes in the order. Let's substitute 30 for c in the equation to check our answer.
Substituting 30 for c into the equation results in a true statement. This means that c = 30 is the correct solution.
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Discussion
Solving Equations With Rational Coefficients
The Multiplication Property of Equality can be used instead of the Division Property of Equality when solving an equation with a rational coefficient. Consider an equation in the form abx = c. 3/5x = 6 The coefficient 35 is a fraction. The equation can be solved by multiplying both sides by the reciprocal of 35. The reciprocal of 35 is 53, so multiply 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
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Example
Vinyl Records, Rotations, and Song Length
A record rotates 190 times while playing a song on the album. It rotates 59 of a full rotation every second that 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
Let's consider the given equation.
5/9 t = 190
Here, the variable is t and its coefficient is the fraction 59. We can isolate the variable by multiplying both sides of the equation by 95, the reciprocal of 59. Let's do it!
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
The solution to the equation is t = 342, which means that the song is 342 seconds long.
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Pop Quiz
Solving Equations
Solve the equations using the Multiplication Property of Equality or the Division Property of Equality. If necessary, round the answer to two decimal places.
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Discussion
Writing Equations
Many real-life situations can be algebraically modeled by equations. These equations usually involve a variable that represents an unknown quantity. Consider modeling the following situation.
|
Diego categorizes his father's book collection by genre. He finds out that there are 8 different genres, each with the same number of books. The collection contains 216 books. How many books are there in each genre? |
This situation can be described in one sentence as follows. The product of the number of genres and the number of books in each genre is equal to the total number of books. The unknown quantity is the number of books in each genre. Let x be the variable representing this unknown quantity. Then, the verbal sentence can be translated into an algebraic equation. The product of the number of genres and the number of books in each genre is equal to the total number of books. ⇓ 8 * x = 200 The equation can now be solved to find the number x of books in each genre. Use the Division Property of Equality.
8x = 216
DivEqn
.LHS /8.=.RHS /8.
8x/8 = 216/8
CrossCommonFac
Cross out common factors
8x/8 = 216/8
SimpQuot
Simplify quotient
x = 216/8
CalcQuot
Calculate quotient
x=27
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Example
Earning Money for a Game System
Maya earns money as a dog walker. She wants to earn $600 for a new video game system. Her neighbor offers her $12 per walk to walk his dog.
a
Write an equation for the number x of times Maya has to walk her neighbor's dog to earn enough money to buy the game system.
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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":["Maya needs to walk the dog exactly 50 times.","Maya needs to walk the dog more than 50 times.","Maya needs to walk the dog less than 50 times."],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
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
We know that Maya gets $ 12 each time she walks the dog. She will get $ 12 x after x walks and she wants these earnings to equal $ 600. The following equation models this situation.
12 x = 600
b
Let's solve the equation we wrote in Part A.
12x = 600 We will isolate x on one side of the equation using inverse operations. Since x is multiplied by 12, let's divide each side by 12 using the Division Property of Equality.
12x = 600
DivEqn
.LHS /12.=.RHS /12.
12x/12 = 600/12
CrossCommonFac
Cross out common factors
12x/12 = 600/12
SimpQuot
Simplify quotient
x = 600/12
CalcQuot
Calculate quotient
x = 50
The solution to the equation is x = 50.
c
In Part B, we found that x = 50 solves our equation. Therefore, Maya needs to walk the dog exactly 50 times.
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Example
Stacking the Decks
Ignacio is sorting t trading cards into decks.
He places 60 cards in each deck and creates 3 decks.
a
Write an equation to find the total number of trading cards t that Ignacio has.
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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":["Ignacio has fewer than 200 cards.","Ignacio has exactly 200 cards.","Ignacio has more than 200 cards."],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
Hint
a
Assign the variable to the unknown quantity.
b
Use the Properties of Equality.
c
What does the variable represent?
Solution
a
Ignacio has t cards. He divides them equally into 3 decks, so each deck contains t 3 cards. Since each deck has 60 cards, we can write the following equation.
t/3 = 60 Note that multiple equivalent equations will also represent this situation. An example of such an equation is t = 3 * 60.
b
Let's solve the equation we wrote in Part A.
t/3 = 60 We will isolate t on one side of the equation using inverse operations. Since t is divided by 3, let's multiply each side by 3 using the Multiplication Property of Equality.
t/3 = 60
MultEqn
LHS * 3=RHS* 3
t/3 * 3 = 60 * 3
FracMultDenomToNumber
a/3* 3 = a
t = 60 * 3
Multiply
Multiply
t = 180
The solution to the equation is t = 180.
c
In Part B, we found that t=180 solves our equation. Therefore, Ignacio has exactly 180 trading cards, which is fewer than 200.
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Closure
How Long Did It Take To Compile the Collection?
We were asked how long it took Kriz to build their record collection. We can write an equation that models the situation and solve it to find this information. Kriz bought 10 records every month until they had 200 records.
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 Kriz to gather their 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
Let m be the number of months Kriz needed to build their record collection. We know that Kriz bought 10 records every month, so, after m months, he had 10 m records. There are 200 records in the collection, so 10 m must be equal to 200.
10 m = 200
b
To solve the equation from Part A, we will isolate m on one side of the equation using inverse operations. Since m is multiplied by 10, let's divide each side by 10 using 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
The solution to the equation is m=20. This means it took 20 months to build their collection.
Extra
Equations With Decimal CoefficientsConsider 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 with integer or fractional coefficients. In this case, this means dividing both sides of the equation by the coefficient 1.6.
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Solving One-Step Equations with Rational Coefficients
Exercises
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We have been given a sentence and we want to translate it into an equation.
Seven times k is equal to twelve.
Every equation has an equals sign and values or expressions on either side of it. Key phrases, such as is,
is equal to,
and equals
tell us about the placement of the equals sign. Let's look for such a keyword in the given sentence and replace it with the equals sign.
Seven timesk is equal to twelve.
⇕
Seven timesk = twelve.
On the left-hand side we have another keyword times.
This word tells us the operation that will be used in our equation, which is multiplication.
Seven times k
7 * k
On the right-hand side, we have the number twelve. Putting these sides together, we have a complete equation.
7 * k = 12
We want to translate the given sentence into an equation.
Forty-one is one and a half times x.
Like every equation, our equation will have an equals sign and values or expressions on either side of it. Key phrases, like is,
is equal to,
and equals
tell us where the equals sign should be placed.
Forty-one is one and a half timesx.
⇕
Forty-one = one and a half timesx.
The left-hand side is the number forty-one. On the right-hand side, we have the keyword times.
This means that we will use multiplication in our equation.
one and a half times x
1.5 * x
Putting the two sides together, we have a complete equation.
41 = 1.5 * x
Let's translate the given sentence into an equation.
The product of n and 5 equals 22.
Like all equations, ours will have an equals sign and values or expressions on either side of it. Key phrases, for example is,
is equal to,
and equals
tell us where to put the equals sign.
The product ofnand5 equals 22.
⇕
The product ofnand5 = 22.
On the left-hand side we have the keyword product.
This word tells us that multiplication will be used in our equation.
The product of n and 5
n * 5
On the right-hand side, we have the number 22. Putting these sides together, we have a complete equation.
n * 5 = 22
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Solution
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["p"],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["p\/2 = 14","14=p\/2"]}}
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We are given the following sentence and we want to translate it into an equation.
The number k divided by eight is equal to four.
Every equation has an equals sign and values or expressions on either side of it. Key phrases, such as is,
is equal to,
and equals
helps us determine where to put the equal sign.
The numberk divided by eight is equal to four.
⇕
The numberk divided by eight = four.
On the left-hand side we have the key phrase divided by.
This tells us the operation that will be used in our equation, which is division.
The numberk divided by eight
k ÷ 8
On the right-hand side, we have the number four. Putting these sides together, we have a complete equation.
k ÷ 8 = 4
We want to translate the given sentence into an equation.
Fourteen is v divided by five.
Like every equation, our equation will have an equals sign and values or expressions on either side of it. Key phrases, like is,
is equal to,
and equals
tell us where the equals sign should be placed.
Fourteen is vdivided by five.
⇕
Fourteen = vdivided by five.
The left-hand side is the number fourteen. On the right-hand side, we have the key phrase divided by.
This means that we will use division in our equation.
v divided by five
v ÷ 5
Putting the two sides together, we have a complete equation.
14 = v ÷ 5
Let's translate the given sentence into an equation.
The quotient of p and 2 equals 13.
Like all equations, ours will have an equals sign and values or expressions on either side of it. Key phrases, for example is,
is equal to,
and equals
tell us where to put the equals sign.
The quotient ofpand2 equals 13.
⇕
The quotient ofpand2 = 13.
On the left-hand side we have the keyword quotient.
This tells us that division will be used in our equation.
The quotient of p and 2
p ÷ 2
On the right-hand side, we have the number 13. Putting these sides together, we have a complete equation.
p ÷ 2 = 13
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Solution
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"r =","formTextAfter":null,"answer":{"text":["-3\/7"]}}
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We have to isolate the variable x on one side to solve the given equation. 4x = 24 In this equation, x is multiplied by 4. To undo this multiplication, we need to divide both sides by 4 because multiplication and division are inverse operations. We can perform this division because of the Division Property of Equality. Let's do it!
Our calculations show that the solution to the given equation is x=6. To check that this solution is correct, let's substitute x=6 into the original equation and simplify.
The left-hand side and right-hand side are equal, so x=6 is the correct solution.
For this equation, we need to isolate u on one side. -2 u = 8 Here, the variable u is multiplied by - 2. Dividing by - 2 is the inverse of multiplying by - 2. Let's use the Division Property of Equality to divide both sides of the equation by - 2. This will isolate u.
The solution to the given equation is u=- 4. Finally, we substitute u=- 4 into the original equation and check if our answer is correct.
Substituting u=- 4 resulted in a true statement. We can conclude that u=- 4 is the correct solution.
This time, we will isolate the variable r to solve the equation. 7r = - 3 We can see that r is multiplied by 7. We will use the Division Property of Equality to divide both sides of the equation by 7. This will undo multiplying by 7 and isolate r.
The solution to the given equation is r=- 37. Now, let's verify that our answer is correct. We will substitute r=- 37 into the original equation and simplify to check that our answer is correct.
This statement is true, so r=- 37 is the correct solution.
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Solution
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"h =","formTextAfter":null,"answer":{"text":["-28"]}}
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We have to isolate the variable a on one side to solve the given equation. a ÷ 4 = 9 In this equation, the variable a is divided by 4. Since the inverse of dividing by 4 is multiplying by 4, we will multiply both sides of the equation by 4. The operations will undo each other and the variable is isolated. Let's use the Multiplication Property of Equality to do what we just said.
Our calculations show that the solution to the given equation is a=36. To check that this solution is correct, we will substitute 36 for x into the original equation and simplify.
The left-hand side and right-hand side are equal. This means that a=36 is the correct solution.
Let's solve this equation. 5 = p/4 We need to isolate the variable p on one side to find its solution. We see that p is divided by 4. In this case, we can eliminate the denominator of the fraction by multiplying both sides of the equation by 4. Recall that we are allowed to do so by the Multiplication Property of Equality.
The solution to the given equation is p=20. We can substitute 20 for p into the original equation and simplify to check that our answer is correct.
This statement is true, so p=20 is the correct solution.
We will solve the given equation. We need to isolate the variable h on one side. h/- 4 = 7 In this equation, h is divided by - 4. Again, we will use the Multiplication Property of Equality. If we multiply both sides of the equation by - 4, the variable h will be isolated. This is because multiplying by - 4 is the inverse of dividing by - 4.
The solution to the given equation is h=- 28. Now, let's verify that our answer is correct. We will substitute h=- 28 into the original equation and simplify to check that our answer is correct.
This statement is true, so h=- 28 is the correct solution.
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Solution
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Let's take a look at the given equation. 2/3q = 16 We will isolate the variable q on one side to solve this equation. We see that q is multiplied by a fraction, 23. In other words, the coefficient of the variable is 23. In this case, we can isolate q by multiplying both sides of the equation by the reciprocal of the coefficient. Recall that the reciprocal of a fraction is found by interchanging the numerator and denominator. Fraction & Reciprocal 2/3 & 3/2 Let's multiply both sides by 32. We are allowed to do it by the Multiplication Property of Equality.
Our calculations show that the solution to the given equation is q=24. To check that this solution is correct, let's substitute 24 for q into the original equation and simplify.
The left-hand side and right-hand side are equal, so q=24 is the correct solution.
Let's now solve this equation. - 5/8k = 10 We will isolate k on one side. Here, the variable k is multiplied by - 58. This is a fraction, so we will multiply both sides of the equation by its reciprocal. Fraction & Reciprocal - 5/8 & - 8/5 Let's multiply the equation by the reciprocal.
The solution to the given equation is k=- 16. We can substitute - 16 for k into the original equation and simplify to check that our answer is correct.
This statement is true, so k=- 16 is the correct solution.
Finally, we will solve the following equation. 4/7g = - 12 We need to isolate the variable. We can do it by multiplying both sides of the equation by the reciprocal of the coefficient. The reciprocal of the coefficient 47 is 74. Fraction & Reciprocal 4/7 & 7/4 Let's solve the equation!
The solution to this equation is g=- 21. Now, let's verify that our answer is correct. We will substitute -21 for g in the original equation and simplify to check that our answer is correct.
This statement is true, so g=-21 is the correct solution.
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Solving One-Step Equations with Rational Coefficients
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