1. Writing and Solving One-Step Addition and Subtraction Equations
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Chapter 6
1.
Writing and Solving One-Step Addition and Subtraction Equations
This lesson aims to teach the fundamentals of solving one-step equations, specifically focusing on addition and subtraction methods. It emphasizes the importance of understanding properties of equality to solve these equations effectively. Real-world scenarios are integrated into the teaching approach, making the learning experience more relatable. For example, the material uses everyday situations like calculating the cost of snacks or determining the number of people who fell sick in a class to illustrate the practical applications of one-step equations. By grasping these basic concepts, learners can tackle a wide range of problems, from personal finance to health statistics, thereby making informed decisions in daily life.
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Lesson Settings & Tools
| | 10 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Writing and Solving One-Step Addition and Subtraction Equations
Slide of 10
Almost all real-life situations can be represented by equations, which makes them very important in math. It is possible to use equations to model real-life situations and predict their outcomes. This lesson will discuss how to represent real-life situations using equations, and how to solve these equations to find a solution.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
How Old Is Diego?
Diego and his younger brother dream of playing college football together on the same team. Right now his younger brother is 12 years old. There is a 3 year difference between their ages.
a Write an equation in terms of x that represents the situation.
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b Solve the equation to find Diego's age.
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Discussion
Equivalent Equations
The variable needs to be isolated on one side of an equation in order to solve the equation. This can be achieved by undoing
certain operations using inverse operations.
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.Concept
Equivalent Equation
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.
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.
A true statement was obtained. Therefore, y=15 is indeed a solution to the original 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
Doing the Dishes Together
Diego and his younger brother continued to talk about their college football dreams. Their abuelo — grandpa — overheard their dream and told them a little secret. He was a college football player! More importantly, he said he had to wash dishes to help pay for college.
Diego's abuelo showed Diego a photo of him doing dishes at home after his playing days were long over. Diego becomes more curious about washing dishes than any football dreams. He asks his abuelo two questions about the night the picture was taken.
- How many plates were washed?
- How long did it take to finish the dishes?
Diego's abuelo really wants to help Diego with math. He writes two equations whose solutions are the answers to Diego's questions. Find the answers to the questions by solving the equations.
a p−5=3
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b m+9=14
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">m<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["5"]}}
Hint
a Use the Addition Property of Equality.
b Use the Subtraction Property of Equality.
Solution
a The variable p must be isolated to solve the equation. On the left-hand side 5 is subtracted from p. This subtraction can be
undoneby adding 5 to both sides by the Addition Property of Equality.
b This time, the variable m has to be isolated. Notice that 9 is added to m. To
undothis addition, subtract 9 from both sides by using the Subtraction Property of Equality.
Pop Quiz
Solving Equations
Solve the equations by using the Addition Property of Equality or the Subtraction Property of Equality.
Discussion
Writing Equations
Some real-life situations can be algebraically modeled by equations. A critical step in doing this is to represent an unknown quantity with a variable. Consider the following situation.
|
In Diego's class, a certain number of people became sick and missed math class. There were 19 people present in class, and Diego's class has 24 people in total. |
The sum of the number of people who fell sick and the number of people who were present is equal to the total number of people in the class.⇓x+19=24
The equation can now be solved to find the number x of people in Diego's class who fell sick. Use the Subtraction Property of Equality.
x+19=24
SubEqn
LHS−19=RHS−19
x+19−19=24−19
Simplify left-hand side
x+19−19=24−19
SubTerms
Subtract terms
x=5
Example
Connecting Math to the Real World
Diego's abuleo gets a great deal of delight from seeing Diego so interested in math.
However, he realized that Diego is having some issues with connecting math to the real world. For this reason, he told Diego that he will buy some snacks and sodas to share if Diego can answer the following question.
|
Some snacks and a few sodas cost, in total, $13. If the sodas cost $7, how much money is spent on snacks? |
a Write an equation in terms of x to represent the situation.
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b Find the value of x for the equation written in Part A.
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c How much money did Diego's abuelo spend on the snacks?
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Hint
a Assign a variable to the unknown quantity. In this case, the unknown quantity is the cost of the snacks.
b Use the Subtraction Property of Equality.
c What does the variable represent?
Solution
a The first thing to do when writing an equation from a verbal description is to assign a variable to the unknown quantity. The given statement is equivalent to the following sentence.
The sum of the money spent on snacksand 7 is equal to 13.
If x is the amount of money spent on the snacks, the equation can be written by following this statement.
x+7=13
b Isolate the x-variable on one side of the equation to solve the equation. Since 7 is added to x, use the Subtraction Property of Equality.
c Remember that x represents the amount of money spent on the snacks. The calculations show that x=6. That means the total cost of the snacks is $6. It looks like Diego is going to enjoy some snacks and soda!
x=6 ⇒ The price of the snacks is $6.
Example
Modelling Cycling Distance With an Equation
Diego's abuelo remembered Diego's goal to become a college football player. He felt so guilty about giving Diego so much junk food! He thinks he should teach Diego about a healthy and active lifestyle. He tells Diego about how he rode his bicycle everyday when he was young. 
Again, Diego's abuelo wanted to give his grandson a math problem about his own history. Diego, when I was young I rode so much you wouldn't believe it. In fact, the difference between the number of kilometers I used to ride and 9 is equal to 3.
a Write an equation in terms of x to represent the situation.
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b Find the value of x for the equation written in Part A.
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c How many kilometers did Diego's abuelo cycle every day?
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Hint
a Assign a variable to the unknown quantity. In this case, the unknown quantity is the number of kilometers Diego's granddad used to cycle per day.
b Use the Addition Property of Equality.
c What does the variable represent?
Solution
a Begin by assigning a variable to the unknown quantity when writing an equation from a verbal description. Let x be the number of kilometers that Diego's abuelo cycled. Then, breakdown the given statement.
The difference between the number ofkilometers I used to bike and 9 is equal to 3.⇓x−9=3
b Isolate the x-variable on one side of the equation to solve it. Notice that 9 is subtracted from x. That means the Addition Property of Equality can be used.
c Recall that x represents the number of kilometers that Diego's abuelo cycled every day. Again, note what x equals. Since x=12, this means that Diego's abuelo cycled 12 kilometers every day.
x=12⇓He cycled 12 kilometers every day.
Closure
Calculating Diego's Age
The challenge presented at the beginning of the lesson can be solved by applying the learned concepts. Recall that the challenge stated that Diego's younger brother is 12 years old and that the difference between Diego's age and his brother's age is 3 years.
The two brothers continued to dream about becoming college football players. Diego then felt inspired by all of his abuelo's hard work. This drove Diego to figure out how to write the equation that determines his age. After all, if he can not do math, how can he succeed as an athlete? He thought.
External credits: @freepik
a Write an equation in terms of x that represents the situation.
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b Solve the equation to find Diego's age.
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Hint
a Assign a variable to Diego's age.
b Use the Addition Property of Equality.
Solution
a The first thing to do is to assign a variable to the unknown quantity. In this case, the unknown quantity is Diego's age, so let it be x. Because Diego's brother is 12 years old, the difference between x and 12 is equal to 3. With this information, the equation can be written.
x−12=3
Diego figured out the equation!
b The x-variable must be isolated to solve the equation. In this case, 12 is subtracted from x. That means it is possible to use the Addition Property of Equality to isolate x.
Writing and Solving One-Step Addition and Subtraction Equations
Exercise 1.1
Solve the given equations.
x+6=13
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d+2=-3
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.69444em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">d<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["-5"]}}
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We are asked to solve the given equation.
x+6=13
When solving equations, we can use inverse operations and Properties of Equality to undo
the operations applied to the variable. Here, 6 is added to the variable x.
x+ 6=13
To undo this operation, we use the inverse operation of addition — subtraction. The Subtraction Property of Equality lets us subtract 6 from both sides of the equation. Then, we will simplify.
The solution to our equation is x = 7.
We are asked to solve the given equation.
d+2=- 3
We can use inverse operations and Properties of Equality to undo
the operations applied to the variable. This lets us isolate the variable on one side and solve the equation. In this case, 2 is added to the variable d.
d+ 2=- 3
We will use the inverse operation of addition — subtraction. By the Subtraction Property of Equality, subtracting 2 from both sides of the equation results in an equivalent equation. This means that the resulting equation has the same solution as the original equation.
The solution to our equation is d = - 5.
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Solve the given equations.
x−2=7
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g−5=8
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.625em;vertical-align:-0.19444em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.03588em;\">g<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["13"]}}
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We need to solve the given equation.
x-2=7
When solving equations, we can use inverse operations and Properties of Equality to undo
the operations applied to the variable. Here, 2 is subtracted from the variable x.
x- 2=7
We use the inverse operation of subtraction to undo this operation. That would be the operation of addition. We can add 2 to both sides of the equation by the Addition Property of Equality. Let's do it! Then, we will simplify.
The solution to our equation is x = 9.
We can start by considering the given equation.
g-5=8
Let's use inverse operations and Properties of Equality to undo
the operations applied to the variable g. We can see that 5 is subtracted from g in this equation.
g- 5= 8
Like in Part A, let's use the inverse operation of subtraction to undo this operation. Again, that would be the operation of addition. The Addition Property of Equality lets us add 5 to both sides of the equation. Then, we will simplify.
The solution to our equation is g = 13.
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Choose all equations that are equivalent to the given equation.
a−8=5
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x−1=7
{"type":"multichoice","form":{"alts":["<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.66666em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">3<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">2<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.66666em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">2<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">0<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.66666em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">\u2212<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">4<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">4<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.66666em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">5<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord\">2<\/span><\/span><\/span><\/span>"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":[1,2]}
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We are asked to choose all the equations that are equivalent to the given equation a - 8 = 5. Two equations are equivalent if they have the same solutions. a - 9 &= 4 a + 2 &= 15 a - 8 &= 4 a - 6 &= 2 By the Addition and Subtraction Properties of Equality, adding or subtracting the same number from both sides of an equation results in an equivalent equation. We can use these properties to show that two equations are equivalent by transforming one into the other.
Equation I: a - 9 = 4
Consider the equation a-9=4. There is a - 9 on the left-hand side. This expression is equivalent to a - 8 - 1. Let's take the given equation a-8=5 and subtract 1 from each side. In other words, we will apply the Subtraction Property of Equality.
Subtracting 1 from each side of the equation a - 8 = 1 results in the equation a - 9 = 4. This means that a - 8 = 5 and a - 9 = 4 are equivalent equations.
Equation II: a +2 = 15
Next, we will consider the equation a + 2 = 15. Note that a + 2 is equivalent to a - 8 + 10. Then, we can take the given equation a - 8 = 5 and add 10 to each side. Let's do it!
This action results in the equation a + 2 = 15. Therefore a-8 = 5 and a + 2 = 15 are equivalent equations by the Addition Property of Equality.
Equation III: a-8=4
Let's consider the next equation, a - 8 = 4. It is not possible for a - 8 to be equal to 5 and 4 at the same time. Given:& a-8= 5 Considered:& a-8= 4 This means that these equations have different solutions. Therefore, they are not equivalent equations.
Equation IV: a - 6 = 2
Finally, let's consider the equation a - 6 = 2. The left-hand side a - 6 is equivalent to a - 8 + 2. Let's add 2 to each side of the equation a - 8 = 5. We want to check if we will get the equation we are considering.
The equations a - 8 = 5 and a - 6 = 7 are equivalent. That is the case because of the Addition Property of Equality. However, there is a 2 instead of a 7 on the right-hand side of our equation. a - 6 = 2 a - 6 = 7 These two equations have different solutions. This means that they are not equivalent equations. Therefore, a - 6 = 2 and a - 8 = 5 are not equivalent equations either.
Conclusion
We found that two of the given equations, a-9=4 and a+2=15, are equivalent to the equation a-8=5.
We are asked to choose all the equations that are equivalent to the given equation x - 1 = 7.
x + 3 &= 2
x+2 &= 10
x-4 &= 4
x+5 &= - 2
Remember that two equations are equivalent when they have the same solutions. We can check if two equations are equivalent by solving them and comparing their solutions. First, let's solve the given equation.
x - 1 = 7
Here, 1 is subtracted from the variable x. We isolate the variable on one side of the equation by undoing
the operations applied to the variable using inverse operations. The inverse operation of subtraction is addition, so we use the Addition Property of Equality to add 1 to both sides of the equation.
The solution to the given equation is x = 8. Now, let's solve the remaining equations, starting with x + 3 = 2. Here, 3 is added to the variable. Then, we will use the Subtraction Property of Equality to subtract 3 from both sides of the equation.
We use the same method to solve the remaining equations.
| Equation | Solution |
|---|---|
| x+3=2 | x = - 1 |
| x + 2 = 10 | x = 8 |
| x -4 = 4 | x = 8 |
| x + 5 = - 2 | x = - 7 |
We can see that two equations have the same solution as the given equation: x + 2 = 10 and x - 4 = 4. This means that only these two equations are equivalent to the given equation.
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Choose all equations that are equivalent to the given equation.
x+2=10
{"type":"multichoice","form":{"alts":["<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.66666em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">6<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">4<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.66666em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">2<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">1<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.66666em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">9<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.66666em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">3<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">2<\/span><\/span><\/span><\/span>"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":[0,2]}
d+8=1
{"type":"multichoice","form":{"alts":["<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.77777em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">d<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">3<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">0<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.77777em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">d<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">0<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">1<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.77777em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">d<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">7<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">0<\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.77777em;vertical-align:-0.08333em;\"><\/span><span class=\"mord mathdefault\">d<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><span class=\"mbin\">+<\/span><span class=\"mspace\" style=\"margin-right:0.2222222222222222em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><\/span><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord text\"><span class=\"mord\">-<\/span><\/span><span class=\"mord\">6<\/span><\/span><\/span><\/span>"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":[2,3]}
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We are asked to choose all the equations that are equivalent to the given equation x + 2 = 10. x + 6 &= 14 x + 2 &= 11 x + 1 &= 9 x + 3 &= 12 Two equations are equivalent if they have the same solutions. By the Addition and Subtraction Properties of Equality, adding or subtracting the same number from both sides of an equation results in an equivalent equation. We can use this to show that two equations are equivalent by transforming one into the other.
Equation I: x + 6 = 14
Consider the equation x+6=14. There is x + 6 on the left-hand side. Note that this is equivalent to x + 2 +4. Let's take the given equation x+2=10 and add 4 to each side. In other words, we will apply the Addition Property of Equality.
Adding 4 to each side of the equation x + 2 = 10 results in the equation x + 6 = 14. This means that the equations x + 2 = 10 and x + 6 = 14 are equivalent equations.
Equation II: x+2=11
Let's consider the next equation, x + 2 = 11. It is not possible for x + 2 to be equal to 11 and 10 at the same time. Given:& x+2= 10 Considered:& x+2= 11 This means that these equations have different solutions. Therefore, they are not equivalent equations.
Equation III: x +1 = 9
Next, we will consider the equation x + 1 = 9. Note that x + 1 is equal to x + 2 - 1. Then, we can take the given equation x + 2 = 10 and subtract 1 from each side. Let's do it!
This action results in the equation x + 1 = 9. Therefore x+2 = 10 and x + 1 = 9 are equivalent equations by the Subtraction Property of Equality.
Equation IV: x+3=12
Finally, let's consider the equation x + 3 = 12. The left-hand side x + 3 is equal to x + 2 + 1. Let's add 1 to each side of the equation x + 2 = 10. We want to check if we will get the equation we are considering.
Thanks to the Addition Property of Equality, we can note that the equations x +2 = 10 and x + 3 = 11 are equivalent. Wait a minute! It is 12 and not 11 on the right-hand side of our equation. This makes things a bit more fun. Let's take a closer look. x + 3 = 12 x + 3 = 11 These two equations have different solutions. This means that they are not equivalent equations. Therefore, x+3=12 and x+2=10 are not equivalent equations either.
Conclusion
We found that two of the given equations, x+6=14 and x+1=9, are equivalent to the equation x+2=10.
We are asked to choose all the equations that are equivalent to the given equation d + 8 = 1.
d + 3 &= 0
d+10 &= 11
d+7 &= 0
d+1 &= - 6
Two equations are equivalent when they have the same solutions. We can check if two equations are equivalent by solving them and comparing their solutions. Let's start by solving the given equation.
d + 8 = 1
Here, 8 is added to the variable d. We can isolate the variable on one side by undoing
the operations applied to the variable using inverse operations. The inverse operation of addition is subtraction, so we will subtract 8 from both sides of the equation using the Subtraction Property of Equality.
The solution to the given equation is d = - 7. Now, let's solve each of the remaining equations. Remember, when a number is added to the variable, we use the Subtraction Property of Equality to isolate the variable on one side.
| Equation | Apply Properties of Equality | Solution |
|---|---|---|
| d+3 = 0 | d + 3 - 3 = 0 - 3 | d = - 3 |
| d + 10 = 11 | d + 10 - 10 = 11 - 10 | d = 1 |
| d + 7 = 0 | d + 7 - 7 = 0 - 7 | d = - 7 |
| d + 1 = - 6 | d + 1 - 1 = - 6 - 1 | d = - 7 |
The equations d + 7 = 0 and d + 1 = - 6 have the same solution as the given equation. This means that these two equations are equivalent to the given equation. The remaining equations have different solutions, so they are not equivalent to the given equation.
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Writing and Solving One-Step Addition and Subtraction Equations
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