2. Solving One-Step Equations in One Variable
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Chapter 1
2.
Solving One-Step Equations in One Variable
This lesson delves into the topic of one-step equations, emphasizing the role of inverse operations and properties of equality in finding solutions. It provides real-life examples to illustrate how these mathematical concepts are not just academic exercises but also practical tools for everyday decision-making. For instance, the material covers scenarios like calculating distances, determining ages, and budgeting. The aim is to equip learners with the skills to solve simple equations quickly and accurately, whether they are students trying to excel in algebra or adults looking to make informed decisions in their daily lives.
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| | 13 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving One-Step Equations in One Variable
Slide of 13
After writing an equation modeling some real-life situation, the next step is to solve it. To do so, it is necessary to apply different Properties of Equality. Along this lesson, these properties will be introduced and applied to equations that can be solved in one step.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
Different Ages of Family Members
Zosia's brother is 10 years older than she is. Also, Zosia's brother is half their father's age.
a If Zosia is 16 years old, write equations for the ages of her brother and father.
b Solve these equations to find their ages.
Discussion
Reflexive, Symmetric, and Transitive Properties of Equality
The Properties of Equality are rules that allow manipulation of an equation in such a way that an equivalent equation is obtained. These properties will be reviewed in sets. The first set of properties is shown below.
Rule
Reflexive Property of Equality
For any real number, the number is equal to itself.
a=a
x+3=8
The sum on the left-hand side can be interpreted as a number, so this sum has to equal 8. This implies that x must be equal to 5.Rule
Symmetric Property of Equality
For all real numbers, the order of an equality does not matter. Let a and b be real numbers.
If a=b, then b=a.
Rule
Transitive Property of Equality
For all real numbers, if two numbers are equal to the same number, then they are equal to each other. Let a, b, and c be real numbers.
If a=b and b=c, then a=c.
{x=5y−305y−30=20⇒x=20
Since this property is an axiom, it does not need a proof to be accepted as true. The Transitive Property of Equality also holds true if a, b, and c are complex numbers.Pop Quiz
Which is the Correct Property?
Select the appropriate property for each example.
Discussion
Isolating the Variable
When solving equations, certain operations usually need to be undone to isolate the variable on one side. For example, consider the following equation.
x+5=7
To isolate x, the addition of 5 on the left-hand side has to be undone. Here is where inverse operations come into play. Concept
Inverse Operations
Inverse operations are two operations that undo one another. For example, adding 6 and subtracting 6 are inverse operations because they cancel each other out. This means that adding 6 to any number and then subtracting 6 results in the original number.
10+6−610 +6−610
Equations are solved by using inverse operations. By the Properties of Equality, any operation performed on one side of an equation must also be performed on the other side of the equation to maintain equality. Consider an example.
x−4=9
This equation can be solved by adding 4 to both sides.
In this case, the subtraction on the left-hand side of the equation can only be eliminated by adding 4 on both sides. The result of applying the Properties of Equality on an equation is an equivalent equation.Discussion
Addition and Subtraction Properties of Equality
Some of the most commonly used inverse operations are addition and subtraction. These operations fall under the Addition Property of Equality and the Subtraction Property of Equality.
Rule
Addition Property of Equality
Adding the same number to both sides of an equation results in an equivalent equation. Let a, b, and c be real numbers.
If a=b, then a+c=b+c.
x−3=5
By adding 3 to both sides of the equation, the variable x can be isolated and the solution to the equation can be found. Rule
Subtraction Property of Equality
Subtracting the same number from both sides of an equation results in an equivalent equation. Let a, b, and c be real numbers.
If a=b, then a−c=b−c.
x+2=7
By subtracting 2 from both sides of the equation, the variable x can be isolated and the solution to the equation can be found. Example
Finding How Much Money Davontay Had
One of Davontay's hobbies is playing the saxophone. He plays in the school band and wants to join a local community band as well. He goes to the music store to buy more reeds for his saxophone since he will be spending more time playing. After spending $20 on reeds, he is left with $140.
a Write an equation for this situation.
b How much money did Davontay originally have?
Answer
a x−20=140
b $160
Hint
a Use a variable for the original amount.
b Use the Properties of Equality to solve the equation set in Part A.
Solution
a It is possible write this situation as an equation using variables. Let x be the original amount of money that Davontay had. It is given that Davontay spent $20. Therefore, after buying the reeds, Davontay had x−20 dollars. This expression is the first part of the desired equation.
x−20
Then, it is given that Davontay has $140 left after spending $20. The equation can be completed by equating the previous expression with 140. x−20=140
b To find how much money Davontay had initially, the equation set in Part A needs to be solved for x. To do so, $20 will be added to both sides of the equation. This is valid by the Addition Property of Equality.
Therefore, Davontay originally had $160.
Example
Finding the Number of Pairs of Pants
Heichi like collecting a particular brand of clothing. He is inspecting his wardrobe before going to the mall. He notices that he has 7 shirts and 4 more shirts than pairs of pants.
a If p is the number of pairs of pants that Heichi has, which equation correctly describes the given situation?
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b How many pairs of pants does Heichi have?
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Hint
a Use variables to describe the situations.
b Use the Properties of Equality to solve the equation set in Part A.
Solution
a To write an equation that describes the given situation, variables need to be assigned to each item.
Number of Pairs of Pants:Number of Shirts: p s
Since Heichi has 4 more shirts than pants, the sum of p and 4 is equal to s.
p+4=s
Also, it is given that Heichi has 7 shirts. Therefore, by the Transitive Property of Equality, the equation can be rewritten using this information.
{p+4=ss=7⇒p+4=7
b To find out how many pairs of pants Heichi has, the equation set in Part A needs to be solved for p. To do so, 4 will be subtracted from both sides of the equation. This is valid by the Subtraction Property of Equality.
Discussion
Multiplication and Division Properties of Equality
The other most common type of inverse operations are the multiplication and division operations. These are valid by the following properties of equality.
Rule
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.
x÷4x÷4×4x=2=2×4=8
Here, by multiplying both sides of the equation by 4, the variable x was isolated and the solution of the equation was found. Note that the Multiplication Property of Equality also holds true if a, b, and c are complex numbers.Rule
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.
5x5x÷5x=10=10÷5=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. Note that the Division Property of Equality also holds true if a, b, and c are complex numbers.Example
Distances From Home
LaShay's hobbies include playing golf with her father. She wants to go to the golf course to practice. She knows that her school is one quarter of the way from her house to the golf course.
Consider that the school is about 2.5 miles from LaShay's house.
a If g is the distance from LaShay's house to the golf course, which equation correctly describes the situation?
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b How far is the golf course from LaShay's house?
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Hint
a Use variables for each distance.
b Use the Properties of Equality to solve the equation set in Part A.
Solution
a To write an equation that describes the given situation, variables need to be assigned to each distance.
Distance to School: Distance to the Golf Course: sg
Since the school is one quarter of the way to the golf course, the value of the division of g by 4 is equal to the value of s.
4g=s
Also, it is given that the distance to school is 2.5 miles. s=2.5
Using the Transitive Property of Equality, it is possible to rewrite the first equation.
{4g=ss=2.5⇒4g=2.5
b To find out how far is the golf course from LaShay's house, the equation set in Part A needs to be solved. To do so, both sides of the equation will be multiplied by 4. This is valid by the Multiplication Property of Equality.
Therefore, the golf course is 10 miles away from LaShay's house.
Example
The Number of Chess Pieces
Zain is in the chess club at their school. During one game, their opponent has double the pieces that Zain has. Zain's opponent has 6 pieces left on the board.
a If z is the number of pieces that Zain has, which equation correctly describes the situation?
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b How many pieces does Zain have left?
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Hint
a Use variables for the number of pieces of each player.
b Use the Properties of Equality to solve the equation set in Part A.
Solution
a An equation can be written if variables are used for the number of pieces each player has left.
Number of Zain’s Pieces: Number of Opponent’s Pieces: zo
Since Zain's opponent has twice the pieces that Zain has, multiplying z by 2 is equal to o.
2z=o
It is given that the opponent has 6 pieces left. Therefore, o is equal to 6. Using the Transitive Property of Equality, it is possible to rewrite the equation.
{2z=oo=6⇒2z=6
b To find the number of pieces that Zain has left, both sides of the equation will be divided by 2. This is valid by the Division Property of Equality.
2z=6
DivEqn
LHS/2=RHS/2
22z=26
CancelCommonFac
Cancel out common factors
22z=26
SimpQuot
Simplify quotient
z=26
CalcQuot
Calculate quotient
z=3
Pop Quiz
Solving Equations Using the Properties of Equality
Find the value of the variable on each equation using the Properties of Equality.
Closure
Finding the Ages of Family Members
The challenge at the beginning of the lesson gave some information about Zosia's family.
- Zosia is 16 years old.
- Zosia's brother is 10 years older than she is.
- Zosia's brother is half the age of their father.
Then, the following exercises were presented.
a Write equations for the ages of Zosia's brother and father.
b Solve these equations to find their ages.
Answer
a {b−10=162f=b
b Brother's Age: 26 years
Father's Age: 52 years
Father's Age: 52 years
Hint
a Use variables for the ages of each person.
b Use the Properties of Equality to solve each equation set in Part A.
Solution
a To write equations with the given information, it is convenient to use variables to represent each person's age.
Zosia’s Age: Zosia’s Brother’s Age: Zosia’s Father’s Age: zbf
Zosia's brother is 10 years older than Zosia. This means that subtracting 10 from b is equal to z.
b−10=z
Also, it is given that Zosia is 16 years old. Using the Transitive Property of Equality, it is possible to rewrite this equation.
{b−10=zz=16⇒b−10=16
Finally, since Zosia's brother's age is half of their father's age, the dividing f by 2 is equal to b.
2f=b
Therefore, these are the two equations for the ages of Zosia's brother and father.
{b−10=162f=b
b To find the age of Zosia's brother, 10 will be added to both sides of the first equation. This is valid by the Addition Property of Equality.
{2f=bb=26⇒2f=26
To find their father's age, both sides of the equation have to be multiplied by 2. This is valid by the Multiplication Property of Equality.
Consequently, Zosia's father is 52 years old.
Solving One-Step Equations in One Variable
Exercises
Looking at their homework, Zain noticed that the letter a makes up one twelfth of the document. If the homework contains 3000 letters, how many letters are not letter a in the document?
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The exercise states that the letter a makes up one twelfth of Zain's homework. If we say that n represents the number of times the letter a appears in the document and that T is the total number of letters in the document, we can then write an equation relating these two values. n = T/12 We need to find how many letters in the 3000-letter document are not the letter a. To do this, we will first find the number of times the letter a is used in the document by using our equation. Then we will be able to subtract this value from the total number of letters.
The letter a appears in the document 250 times. As the document has a total of 3000 letters, the number of letters that are not a is the difference between this number and 250. 3000- 250= 2750 Therefore, 2750 letters in the document are not a.
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Davontay brought half of the board games to this week's game night. He brought 4 games. Diego said that one half of 4 is 2, so there were 2 different board games available at game night. Help Diego find the correct number of board games available at the game night.
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The exercise says that Davontay brought half of the total board games. We will write this information as an algebraic expression. Let g be the total number of board games available at the game night. 1/2g We also know that Davontay brought 4 games. With this in mind, we can use the above expression to write an equation. 1/2g = 4 We now have an equation that relates the number of board games brought by Davontay to the total number of board games at the game night. Let's solve the equation for g to find the total number of board games available.
We can see that there were 8 total board games available at game night this week.
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Determine the values indicated below.
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To find the value of y-3, we have to solve the equation by isolating y.
Now we can find the value of y-3 by substituting 22 for y in the expression.
The value of y-3 is 19.
Let's solve the equation to find z.
Since we now know that z=-24, we can find the value of z+2 by substituting -24 for z in the expression.
The value of z+2 is -22.
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Solving One-Step Equations in One Variable
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