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CHAPTER 1: ANALYZE AND SOLVE LINEAR EQUATIONS (3 WEEKS) ............................................................. 1 SECTION 1.1: SIMPLIFY AND SOLVE LINEAR EQUATIONS IN ONE VARIABLE, APPLICATIONS ........................................... 3
1.1a Classwork: Review Simplify Linear Expressions ................................................................................................... 4 1.1a Homework: Review Simplify Linear Expressions ................................................................................................... 7 1.1b Anchor Problem: Chocolate Problem .................................................................................................................... 9 1.1c Classwork: Solve Linear Equations in One Variable (simplify by combining terms) .......................................... 10 1.1c Homework: Solve Linear Equations in One Variable (simplify by combining terms) ......................................... 13 1.1d Classwork: Solve Linear Equations in One Variable (simplify using distribution) ............................................. 15 1.1d Homework: Solve Linear Equations in One Variable (simplify using distribution) ............................................ 17 1.1e Classwork: Applications Part 1 ............................................................................................................................ 21 1.1e Homework: Applications Part 1 ........................................................................................................................... 24 1.1f Classwork: Solve Linear Equations in One Variable (simplify with variables on both sides) ............................. 27 1.1f Homework: Solve Linear Equations in One Variable (simplify with variables on both sides) ............................. 29 1.1g Classwork: Solve Linear Equations in One Variable (multi-step) ....................................................................... 30 1.1g Homework: Solve Linear Equations in One Variable (multi-step) ...................................................................... 33 1.1h Classwork: Applications Part 2 ............................................................................................................................ 35 1.1h Homework: Applications Part 2 ........................................................................................................................... 37 1.1i Self-Assessment: Section 1.1 ................................................................................................................................. 39
SECTION 1.2: SOLVE LINEAR EQUATIONS IN ONE VARIABLE (SPECIAL CASES) .............................................................. 40 1.2a Anchor Problem: Chocolate Problem ................................................................................................................. 41 1.2a Homework: Review of Multi-step Solving ............................................................................................................ 42 1.2b Classwork: Solve Linear Equations in One Variable (special cases) .................................................................. 43 1.2b Homework: Solve Linear Equations in One Variable (special cases) ................................................................. 47 1.2c Self-Assessment: Section 1.2 ................................................................................................................................. 49
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Chapter 1: Analyze and Solve Linear Equations (3 weeks) Utah Core Standard(s):
• Solve linear equations in one variable. (8.EE.7) a) Give examples of linear equations in one variable with one solution, infinitely many solutions, or no
solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form 𝑥 = 𝑎,𝑎 = 𝑎, or 𝑎 = 𝑏 results (where a and b are different numbers).
b) Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
Academic Vocabulary: linear expression, simplify, linear equation, solve, solution, like terms, distributive property, no solution, infinitely many solutions Chapter Overview: This chapter begins with a review of simplifying and writing expressions and then moves into solving multi-step linear equations in one variable. The chapter includes equations with one solution, no solution, and infinitely many solutions. Students use algebra tiles to model, simplify, and solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms. While working with the concrete representation of an equation, students are simultaneously manipulating the symbolic representation of the equation. By the end of the chapter, students should be fluently solving multi-step equations represented symbolically. They should feel comfortable with the laws of algebra that allow them to simplify expressions and the properties of equality that allow them to transform a linear equation into its simplest form, thus revealing the solution if there is one. While solving, students will become comfortable with the inverse operations that allow them to transform a linear equation into its simplest form. An important feature of allowable operations on equations is that they can be reversed. This chapter utilizes error analysis to highlight common mistakes that are made when solving equations. In the last section of the chapter, students look at equations with infinitely many or no solutions. They analyze what it is about the structure of the equation and the solving outcome that results in one solution, infinitely many solutions, or no solution. Applications are interwoven throughout the chapter in order that students realize the power of being able to write and solve a linear equation to solve real world problems. The ability to be able to solve real world problems by writing and solving linear equations gives purpose to the skills students are learning in this chapter. Connections to Content: Prior Knowledge In previous coursework, students used properties of operations to generate equivalent expressions, including those that require expansion using the distributive property. Students solved one- and two-step equations. Students have solved real-life mathematical problems using numerical and algebraic expressions and equations. Future Knowledge As students move on in this course, they will begin to study linear functions. They should understand that when working with linear equations in one variable, the variable takes on a specific value as studied here. As they move to the study of linear functions, they will see that functions can take on an infinite number of values. They will analyze and solve pairs of simultaneous linear equations, including systems with one solution, no solution, and infinitely many solutions. They will also write and solve linear equations in two-variables to solve real-world problems.
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MATHEMATICAL PRACTICE STANDARDS (emphasized):
Make sense of problems and persevere in solving them.
At Discovery Preschool, parents who have two students enrolled get a discount on the second child. The second child’s tuition is 10 dollars less per day than the first child’s. If Tess has her two children enrolled for 5 days and her total bill for both children is $200, how much does she pay each day for her second child to attend daycare? Consider the following situation. Then answer the questions below. Include any pictures, tables, equations, or graphs that you used to solve the problem. Two students, Arthur and Oliver, each have some chocolates. They know that they each have the same number of chocolates. Arthur has two tubs of chocolates, one bag of chocolates, and twenty-five remaining chocolates. Oliver has two tubs of chocolates, two bags of chocolates, and seven remaining chocolates. Can you determine how many chocolates are in a tub? Can you determine how many chocolates are in a bag? To the extent possible, figure out how many chocolates are in a tub and how many chocolates are in a bag.
Reason abstractly and quantitatively.
Write the word problem that goes with the equation. Then solve for the unknown information and verify your answer.
A Trip to the Fair Cost of a pony ride: b Cost to ride the Ferris wheel: !
!𝑏
Cost to bungee jump: 2𝑏 + 5
Construct viable arguments and critique the reasoning of others.
Error Analysis: Ricardo solved the following equation incorrectly. Circle the mistake and describe the mistake in words. Then, solve the equation correctly.
Combine like terms (2x, x, and 5x).
Divide both sides by 7. What is it about the structure of an expression that leads to one solution, infinitely many solutions, or no solution? Provide examples to support your claim.
Model with Mathematics.
You burn approximately 230 calories less per hour if you ride your bike versus go on a run. Lien went on a two-hour run plus burned an additional 150 calories in his warm-up and cool down. Theo went on a 4 hour bike ride. If Lien and Theo burned the same amount of calories on their workouts, approximately how many calories do you burn an hour for each type of exercise? Write the word problem that goes with the equation in the problem below. Then solve for the unknown information and verify your answer. Fast Food Calories Calories in a Big Mac: c Calories in a Grilled Chicken Sandwich: c – 200
Calories in a 6-piece chicken nugget:
Look for and make use of structure
Write an expression for the amount of money Peter earned last week. Peter is paid p dollars per hour he works. For every hour he works over 40 hours, he is paid an additional $10 per hour. Peter worked 46 hours last week. Consider the expression . Write 3 different expressions that if set equal to would result in the equation having infinite solutions.
13 4 (2 5) 332
b b b⎛ ⎞+ + + =⎜ ⎟⎝ ⎠
2 5 56x x x+ + =7 56x =8x =
2c
( 200) 11752cc c+ − + =
4 12a − 4 12a −
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Section 1.1: Simplify and Solve Linear Equations in One Variable, Applications Section Overview: This section begins with a review of writing and simplifying algebraic expressions. Then students move into writing and solving linear equations in one variable. Students start with linear equations whose solutions require collecting like terms. They then move onto equations whose solution requires the use of the distributive property and collecting like terms. Following these lessons, students will write word problems to match equations and they will write equations to match word problems. The equations will be in the same form that students studied in prior lessons (those whose solutions require the use of the distributive property and collecting like terms). As students continue in this section, they will move to equations that contain variables on both sides of the equation. Finally, they will put everything together and fluently solve multi-step equations with rational coefficients and variables on both sides that require expanding expressions using the distributive property and collecting like terms. In the last lesson, students will again write word problems to match equations and they will write equations to match word problems. The equations can take on any form of equation studied in this chapter. Concepts and Skills to Master: By the end of this section, students should be able to:
• Understand the meaning of linear expression and linear equation. • Simplify linear expressions, including those requiring expanding using the distributive property and
collecting like terms. • Write and simplify linear expressions that model real world problems. • Translate between the concrete and symbolic representations of an expression and an equation. • Solve multistep linear equations with rational coefficients, including equations whose solutions require
expanding expressions using the distributive property and collecting like terms. • Write and solve multistep linear equations that model real world problems.
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1.1a Classwork: Review Simplify Linear Expressions In this lesson, we will study linear expressions. A linear expression is a mathematical phrase consisting of numbers, variables (symbols that represent numbers), and arithmetic operations that describes a mathematical or real-world situation. The following are all examples of linear expressions:
• 3𝑥 − 5 • 2𝑥 − 𝑥 − 17 • 3𝑥 • 6 2𝑥 − 5 + 11 • 3𝑥 + 7𝑥 − 3+ 2𝑥 • 25
It is often the case that different linear expressions have the same meaning. For example, 𝑥 + 𝑥 and 2𝑥 have the same meaning. The linear expression 𝑥 + 𝑥 has been simplified to 2𝑥. If a linear expression is in the form 𝐴𝑥 + 𝐵 where 𝐴 and 𝐵 are numbers and 𝑥 represents an unknown, then the linear expression is simplified. In the following examples, we will simplify linear expressions to the form 𝐴𝑥 + 𝐵. We will start by using tiles to model and simplify linear expressions. Key for Tiles: = 1
= x
= –1
= –x
Remember that a positive tile and a negative tile can be combined to find a zero pair or add to zero. + = 0
+ = 0
1. The following is a model of the expression
Find zero pairs, and write the simplified form of this expression. 2𝑥 + −2 or 2𝑥 − 2
5 3 6 4x x+ − − +
4 –6 –3x 5x
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2. Model the expression . Then find zero pairs and simplify the expression. When students begin modeling using the tiles, it may help to have them lay out the tiles as a straight transliteration of the equation and then commute the tiles so that like terms are collected before drawing the model on their paper. This will make it easier for them to find zero pairs and simplify the expression. 𝑥 + 2
3. The following is a model of the expression 3(𝑥 + 1)
Write the simplified form of this expression. 3𝑥 + 3
Directions: Model and simplify each expression.
4.
𝑥 + 4
5. 𝑥 + −9 or 𝑥 − 9
Directions: Simplify each expression.
6. −4𝑥 + 3+ 5(2𝑥 − 1) 6𝑥 + (−2)
8.
32 𝑎 + (−6.8)
10. Write three expressions that are equivalent to 6𝑥 + 12. Possible expressions: 6 𝑥 + 2 , 2 3𝑥 +6 , 5𝑥 + 𝑥 + 12
7. 3𝑥 + 14
9. 2.5𝑦 + 4.25
4 3 5 1x x− + + + −
2( 2)x x+ − 3 3 4( 3)x x− + −
1 4.3 2.52a a− + −
12 ( 2) 4x x− − +
0.5( 1.5) 2 5y y− + +
x + 1
x + 1
x + 1
There are 3
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Linear expressions can be used to model many real world contexts. For example, if you buy 3 pieces of chicken that cost x dollars each, a slice of pie for $3, and a drink for $2, we can write an expression that represents the total amount you spent: 3𝑥 + 3+ 2. The simplified form of this expression would be 3𝑥 + 5. Directions: Write a linear expression that models each of the following situations. Then, simplify the
expression.
11. An expression that represents the total amount made in one week. A salesperson gets a base salary of $300 per week plus $20 for each item they sell. They sell x items in one week. The salesperson also spends $40 a week on travel expenses. 300+ 20𝑥 − 40 260+ 20𝑥
12. An expression that represents the total amount spent. Sara bought 3 baby outfits and one bottle of baby lotion. The baby lotion costs x dollars. Each outfit costs $2 more than the baby lotion. 3 𝑥 + 2 + 𝑥 4𝑥 + 6
13. An expression that represents the amount of money remaining. Tim took his friends to the movies. He started with $40 and bought 3 movie tickets that each cost x dollars. He also bought one tub of popcorn that cost $5.75. 40− 3𝑥 − 5.75 34.25− 3𝑥
14. An expression that represents the perimeter of the basketball court. In the NBA, the length of the basketball court is 44 feet longer than the width. The width of the court is w. 2𝑤 + 2(𝑤 + 44) 4𝑤 + 88
This is a good problem for discussing how the structure of this expression in its original form reveals more about the context than the simplified form of the expression.
15. An expression for the amount of money Peter earned last week. Peter is paid p dollars per hour he works. For every hour he works over 40 hours, he is paid an additional $10 per hour. Peter worked 46 hours last week. 40𝑝 + 6(𝑝 + 10) 46𝑝 + 60
This is a good problem for discussing how the structure of this expression in its original form reveals more about the context than the simplified form of the expression.
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1.1a Homework: Review Simplify Linear Expressions
1. The following is a model of the expression −6𝑥 + 2𝑥 − 5+ 2 Find zero pairs, and write the simplified form of this expression. −4𝑥 − 3
2. Model the expression 5𝑥 + 2− 𝑥 − 4. Then find zero pairs and simplify the expression. 4𝑥 − 2
3. The following is a model of the expression 2(3𝑥 − 1)
Write the simplified form of this expression. 6𝑥 − 2 Directions: Model and simplify each expression.
4. 2 𝑥 + 3 + 4𝑥 6𝑥 + 6
5. 4+ 2 𝑥 − 2 − 4 2𝑥 − 4
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Directions: Simplify each expression.
6. 6 2𝑥 − 4 − 3𝑥 9𝑥 − 24
7. 4− 2𝑥 + 3 + 5𝑥 3𝑥 + 1
8. −6𝑥 − 4+ 5𝑥 + 7 −𝑥 + 3
10. Write three expressions that are equivalent to 3𝑥 + 15. Possible expressions: 3 𝑥 + 5 ,−3 −𝑥 − 5 , 2𝑥 + 𝑥 + 15, 7𝑥 +−4𝑥 − 4+ 19
9. !!4𝑥 + 12 + 2𝑥
4𝑥 + 6
Directions: Write a linear expression that models each of the following situations. Then, simplify the expression.
11. An expression that represents the total cost of the concert tickets. Master Tickets charges $35 for each concert ticket, plus an additional $2 service fee for each ticket sold. x tickets were purchased. 35𝑥 + 2𝑥 37𝑥
12. The total number of baseball cards Jayant has. Jayant bought 3 packs of baseball cards. He already has 35 baseball cards in his collection. Each pack of cards has x cards in it. 3𝑥 + 35
13. An expression that represents the amount of money remaining. Lisa went shopping for decorations for
her birthday party. She started with $20. She bought 4 bags of balloons that cost x dollars each, 2 rolls of streamers that cost $2.25 each, and 1 pack of hats that cost $4.50. 20− 4𝑥 − 2 2.25 − 4.50 11− 4𝑥
14. An expression that represents the perimeter of the tennis court. An official tennis court has a width that is 42 feet shorter than the length. The length of the court is l. 2𝑙 + 2 𝑙 − 42 4𝑙 − 84
15. An expression for the total amount spent. At Six North Shoe Store, a pair of boots is $30 more than a
pair of shoes. Emily purchased 2 pairs of boots and 3 pairs of shoes. The cost of one pair of shoes is s. 2 𝑠 + 30 + 3𝑠 5𝑠 + 60
16. An expression for the amount of money Naja earned last week. Naja is paid p dollars per hour she works. For every hour she works over 40 hours, she is paid time and a half which means she is paid 1.5 times her normal hourly rate. She worked 50 hours last week. 40𝑝 + 10(1.5𝑝) 55𝑝
This is a good problem for discussing how the structure of this expression in its original form reveals more about the context than the simplified form of the expression.
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1.1b Anchor Problem: Chocolate Problem Consider the following situation. Then answer the questions below. Include any pictures, tables, equations, or graphs that you used to solve the problem. Two students, Theo and Lance, each have some chocolates. They know that they have the same number of chocolates. Theo has four full bags of chocolates and five loose chocolates. Lance has two full bags of chocolates and twenty-nine loose chocolates. Can you determine how many chocolates are in a bag? Can you determine how many chocolates each child has? Teacher Notes: 1. Start by engaging students by presenting the situation in an encouraging way. 2. Make sure that students understand that the bags are full and that each bag has the same number of
chocolates in it. 3. Have the students work in pairs or groups to answer the problem. Anticipate the correct answer for
chocolates in a bag (12 chocolates) and be ready to help students who don’t get that answer. 4. Have student share their work. First, have some show their thinking about the bags. Choose different
methods such as a picture, table, or equation and make connections between them. 5. Have students share how they determined how many chocolates each child has.
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1.1c Classwork: Solve Linear Equations in One Variable (simplify by combining terms) In the previous section, we studied linear expressions. When we set two linear expressions equal to each other, we have created a linear equation. A linear equation is a statement that two linear expressions are equal to each other. Let’s look at an example from the previous section: A salesperson gets a base salary of $300 per week plus $20 for each item they sell. They sell x items in one week. The salesperson also spends $40 a week on travel expenses. We can write an expression that represents the total amount made in one week: 300+ 20𝑥 − 40
If we know that the salesperson made $860 last week, we can set these two expressions equal to each other to create a linear equation:
300+ 20𝑥 − 40 = 860 We can then solve this equation to determine how many items the salesperson sold last week. When this equation is solved, the solution is 𝑥 = 30, which means the salesperson sold 30 items last week. Verify this by plugging in 30 for x in the equation above. 300+ 20 30 − 40 860 It is important to note that when we create an equation, the two expressions on either side of the equal sign might not actually always be equal; that is, the equation might be true for some values of x and false for others. A solution to an equation is a number that makes the equation true when substituted for the variable. In the example of the salesperson, the linear equation is true for one value of x (𝑥 = 30) which represents the number of items the salesperson would have to sell to earn $860. The solution to this equation is 𝑥 = 30. In the first section of this chapter, we will look at examples of linear equations that are true for one value of x as we saw in the sales example above. We will learn how to model and solve these types of linear equations. It might also be the case that an equation has no solutions or an equation may have infinite solutions (every number is a solution). We will examine these two cases is section 1.2.
Reiterate for students that a number is considered a linear expression so in this case, 860 is a linear expression.
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1. The following is a model of the equation . Create this model with your tiles and solve the equation, showing your solving actions (steps) below.
a. Solving Actions:
Students should show their solving actions in this space as they manipulate the tiles. 𝑥 = 4
b. Verify your solution in the space below. Students can verify the solution by substituting into the equation. Alternatively, they can label the pieces of the model above by writing the values on each of the tiles and verifying that the two sides are equal.
Directions: Model and solve the following equations. Show your solving actions and verify your solution.
2. 𝑥 = 1
3. 𝑥 = 3
4. 𝑥 = 4
5 8 2 4x x− − =
5 8 2 4x x− − =
4 3 1 6x x+ − =
10 3 4x x= − + +
10 2 3 4 5x x= − − + +
–
–
–
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5. The following is a model of an equation.
Write the symbolic representation for this model. Solve the equation. 3𝑥 + 𝑥 + 6 = 2 𝑥 = −1 Directions: Solve the following equations without the use of the tiles.
6. 𝑥 = 6
7. −6 = 𝑚
8. 𝑏 = −20
9. − !!= − !
!+ 6𝑟
𝑟 =19
10. Error Analysis: Ricardo solved the following equation incorrectly. Circle the mistake and describe the
mistake in words. Then, solve the equation correctly.
Combine like terms (2x, x, and 5x).
Divide both sides by 7.
11. Write an equation that contains like terms and has a solution of x = 2. Answers will vary. One possible solution: 3𝑥 + 5𝑥 = 16. Students may construct their equation by trial and error. Another approach is to build the equation. Here is how the equation shown above was built: 𝑥 = 2 3𝑥 = 6 Multiply both sides by 3 3𝑥 + 5𝑥 = 16 Add equivalent expressions for 10 to both sides (5x on the left and 10 on the right)
*Students should recognize that the same properties of equality that allow them to solve an equation are the same properties of equality that allow them to build an equation. They should also recognize that they can replace expressions in an equation with equivalent expressions and not change the nature of the original equation. You may wish to have students solve their neigbor’s equation to verify that the solution is 2.
7 5 3 9x x− + + = − 17 5 3m m= + −
0.5 2 50b b+ = −
2 5 56x x x+ + =
7 56x =
8x =
Ricardo did not realize that the coefficient in front of x is 1. When combined, the result is 8x. The correct solution is x = 7.
When students solve this equation using the tiles, they will most likely start by combining the 3x and x. The next logical step is to subtract 6 from both sides; however when they try to subtract 6 from the right side, there are not enough tiles. Students can create zero pairs on to the right side without changing the equation. By creating 4 zero pairs, students can now remove 6 tiles from the right side. Alternatively, students can choose to add 6 negative tiles to both sides. It should be made transparent to students that both methods are valid and that you are essentially doing the same thing.
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1.1c Homework: Solve Linear Equations in One Variable (simplify by combining terms) Directions: Solve the following equations. Verify your solutions. 1. 𝑥 = 6
2. 𝑥 = 3
3. 𝑎 = −5
4. 𝑥 = −4
5. 𝑡 = 8
6.
𝑥 = 45
7. 𝑏 = 20
8. 𝑦 = −29
9.
𝑟 = 3
10. The following is a model of an equation.
Write the symbolic representation for this model. Solve the equation. 2𝑥 + −5𝑥 + 5 = −7 𝑥 = 4
7 3 24x x− = 2 9 17 4x x− + = − 5 8 ( 2 ) 7a a+ + − = −
2 7 4 17x x− − − = 6 4 3 2t t= − − 2 1 95 5x x− =
1.3 0.7 12b b− = 2.5 0.4 0.1y y− = + 1 1 13 14 4 2
r r= − +
–
–
–
–
–
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Directions: In #11 – 12, a common mistake has been made. Circle the mistake and describe the mistake in words. Then, solve the equation correctly. 11.
Combine like terms (2x and 4x).
Combine like terms (6x and –2).
Divide by 4.
Explanation of Mistake: 6x and –2 are not like terms.
Solve Correctly: 𝑥 = !!!
12.
Combine like terms (3x and –8x).
Add 5 to both sides.
Divide both sides by 5.
Explanation of Mistake: When 3x and –8x are combined, the result is –5x, not 5x. Solve Correctly: 𝑥 = −3
13. Write an equation that contains like terms and has a solution of x = 1. Answers will vary. One possible solution: 3𝑥 − 𝑥 + 5 = 7 14. Write an equation that contains like terms and has a solution of x = –3. Answers will vary. One possible solution: 2𝑥 + 3− 4𝑥 − 1 = 8 Students can exchange equations and solve.
2 4 2 20x x+ − =
6 2 20x − =
4 20x =
5x =
3 8 5 10x x− − =
5 5 10x − =
5 15x =
3x =
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1.1d Classwork: Solve Linear Equations in One Variable (simplify using distribution) 1. The following is a model of the equation . Create this model with your tiles and solve the
equation, showing your solving actions below.
Solving Actions:
𝑥 = 3
Directions: Model and solve the following equations.
2. 3. 4. 𝑥 = 2 𝑥 = 2 𝑥 = −2
As in the previous problem, there are two valid ways to solve these problems. Talk about when it makes sense to start with distributing and when it makes sense to start by dividing.
3( 1) 12x + =
3( 1) 12x + =
2( 5) 14x + = 2(3 1) 2 10x x+ − = 3( 2) 12x − = −
There are two valid ways to solve this problem and students should be exposed to both. The first way is to distribute the 3 to the (𝑥 + 1) and continue solving. Alternatively, one can start by dividing both sides by 3. Show students on the model what happens when you divide both sides by 3. Talk about when it makes sense to start with distributing and when it makes sense to start by dividing.
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5. The following is a model of an equation.
Write the symbolic representation for this model. Solve the equation. 4 𝑥 − 4 = 8 𝑥 = 6 Directions: Solve the following equations without the use of the tiles.
6. 𝑥 = −5
7. 𝑥 = −3
8. 𝑎 = 2
9. 𝑡 = −10
10. Write an equation that contains parentheses and has a solution of x = 5. Answers will vary. One possible solution: 2 𝑥 + 3 = 16. Students may construct their equation by trial and error. Another approach is to build the equation. Here is how the equation shown above was built: 𝑥 = 5 𝑥 + 3 = 8 2(𝑥 + 3) = 16 *Students should recognize that the same properties of equality that allow them to solve an equation are the same properties of equality that allow them to build an equation. They should also recognize that they can replace expressions in an equation with equivalent expressions and not change the nature of the original equation.
2( 1) 8x− + = 3( 4) 8 13x− − − =
5 2(3 1) 15a+ − = 1 (2 4) 82t + = −
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1.1d Homework: Solve Linear Equations in One Variable (simplify using distribution) 1. The following is a model of an equation.
Write the symbolic representation for this model. Solve the equation. 3 2𝑥 + 1 − 2𝑥 = 15 𝑥 = 3 2. The following is a model of an equation.
Write the symbolic representation for this model. Solve the equation. 2 𝑥 − 5 + 3𝑥 = −7
𝑥 =35
– – – – –
–
–
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Directions: Solve the following equations. Verify your solutions. 1. 𝑥 = 3
2. 𝑥 = −4
3. 𝑥 = −4
4. 𝑥 = −8
5. 𝑡 = −1
6. 𝑥 = 3
7. 𝑎 = 12
8. 𝑥 = 2
9. 𝑥 = 15
10. 𝑥 = 10
11.
𝑥 = −3
12. 𝑏 = 2
13. 𝑥 = −2
14. 0.2 10𝑡 − 4 − 𝑡 = 1.2 𝑡 = 2
15.
𝑥 = 21
3(4 2) 30x − = 4(2 2 ) 24x+ = − 16 2(4 8)x− = +
3( 10) 5 11x + + = 3 2 5 1t t t− + − = − 24 2(1 5 ) 4x− = − +
2( 3) 4 18a a− + + = 5 3( 4) 28x x+ + = 4 3( 2) 21x x− − =
0 2( 5) 3x x= − + + 1 ( 6) 13x + =
5 4(2 5) 3 15b b− − + =
10 3( 2) 2(5 1)x x= − − − 126 ( 7) 227
x= − − − +Draf
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Directions: In the following problems, a common mistake has been made. Circle the mistake and describe the
mistake in words. Then, solve the equation correctly.
16.
Distribute the 2.
Combine like terms (7 and 4).
Subtract 11 from both sides.
Divide both sides by 6.
Simplify the fraction.
Explanation of Mistake: The 2 was only distributed to the 3x and not to the entire quantity in the parentheses. Solve Correctly: 𝑥 = −2
17.
Distribute the –2.
Combine like terms (–6 and 5).
Add 1 to both sides.
Divide both sides by –2. Explanation of Mistake: When the –2 was distributed to (y – 3), the result should have been –2y + 6.
Solve Correctly: 𝑦 = 4
7 2(3 4) 3x+ + =
7 6 4 3x+ + =
11 6 3x+ =
6 8x = −
86
x −=
43
x −=
2( 3) 5 3y− − + =
2 6 5 3y− − + =
2 1 3y− − =
2 4y− =
2y = −
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18.
Distribute the negative sign.
Combine like terms (5 and –7).
Add 2 to both sides.
Divide both sides by –2. Explanation of Mistake: The negative sign was only distributed to the 2x and not the –7.
Solve Correctly: 𝑥 = −1
19.
Distribute the
Subtract 5 from both sides.
Divide both sides by 3.
Explanation of Mistake: To undo the !
! in front of the x, you need to divide by !
! which is the same as
multiplying by 3, not dividing by 3.
Solve Correctly: 𝑥 = 18
20. Write an equation that contains parentheses and has a solution of m = –6. Answers will vary. One possible solution: 3 2𝑚 − 3 = −45 Students may construct their equation by trial and error. Another approach is to build the equation. Here is how the equation shown above was built: 𝑚 = −6 2𝑚 = −12 Multiply both sides by 2 2𝑚 − 3 = −15 Subtract 3 from both sides 3(2𝑚 − 3) = −45 Multiply both sides by 3. *Students should recognize that the same properties of equality that allow them to solve an equation are the same properties of equality that allow them to build an equation. They should also recognize that they can replace expressions in an equation with equivalent expressions and not change the nature of the original equation. You may wish to have students solve their neigbor’s equation to verify that the solution is -6.
5 (2 7) 14x− − =
5 2 7 14x− − =
2 2 14x− − =
2 16x− =
8x = −
1 ( 15) 113x + =
1 5 113x + = 1
3
1 63x =
2x =
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1.1e Classwork: Applications Part 1 Directions: Write the word problem that goes with the equation in each problem. Then solve for the unknown
information and verify your answer.
1. The Cost of Lunch Cost of a pear: x Cost of juice: 1.5x Cost of a sandwich: x + 1.40
The cost of a lunch that consists of a pear, a juice, and a sandwich is $3.50. The juice is 1.5 times the cost of the pear and the sandwich costs $1.40 more than the pear. What is the cost of each item in the lunch? The cost of a pear is $0.60. The cost of a juice is $0.90. The cost of a sandwich is $2. Ask students how they can verify that their answer is correct.
2. Ages Talen’s age now: t Peter’s age now: 8𝑡 + 3
The sum of Talen’s and Peter’s ages is 39. Peter’s age is 3 more than eight times Talen’s age. How old is each person? Talen is 4 years old and Peter is 35 years old. Ask students how they can verify that their answer is correct.
3. Angles 𝑚∠𝐴: a 𝑚∠𝐵: 4𝑎 − 5
Two angles are complementary. ∠𝐵 is 5 less than 4 times 𝑚∠𝐴. What is the measure of each angle? The 𝑚∠𝐴 is 19 degrees and the 𝑚∠𝐵 is 71 degrees. Ask students how they can verify that their answer is correct.
1.5 ( 1.40) 3.50x x x+ + + =
(8 3) 39t t+ + =
(4 5) 90a a+ − =
For many of these problems, it will help students to first draw a picture of the situation.
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4. Rectangles Width of a rectangle: 𝑤 Length of a rectangle: (𝑤 + 5) 𝑤 + 𝑤 + 5 + 𝑤 + 𝑤 + 5 = 46 ft. The perimeter of a rectangle is 46 ft. The length is 5 feet longer than the width. What are the dimensions of the rectangle? The width of the rectangle is 9 ft. and the length of the rectangle is also 14 ft. Ask students how they can verify that their answer is correct.
5. Triangles 𝑚∠𝐴: 𝑥 𝑚∠𝐵:3𝑥 𝑚∠𝐶: 𝑥 − 20 𝑥 + 3𝑥 + 𝑥 − 20 = 180 The sum of the angles in a triangle is 180 degrees. The 𝑚∠𝐵 is three times greater than the 𝑚∠𝐴 and the 𝑚∠𝐶 is 20 degrees less than the 𝑚∠𝐴. What is the measure of each angle? The 𝑚∠𝐴 is 40 degrees, the 𝑚∠𝐵 is 120 degrees, and the 𝑚∠𝐶 is 20 degrees. Ask students how they can verify that their answer is correct.
6. Making Lemonade Cups of water to make a batch of lemonade: c Cups of sugar to make a batch of lemonade: !
!𝑐
Cups of lemon juice to make a batch of lemonade: !!𝑐
Marcia uses a total of 14 cups of water, sugar, and lemon juice to make a batch of lemonade. How many cups of each of the ingredients does she use? Marcia used 8 cups of water, 2 cups of sugar, and 4 cups of lemon juice. Ask students how they can verify that their answer is correct.
7. Fast Food Calories Calories in a Big Mac: c Calories in a Grilled Chicken Sandwich: c – 200
Calories in a 6-piece chicken nugget:
The total number of calories in a Big Mac, Grilled Chicken Sandwich, and 6-piece chicken nugget is 1,175. How many calories are in each food item? There are 550 calories in a Big Mac, 350 calories in a Grilled Chicken Sandwich, and 275 calories in a 6-piece chicken nugget. Ask students how they can verify that their answer is correct.
1 1 144 2
c c c+ + =
2c
( 200) 11752cc c+ − + =
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Directions: Write an expression for each unknown quantity in the word problem. Then write an equation for
each problem. Then solve your equation and answer the question in a complete sentence.
8. In a triangle, angle A is 3 times larger than angle B. Angle C is 20 degrees larger than Angle B. The sum of the angles is 180 degrees. What is the measure of each angle?
𝑚∠𝐴:3𝑏 𝑚∠𝐵: 𝑏 𝑚∠𝐶: 𝑏 + 20 3𝑏 + 𝑏 + 𝑏 + 20 = 180
9. The width of a rectangle is five less than three times the length of the rectangle. If the perimeter of the rectangle is 70 ft., what are the dimensions of the rectangle?
width of the rectangle: 3𝑙 − 5 length of the rectangle: 𝑙 2 3𝑙 − 5 + 2𝑙 = 70
10. The sum of three consecutive integers is 84. Find the three integers. integer 1: 𝑛 integer 2: 𝑛 + 1 integer 3: 𝑛 + 2 𝑛 + 𝑛 + 1 + 𝑛 + 2 = 84
11. Afua got an 90% on her first math exam, a 76% on her second math exam, and a 92% on her third math exam. What must she score on her fourth exam to have an average of 88% in the class? Score on Exam 1: 90% Score on Exam 2: 76% Score on Exam 3: 92% Score on Exam 4: 94% 90+ 76+ 92+ 𝑥
4 = 88
12. Kelly works 40 hours a week as a nurse practitioner. She makes time and a half for every hour she works over 40 hours. If she worked 60 hours one week and made $2100, what is her hourly wage? Regular Hourly Pay: 𝑥 Overtime Hourly Pay: 1.5𝑥 Wages at Regular Hourly Pay: 40𝑥 Wages at Overtime Hourly Pay: 20(1.5𝑥) Total Wages: 40𝑥 + 20 1.5𝑥 = 2100
The 𝑚∠𝐵 is 32 degrees, the 𝑚∠𝐴 is 96 degrees, and the 𝑚∠𝐶 is 52 degrees.
The length of the rectangle is 10 feet and the width of the rectangle is 25 feet.
The three integers are 27, 28, and 29.
Afua must score a 94% to have an average of 88% in the class.
Kelly’s regular hourly pay is $30 and her overtime hourly pay is $45.
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1.1e Homework: Applications Part 1 Directions: Write the word problem that goes with the equation in each problem. Then solve for the unknown information and verify your answer. For some of the problems, it may help to start by drawing a picture. 1. Angles
𝑚∠𝐴: a 𝑚∠𝐵: 2a
Two angles are supplementary. The 𝑚∠𝐵 is twice the 𝑚∠𝐴. What is the measure of each angle?
The 𝑚∠𝐴 is 60 degrees and the 𝑚∠𝐵 is 120 degrees. Ask students how they can verify that their answer is correct.
2. Ages
Felipe’s age now: f Felipe’s sister’s age: 𝑓 − 6 Felipe’s mom’s age: 3𝑓 − 9 𝑓 + 𝑓 − 6 + 3𝑓 − 9 = 60 The sum of the ages of Felipe, his sister, and his mom is 60. Felipe’s sister is 6 years younger than Felipe. His mom’s age is 9 less than 3 times his age. How old is each person? Felipe is 15 years old, his sister is 9 years old, and his mom is 36 years old. Ask students how they can verify that their answer is correct.
3. A Trip to Disneyland Miles driven on the first day: m Miles driven on the second day: 2𝑚 Miles driven on the third day: 2𝑚 + 50
Kelly and her family drove to Disneyland which is 650 miles from their house. They drove twice as far on the second day than they did on the first. They drove 50 miles more on the third day than they did on the second day. How many miles did they drive each day? They drove 120 miles on the first day, 240 miles on the second day, and 290 miles on the third day. Ask students how they can verify that their answer is correct.
4. Triangles 𝑚∠𝑋: 𝑥 𝑚∠𝑌: 𝑥 𝑚∠𝑍:3𝑥
The sum of the angles in a triangle is 180 degrees. The 𝑚∠𝑋 and 𝑚∠𝑌 are congruent. The 𝑚∠𝑍 is three times the 𝑚∠𝑌. What is the measure of each angle? ∠𝑋 and ∠𝑌 both have a measure of 36 degrees and the 𝑚∠𝑍 is 108 degrees. Ask students how they can verify that their answer is correct.
2 180a a+ =
2 (2 50) 650m m m+ + + =
(3 ) 180x x x+ + =
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5. The Cost of Clothes Cost of a shirt: c Cost of a pair of jeans: 𝑐 + 12
The cost of 3 shirts and 1 pair of jeans is $124. A pair of jeans costs 12 dollars more than a shirt. How much does each item cost? A shirt costs $28 and a pair of jeans costs $40. Ask students how they can verify that their answer is correct.
6. A Trip to the Fair # of tickets for a pony ride: b # of tickets to ride the Ferris wheel: !
!𝑏
# of tickets to bungee jump: 2𝑏 + 5
It takes 33 tickets for three pony rides, 4 rides on the Ferris Wheel, and 1 bungee jump. The number of tickets it takes to ride on the Ferris Wheel is half the number of tickets it takes to take a pony ride. The number of tickets it takes for a bungee jump is 5 more than twice the number of tickets for a pony ride. How many tickets does it take for each activity at the fair? The number of tickets needed for a pony ride is 4, the number of tickets for a ride on the Ferris Wheel is 2, and the number of tickets for a bungee jump is 13.
Directions: Write an expression for each unknown quantity in the word problem. Then write an equation for each problem. Then solve your equation and answer the question in a complete sentence.
7. At Shoes for Less, a pair of shoes is $15 less than a pair of boots. Cho purchased three pairs of shoes and
two pairs of boots for $120. How much does a pair of boots cost? Cost of a pair of boots: 𝑏 Cost of a pair of shoes: 𝑏 − 15 3 𝑏 − 15 + 2𝑏 = 120
8. The length of a rectangle is 4x and the width is 5 ft. If the perimeter of the rectangle is 34 ft., find the length of the rectangle. Length of Rectangle: 4𝑥 Width of Rectangle: 5 ft 2 4𝑥 + 2 5 = 34
9. Central Lewis High School has five times as many desktop computers as laptops. The school has a total of 360 computers. How many of each type of computer does the school have? Number of laptops: 𝑙 Number of desktops: 5𝑙 5𝑙 + 𝑙 = 360
3 ( 12) 124c c+ + =
13 4 (2 5) 332
b b b⎛ ⎞+ + + =⎜ ⎟⎝ ⎠
A pair of shoes costs $18 and a pair of boots costs $33.
The length of the rectangle is 12 ft.
The school has 60 laptops and 300 desktops.
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10. The sum of three consecutive integers is –36. Find the three integers. Integer 1: 𝑛 Integer 2: 𝑛 + 1 Integer 3: 𝑛 + 2 𝑛 + 𝑛 + 1 + 𝑛 + 2 = −36
11. Hamir got a 98%, 87%, 92%, and 92% on his first four math tests. What score must he get on the fifth test to get a 93% in the class? Exam 1: 98 Exam 2: 87 Exam 3: 92 Exam 4: 92 Exam 5: x 98+ 87+ 92+ 92+ 𝑥
𝑥 = 93
12. At Discovery Preschool, parents who have two students enrolled get a discount on the second child. The second child’s tuition is 10 dollars less per day than the first child’s. If Tess has her two children enrolled for 5 days and her total bill for both children is $200, how much does she pay each day for her second child to attend daycare? 1st child daily tuition: 𝑡 2nd child daily tuition: 𝑡 − 10 5𝑡 + 5 𝑡 − 10 = 200
The three consecutive integers are -13, -12, and -11.
Hamir must get a 96% on his fifth test.
Tess pays $15 a day for her second child to attend daycare.
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1.1f Classwork: Solve Linear Equations in One Variable (simplify with variables on both sides)
1. The following is a model of the equation: . Create this model with your tiles and solve the equation, showing your solving actions below.
a. Solving Actions:
𝑥 = 5
b. How can you verify your solution? Students can verify the solution by substituting into the equation. Alternatively, they can label the pieces of the model above by writing the values on each of the tiles and verifying that the two sides are equal.
Directions: Model and solve the following equations.
2. 3. 4. 𝑥 = 4 𝑥 = 4 𝑥 = 5
5 2 3 12x x+ = +
5 2 3 12x x+ = +
2 4x x= + 3 3 2 7x x+ = + 10 2 5x x+ = +
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5. 6. 7. 𝑥 = −4 𝑥 = 2 𝑥 = 3
Directions: Solve the following equations without the use of the tiles.
8. 𝑥 = 6
9. 𝑥 = 6
10. 𝑥 = 6
11. 𝑥 = 0.2
12. 𝑥 = 1 13.
𝑥 = −20 14. Write an equation that contains variables on both sides and has a solution of x = 3. Answers will vary. One possible solution: 2𝑥 − 4𝑥 = 6− (3𝑥 + 𝑥) Students may construct their equation by trial and error. Another approach is to build the equation. Here is how the equation shown above was built: 𝑥 = 3 Given 2𝑥 = 6 Multiply both sides by 2. 2𝑥 − 4𝑥 = 6− 4𝑥 Subtract 4x from both sides. 2𝑥 − 4𝑥 = 6− (3𝑥 + 𝑥) Replace 4x on the right side of the equation with an equivalent expression. *Students should recognize that the same properties of equality that allow them to solve an equation are the same properties of equality that allow them to build an equation. They should also recognize that they can replace expressions in an equation with equivalent expressions and not change the nature of the original equation. You may wish to have students solve their neigbor’s equation to verify that the solution is 3.
5 3x x+ = − − 4 2 12x x= − + 2 5 6 5x x− = − +
4 2 12x x= + 5 3 27x x+ = +
4 6 3 36x x− + = − 0.7 0.6 0.2 0.42x x− = − −
8 4 4x x− = 1 48 123 3x x− = +
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1.1f Homework: Solve Linear Equations in One Variable (simplify with variables on both sides) Directions: Solve the following equations. Verify your solutions. 1. 𝑥 = 2
2. 4 = 𝑥
3. 𝑥 = 3
4. 𝑥 = 2
5. 𝑥 = 2
6. 𝑎 = 4
7. 𝑏 = 2
8. 4 = 𝑦
9.
𝑥 = 24
Directions: In the following problems, a common mistake has been made. Circle the mistake and describe the
mistake in words. Then, solve the equation correctly. 10.
2𝑥 − 8 = 20 Combine like terms (4x and -2x)
Add 8 to both sides.
Divide both sides by 2.
Explanation of Mistake: The 4x and -2x are on different sides of the equation so cannot be combined. If students think about what really happened in this step, they subtracted 2x from the left side and added 2x to the right side so they added different quantities to both sides, breaking one of the laws of equality.
Solve Correctly: 𝑥 = !"
!
11.
Add 2𝑥 to both sides.
Divide both sides by 8.
Simplify the fraction.
Explanation of Mistake: The 4 was moved over to the other side of the equation. If students think about what really happened in this step, they subtracted 4 from the left side and added 4 to the right side so they added different quantities to both sides, breaking one of the laws of equality.
Solve Correctly: 𝑥 = − !
!
14. Write an equation that contains variables on both sides and has a solution of c = –4 . Answers will vary. One possible equation: 𝑐 + −3 = 1+ 2𝑐
3 2 2x x= + 3 5 4 1x x+ = + 6 3 3 12x x+ = +
3 8 2 10x x+ = + 5 3 3 7x x+ = + 5 5 7 2a a− = +
3 6 8 4b b− = − 3 12 3 12y y− + = − 1 13 94 2x x− = − +
4 8 2 20x x− = − +
2 12x =
6x =
6 4 2x x+ = −
8 4x =
48
x =
12
x =
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1.1g Classwork: Solve Linear Equations in One Variable (multi-step)
1. The following is a model of the equation . Create this model with your tiles and solve the equation, showing your solving actions below.
a. Solving Actions:
𝑥 = 3
b. How can you verify your solution?
Students can verify the solution by substituting into the equation. Alternatively, they can label the pieces of the model above by writing the values on each of the tiles and verifying that the two sides are equal.
5 2 4 8x x x+ + = +
5 2 4 8x x x+ + = +
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2. The following is a model of the equation . Create this model with your tiles and solve the equation, showing your solving actions below.
a. Solving Actions:
𝑥 = 1
b. How can you verify your solution? Model and solve the following equations.
3. 𝑥 = 3
4. 𝑥 = 3
5. 𝑥 = 4
7 9 4 2( 5)x x x+ − = +
7 9 4 2( 5)x x x+ − = +
2( 3) 5 3x x+ = − 8 3 2 3 12x x x+ − = + 10 2 3 3(2 2)x x x+ − = +
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6. The following is a model of an equation.
Write the symbolic representation for this model. Solve the equation. 3 𝑥 − 2 + (−𝑥) = 2 2𝑥 + 3 𝑥 = −6 Directions: Solve the following equations without the use of the tiles.
7. 𝑥 = 1
8. 𝑎 = 7
9. 𝑥 = 8
10. 6 = 𝑡
11. 𝑥 = 9
12. 𝑥 = −6
13.
𝑥 = 2
14.
6 = 𝑥
15. Write an equation that has parentheses on at least one side of the equation, variables on both sides of the equation, and a solution of x = 4. Answers will vary. One possible equation: 2 𝑥 + 3 + 12 = 14+ 3𝑥
9 4 7 2x x− = − 2( 2) 11a a+ = +
5 4 18 3 2x x x− + = + 2 21 3( 5)t t+ = +
7 2 4 7 2 4x x x+ − = + + 4(1 ) 3 2( 1)x x x− + = − +
1 (12 16) 10 3( 2)4
x x+ = − − 2 4 23 2 5xx− =+
– –
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1.1g Homework: Solve Linear Equations in One Variable (multi-step)
1. The following is a model of an equation.
Write the symbolic representation of the equation for this model. Solve the equation. 3 2𝑥 − 1 + −2𝑥 = −2𝑥 − 15 𝑥 = −2 Directions: Solve the following equations. Verify your solutions.
2. 𝑥 = 3
3. 𝑐 = 2
4.
𝑥 = −112
5. 𝑥 = 2
6. 7 = 𝑥
7. 𝑥 = 7
3 9x x x+ = + 4 4 10c c+ = + 3(4 1) 2(5 7)x x− = −
3 10 2 2( 8)x x x+ + = + 2( 8) 2(2 1)x x+ = + 4( 3) 26x x x+ = + +
–
–
–
–
–
–
–
–
–
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8. 𝑎 = 17
9. 𝑐 = 1
10. 𝑥 = 1
11.
𝑥 =65
12. −5 = 𝑥
13.
𝑦 =−32
14.
𝑥 = −95
15.
𝑥 = 2
16.
𝑛 = 1
17. Write an equation that has parentheses on at least one side of the equation, variables on both sides of the equation, and a solution of x = 2. Answers will vary. One possible equation: 2 3𝑥 − 4 + 10 = 4− 2𝑥 + 7𝑥
3 5( 2) 6( 4)a a a+ − = + 13 (2 2) 2( 2) 3c c c− + = + + 2(4 1) 2 9 1x x x+ − = −
2 (2 2) 2( 3)x x x− + = + + 3( 6) 4( 2) 21x x− = + − 3( 7) 2( 9)y y y+ = + −
4( 3) 6( 5)x x− − = + 1 (12 2 ) 4 5 2( 7)2
x x x− − = + − 1 (12 4) 14 102
n n− = −
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1.1h Classwork: Applications Part 2 Directions: Write the word problem that goes with the equation in each problem. Then solve for the unknown
information and verify your answer. 1. Birthday Parties
Number of friends at birthday party: f Cost of party at Boondocks: Cost of party at Raging Waters:
The cost of a birthday party at Boondocks is 8 dollars per person plus $60 to rent the party room. The cost of a party at Raging Waters is 20 dollars per person. How many friends would you invite to your party so that the cost of the parties is the same? If you invite 6 friends to your party the cost of the parties will be the same. Note: You could ask students additional questions about this problem such as: How much will each party cost when 6 friends are invited? Which option is cheaper if you are inviting 7 friends to your party? How much cheaper is it?
2. A Number Trick
Starting number: n Lily’s number: Kali’s number:
Lily starts with a number, adds 5 to it, and multiplies the result by 3. Kali starts with a number and subtracts 5 from it. When they are finished, they realize they have they are left with the same number. What number did the girls start with? Both girls started with the number -10.
3. Savings Number of weeks: w Sophie’s Money: Raphael’s Money:
Sophie currently has $300 and is spending money at a rate of $40 per week. Raphael currently has $180 and is saving money at a rate of $20 per week. After how many weeks will Sophie and Raphael have the same amount of money? After 2 weeks, Sophie and Raphael will have the same amount of money. Note: You can ask students additional questions such as: How much money will each of them have after 2 weeks? Who will have more after 2 weeks? If Sophie continues to spend at this rate, when will she run out of money? If Raphael continues to save at this rate, how long will it take for him to have $1,000?
8 60f +20 f
8 60 20f f+ =
3( 5)n +5n −
3( 5) 5n n+ = −
300 40w−180 20w+
300 40 180 20w w− = +Draft: F
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Directions: Write an expression for each unknown quantity in the word problem. Then write an equation for
each problem. Then solve your equation and answer the question in a complete sentence. 4. Horizon Phone Company charges $15 a month plus 10 cents per text. G-Mobile charges a flat rate of $55
per month with unlimited texting. At how many texts would the two plans cost the same? Which plan is the better deal if you send 200 texts per month? Number of texts sent per month: t Horizon monthly charge: 15+ 0.1𝑡 G-Mobile monthly charge: 55 15+ 0.1𝑡 = 55
5. The enrollment in dance class is currently 80 students and is increasing at a rate of 4 students per term. The enrollment in choir is 120 students and is decreasing at a rate of 6 students per term. After how many terms will the number of students in dance equal the number of students in choir? How many students will be in each class? Number of terms: t Number of students in dance: 80+ 4𝑡 Number of students in choir: 120− 6𝑡 80+ 4𝑡 = 120− 6𝑡
6. You burn approximately 230 calories less per hour if you ride your bike versus go on a run. Lien went on a two-hour run plus burned an additional 150 calories in his warm-up and cool down. Theo went on a 4 hour bike ride. If Lien and Theo burned the same amount of calories on their workouts, approximately how many calories do you burn an hour for each type of exercise? Calories: c Calories burned biking: 𝑐 − 230 Calories burned running: 𝑐 Calories burned by Lien: 2𝑐 + 150 Calories burned by Theo: 𝑐 − 230 2𝑐 + 150 = 4(𝑐 − 230)
At 400 texts, the two plans cost the same. If you send 200 texts per month, Horizon is the better deal.
After 4 terms, the number of students in dance will equal the number of students in choir. The number of students in each class will be 96.
You burn approximately 305 calories per hour riding a bike and approximately 535 calories per hour running.
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1.1h Homework: Applications Part 2 Directions: Write the word problem that goes with the equation in each problem. Then solve for the unknown information and verify your answer. 1. Fixing Your Car
Number of hours it takes to fix your car: h Cost of Mike’s Mechanics: Cost of Bubba’s Body Shop: 25ℎ
The cost to fix your car at Mike’s Mechanics is $15 per hour plus an initial fee of $75. The cost toyou’re your car at Bubba’s Body Shop is $25. At how many hours will both shops charge the same to fix your car? At 7.5 hours the two shops will charge the same amount to fix your car. Note: You can ask students additional questions for this information or challenge them to come up with some of their own: Which shop would be a better deal if it took 4 hours to fix your car? 10 hours to fix your car?
2. World Languages
Number of years: t Number of Students Enrolled in French: Number of Students Enrolled in Spanish: 160− 9𝑡 = 85+ 6𝑡 The number of students currently enrolled in French class is 160 and is decreasing at a rate of 9 students per year. The number of students currently enrolled in Spanish class is 85 and is increasing at a rate of 6 students per year. After how many years will the number of students in the classes be the same? At 5 years, the number of students in the two classes will be the same.
3. Downloading Music Number of songs downloaded: s Monthly cost of downloading songs at bTunes: Monthly cost of downloading songs at iMusic:
The monthly cost of downloading songs at bTunes is $0.99 per song. iMusic charges a monthly fee of $10 plus an additional $0.79 per song. At how many songs would the two plans cost the same? If you download 50 songs, the two plans will cost the same. Note: You can ask students additional questions for this information or challenge them to come up with some of their own: How much does it cost on each plan to download 50 songs? Which is a better deal if you download 20 songs per month? 100 songs per month?
15 75h +
15 75 25h h+ =
160 9t−85 6t+
0.99s10 0.79s+
0.99 10 0.79s s= +
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Directions: Write an expression for each unknown quantity in the word problem. Then write an equation for each problem. Then solve your equation and answer the question in a complete sentence. 4. Underground Floors charges $8 per square foot of wood flooring plus $150 for installation. Woody’s
Hardwood Flooring charges $6 per square foot plus $200 for installation. At how many square feet of flooring would the two companies charge the same amount for flooring? If you were going to put flooring on your kitchen floor that had an area of 120 square feet, which company would you choose?
Number of square feet: s Cost of Underground Floors: 8𝑠 + 150 Cost of Woody’s Hardwood Flooring: 6𝑠 + 200 8𝑠 + 150 = 6𝑠 + 200
5. Owen and Charlotte’s mom give them the same amount of money to spend at the fair. They both spent all of their money. Owen goes on 8 rides and spends $5 on pizza while Charlotte goes on 5 rides and spends $6.50 on pizza and ice cream. How much does each ride cost? Cost per ride: r Amount Owen spends: 8𝑟 + 5 Amount Charlotte spends: 5𝑟 + 6.50 8𝑟 + 5 = 5𝑟 + 6.50
6. Ashton and Kamir are arguing about how a number trick they heard goes. Ashton tells Andrew to think of a number, multiply it by five and subtract three from the result. Kamir tells Andrew to think of the same number add five and multiply the result by three. Andrew says that whichever way he does the trick he gets the same answer. What was the number? Ashton: 5𝑛 − 3 Ashton: 3(𝑛 + 5) 5𝑛 − 3 = 3(𝑛 + 5)
At 25 square feet, the cost of the two companies is the same. If you were putting flooring in your kitchen, the company that is the better deal is Woody’s. Note: You may wish to have a conversation with students about other factors that come into play when making a decision like this: who has a better reputation, how long it will take to get the work done, etc.
The cost of each ride is $0.50.
The number is 9.
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1.1i Self-Assessment: Section 1.1 Consider the following skills/concepts. Rate your comfort level with each skill/concept by checking the box that best describes your progress in mastering each skill/concept.
Skill/Concept Beginning Understanding
Developing Skill and
Understanding
Deep Understanding, Skill Mastery
1. Understand the meaning of linear expression and linear equation.
2. Simplify linear expressions, including those requiring expanding using the distributive property and collecting like terms.
3. Write and simplify linear expressions that model real world problems.
4. Translate between the concrete and symbolic representations of an expression and an equation.
5. Solve multistep linear equations with rational coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
6. Write and solve multistep linear equations that model real world problems.
This self-assessment is meant to be used as a formative assessment and is one way to assess how students are doing toward mastery of the skills and concepts in a particular section. Teachers should guide students through this self-assessment by asking probing questions of the students. For example, to assess a student’s understanding of the first skill/concept in the chart above, a teacher may ask the students if they can define linear expression and linear equation in their own words. They may ask the students to give examples of each or put a list of expressions and equations on the board and ask students to classify each as an expression or an equation. For the second skill/concept, you may put a list of expressions on the board and ask students if they can simplify all of the expressions. You may also put an expression on the board and ask students to come up equivalent expressions. You may ask them to circle the expressions that they feel they would have a difficult time simplifying and work with the teacher or a peer on these problems. A similar process can be used to formatively assess all of the skills/concepts in this section.
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Section 1.2: Solve Linear Equations in One Variable (special cases) Section Overview: In this section students will extend their understanding of linear equations in one variable with unique solutions to linear equations in one variable with either no solution or infinitely many solutions (special cases). Students start the section with a problem that gives them an opportunity to explore these concepts. They will use models and skills developed and solidified in section 1.1 to recognize the structure and then solve special cases of linear equations. Students should come to understand that the structure of an equation as well as the solving outcome (𝑥 = 𝑎, 𝑎 = 𝑏, or 𝑎 = 𝑎 where a and b are different numbers) indicates if the equation has one, no, or infinitely many solutions (is true for a unique value, no value, or all values of the variable.) Concepts and Skills to Master: By the end of this section, students should be able to:
1. Solve multi-step linear equations that have one solution, infinitely many solutions, or no solution. 2. Understand what it is about the structure of a linear equation that results in equations with one solution,
infinitely many solutions, or no solutions. 3. Identify and provide examples of equations that have one solution, infinitely many solutions, or no
solutions.
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1.2a Anchor Problem: Chocolate Problem Consider the following situation. Then answer the questions below. Include any pictures, tables, equations, or graphs that you used to solve the problem. Two students, Arthur and Oliver, each have some chocolates. They know that they have the same number of chocolates. Arthur has two tubs of chocolates, one bag of chocolates, and twenty-five loose chocolates. Oliver has two tubs of chocolates, two bags of chocolates, and seven loose chocolates. Can you determine how many chocolates are in a tub? Can you determine how many chocolates are in a bag? To the extent possible, figure out how many chocolates are in a tub and how many chocolates are in a bag. Teacher Notes:
1. Start by engaging students by presenting the situation in an encouraging way. 2. Make sure that students understand that the tubs and bags are full and that the tubs have the same
number of chocolates in each and the bags have the same number of chocolates in each. 3. Have the students work in pairs or groups to answer the problem. Anticipate the correct answer for
chocolates in a bag (18 chocolates) and be ready to help students who don’t get that answer. Anticipate that you will get many answers for chocolates in a tub such as it can’t be done, it could be any number, or a specific number such as 10. It is most beneficial to have several different numbers for the tub that you will be able to check as a class to see that they all work.
4. Have student share their work. First, have some show their thinking about the bags. Choose different methods such as a picture, table, or equation and make connections between them.
5. Then have several that chose different values for the tub share their values and show why it works or test to see that it works in the situation. Use this to lead into a discussion of how it works for any number meaning that there are infinitely many solutions.
6. Possible extension: Time permitting have students come up with other situations that may lead to infinitely many solutions.
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1.2a Homework: Review of Multi-step Solving Directions: Solve the following equations. Show all your work. 1.
𝑥 = 15
2. 𝑦 = 3
3. 𝑔 = −2
4. 𝑑 = 6
5. 𝑡 = 2
6. 𝑘 = −2
7.
𝑠 = −1
8.
𝑥 = 30
9. !!4𝑥 − 16 = 12
𝑥 = 10
10. Like Arthur and Oliver, Abby and Amy also say that they have the same number of chocolates. Abby has
one full tub of chocolates and 21 remaining chocolates. Amy has one full tub of chocolates and 17 remaining chocolates. Each tub has the same number of chocolates. Can you determine how many chocolates are in a tub? Why or why not? If so, how many chocolates are a tub?
Since the equation that matches this problem is 𝑏 + 21 = 𝑏 + 17 and this equation has no solution, it is not possible to determine the number of chocolates in each tub. As a matter of fact, it is not mathematically possible that the tubs have the same number of chocolates.
7( 3) 3 3( 2)x x x− = + − 8 19 5y − = 8 9 4 1g g+ = +
2( 4 3) 42d− + = − 0.5 0.25 2t t+ = + 7 2( 6) 22k k− + = −
1 34 82 4
s s⎛ ⎞ ⎛ ⎞+ = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠20 105x+ =
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1.2b Classwork: Solve Linear Equations in One Variable (special cases)
1. Consider the following model:
a. Using the model, make some observations.
Students may make observations that the two sides contain the same number of x tiles and the same number of unit tiles but that they are grouped differently.
b. Write the symbolic representation for this model and then solve the equation you wrote. 2 𝑥 + 3 = 2𝑥 + 6 6 = 6
c. What happened when you solved the equation? When this equation was transformed into its simplest form, the result was 𝑎 = 𝑎.
d. What is it about the structure of the equation that led to the solution? Since both sides of the equation are equivalent expressions, this equation would be true for all values of x. To help to solidify what is happening here, select multiple values for x and have students substitute them into the symbolic or concrete model of the representation and observe what happens. This will guide them toward the conclusion that this equation has infinitely many solutions.
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2. Consider the following model:
a. Using the model, make some observations.
Students may observe that the left and right side both have 2 x tiles and the right side 8 unit tiles while the right side has 4 unit tiles.
b. Write the symbolic representation for this model and then solve the equation you wrote. 2𝑥 + 8 = 2𝑥 + 4 8 = 4 The original claim that these two expressions are equal to each other is false; therefore there are no values of x that make this equation true.
c. What happened when you solved the equation? When this equation was transformed into its simplest form, the result was 𝑎 = 𝑏.
d. What is it about the structure of the equation that led to the solution? The original claim that these two expressions are equal to each other is false; therefore there are no values of x that make this equation true. To help to solidify what is happening here, select multiple values for x and have students substitute them into the symbolic or concrete model of the representation and observe what happens. This will guide them toward the conclusion that this equation has no solution.
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Directions: Solve the following equations. If there is one solution, state what the solution is. Otherwise, state if there are infinitely many solutions or no solution. 3. no solution
4. 5𝑥 − 10 = 10− 5𝑥 𝑥 = 2
5. infinitely many solutions
6. infinitely many solutions
7. no solution
8. no solution
9. no solution
10. infinitely many solutions
11. What is it about the structure of an expression that leads to one solution, infinitely many solutions, or no solution? Provide examples to support your claim.
When an equation is transformed into its simplest form and the result is 𝑥 = 𝑎, then the equation is true for one value of x. This value is a solution of the equation. Examples may vary When an equation is transformed into its simplest form and the result is 𝑎 = 𝑎, then the equation is true for all values of x. In the original equation, both sides of the equation are the same expression. Examples may vary When an equation is transformed into its simplest form and the result is 𝑎 = 𝑏, then the equation is true for no values of x. The original claim that these two expressions are equal to each other is false; therefore there are no values of x that make this equation true. Examples may vary
Directions: Without solving completely, determine the number of solutions of each of the equations.
1 1x x− = +
4( 3) 10 6( 2)m m m− = − + 4( 4) 4 16x x− = −
2 5 2( 5)x x− = − 3 3 4x x= −
3 5 2 5(2 )v v v+ + = + 5 (4 8) 5 4 8a a− + = − −
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12. infinitely many solutions
13. no solution
14. no solution
15. one solution
16. infinitely many solutions
17. Consider the expression . Write 3 different expressions that if set equal to would result in
the equation having infinite solutions. Possible answers: 4(𝑎 − 3) 2(2𝑎 − 6) 2𝑎 + 2𝑎 − 12
18. Consider the expression . Write 3 different expressions that if set equal to would result in the
equation having no solution. Possible answers: 𝑥 + 2 𝑥 𝑥 − 7
19. Consider the expression . Write 3 different expressions that if set equal to would result in the equation having one solution. Possible answers: 5𝑥 15 3𝑥 − 7
20. Determine whether the equation 7𝑥 = 5𝑥 has one solution, infinitely many solutions, or no solution. If it has one solution, determine what the solution is. This equation has one solution 𝑥 = 0. This may confuse some students as at first glance it may appear as though the equation has no solution. Talk them through how this equation is different than the ones that they have seen that have no solution.
6 3 3(2 1)a a− = − 5 2 5x x− = 8 2 4 6 1x x x− + = −
5 2 3 8m m+ = − 2(3 12) 3(2 8)a a− = −
4 12a − 4 12a −
1x + 1x +
2 6x + 2 6x +
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1.2b Homework: Solve Linear Equations in One Variable (special cases) Directions: Without solving completely, determine the number of solutions of each of the equations. 1.
no solution 2.
infinitely many solutions 3.
infinitely many solutions
4. one solution
5. no solution
Directions: Solve the following equations. If there is one solution, state what the solution is. Otherwise, state if there are infinitely many solutions or no solution. Show all your work. 6. infinitely many solutions
7. 𝑎 = −24
8. no solution
9. −2 = 𝑥
10. infinitely many solutions
11. no solution
211x x− = 3( 3) 3 9m m− = − 5 5x x− = − +
4 12 4 12m m− + = + 3( 2) 3 6x x− + = − +
3 1 3( 1) 4x x+ − − = 3( 6) 2( 6) 6a a+ − − =
3( 4) 3 4r r− = − 2( 1) 3 4x x+ = +
3 (4 2) 3 4 2b b− − = − + 3 (4 2) 3 4 2b b− − = − −Draf
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12. infinitely many solutions
13. infinitely many solutions
14. 𝑎 = −9
15.
infinitely many solutions
Directions: Fill in the blanks of the following equations to meet the criteria given. In some cases, there may be more than one correct answer.
16. An equation that yields one solution:
One possible equation: 8𝑥 + 5 = 6𝑥 + 10
17. An equation that yields no solution: One possible equation: 8𝑥 + 4 = 8𝑥 + 10
18. An equation that yields infinitely many solutions:
One possible equation: 8𝑥 + 24 = 2(4𝑥 + 12)
Directions: Create your own equations to meet the following criteria. 19. An equation that yields one solution of x = 5. Equations will vary. You can have students swap equations with a neighbor.
20. An equation that yields no solution. Equations will vary. You can have students swap equations with a neighbor. 21. An equation that yields infinitely many solutions.
Equations will vary. You can have students swap equations with a neighbor.
22. Challenge: Can you think of an equation with two solutions? One possible equation: 𝑥! = 4
2 5 6 (3 6)y y y− + = − − 1 7 12 11 6f f f+ = + − −
12 8 6 6a a+ = − ( )1 6 10 3 52
m m− = −
8 10x x+ = +
8 10x x+ = +
( )8 24x x+ = +
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1.2c Self-Assessment: Section 1.2 Consider the following skills/concepts. Rate your comfort level with each skill/concept by checking the box that best describes your progress in mastering each skill/concept.
Skill/Concept Beginning Understanding
Developing Skill and
Understanding
Deep Understanding, Skill Mastery
1. Solve multi-step linear equations that have one solution, infinitely many solutions, or no solution.
2. Understand what it is about the structure of a linear equation that results in equations with one solution, infinitely many solutions, or no solutions.
3. Identify and provide examples of equations that have one solution, infinitely many solutions, or no solutions.
This self-assessment is meant to be used as a formative assessment and is one way to assess how students are doing toward mastery of the skills and concepts in a particular section. Teachers should guide students through this self-assessment by asking probing questions of the students. For example, to assess a student’s understanding of the first skill/concept in the chart above, a teacher may put a variety of equations on the board (including those with one solution, no solution, and infinitely many solutions) and ask students to solve them, noting which ones they have difficulty with. Students can come up with their own equations that would have one solution, no solution, or infinitely many solutions and swap with a peer to verify. A similar process can be used to formatively assess all of the skills/concepts in this section.
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