1. Topic 2: Algebraic Expressions/Equations Lesson 6
Corresponds to Section 1.5

2. Learning Targets
Define the following vocabulary terms: open sentence, equation, solving an equation, solution, replacement set, set, element, solution set, identity
Use a replacement set to solve an equation
Solve an equation with one variable
Solve an equation with multiple variables
Create an equation with a unique solution, no solution, and many solutions

3. Open Sentence vs Equation
Open Sentence: Mathematical statement that contains expressions and symbols EX: 3𝑥+7
Equation:
Mathematical sentence that contains an equal sign
EX: 3𝑥+7=13
Key Question: Compare and Contrast Expressions and Equations. Connect it to a non-mathematical idea

4. Components of an Equation
Solving an Equation:
The process of finding a value(s) for a variable that makes the sentence true.
EX: 𝑥+5=8
Replacement Set:
A set of numbers where replacements of a variable may be chosen
EX: {2, 4, 6, 8}
Set: a collection of objects or numbers
EX: {2, 5, 9, 10, 11}
EX: …−2, −1, 0, 1, 2, …
Element: each object/member in the set
EX: 2 is an element of the set {2, 5, 9, 10, 11}
EX: -10 is an element of the set {…−2, −1, 0, 1, 2, …}
Equation
Solution:
A specific value that replaces the variable to make the sentence true.
EX: 𝑥=3
Solution Set:
Set of elements from the replacement set that make an equation true.
EX: {3}

5. Example 1: Replacement Set
Find the solution set of the equation 2𝑞+5=13 if the replacement set is {2, 3, 4, 5, 6}
1. Create a Table
2. Replace 𝑞 with each value in the replacement set
Solution Set: {4} 𝒒
𝟐𝒒+𝟓=𝟏𝟑
True or False? 2
2 2 +5=9≠13
False 3
2 3 +5=11≠13 4
2 4 +5=13
True 5
2 5 +5=15≠13 6
2 6 +5=17≠13

6. Example 2: Solving Equations with One Variable
Solve the equation 7− 4 2 −10 +𝑛=10
1. 7− 16−10 +𝑛=10
2. 7− 6 +𝑛=10
3. 1+𝑛=10
4. 𝑛=9
5. This equation has one unique solution of 9.

7. Example 3: Solving Equations with One Variable
Solve the equation 𝑛 =5𝑛+(10−3)
1. 𝑛 5 +6=5𝑛+(7)
2. 5𝑛+6=5𝑛+7
3. No matter what value we substitute for 𝑛, the equation will never be true. Thus, the equation has no solution.

8. Example 4: Solving Equations with One Variable
Solve the equation 2∙5−8 3ℎ+6 = 2ℎ+ℎ +6 2
1. 10−8 3ℎ+6 = 2ℎ+ℎ +6 2
2. 2 3ℎ+6 = 2ℎ+ℎ +6 2
3. 6ℎ+12= 2ℎ+ℎ +6 2
4. 6ℎ+12= 3ℎ +6 2
5. 6ℎ+12=6ℎ+12
6. Since both sides are the same, this equation’s solution is all real numbers. In other words, any number will make it true.

9. Identity An equation that is true for every value of the variable.
EX: 6ℎ+12=6ℎ+12

10. Example 5: Solve the equation 3 𝑏+1 −5=3𝑏−2 1. 3𝑏+3−5=3𝑏−2
2. 3𝑏−2=3𝑏−2
3. Solution Set: All Real Numbers
4. This is an identity

11. Example 6: Solve the equation 2𝑑+ 2 3 −5 =10 5−2 +𝑑 12÷6
1. 2𝑑+ 8−5 =10 5−2 +𝑑(12÷6)
2. 2𝑑+ 3 =10 5−2 +𝑑(12÷6)
3. 2𝑑+3=10 3 +𝑑(12÷6)
4. 2𝑑+3=10 3 +𝑑(2)
5. 2𝑑+3=30+2𝑑
6. 2𝑑+3=2𝑑+30
7. Solution Set: No Solutions

12. Example 7: 2 Variables(pg 35 #5)
Mr. Hernandez pays $10 a month for movies delivered by mail. He can also rent movies in the store for $1.50 per title.
A) Write an equation that describes the totally amount Mr. H pays each month. Let 𝐶 be the total cost and 𝑚 be the number of movies.
C = 1.5m + 10

13. Example 7 continued(pg 35 #5)
Mr. Hernandez pays $10 a month for movies delivered by mail. He can also rent movies in the store for $1.50 per title.
B) How much did Mr. H spend this month if he rents 3 movies from the store?
C = 1.5(3) + 10 = $14.50

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