Input/Model
(Teacher Presents)
Directions: Solve for the variable.
1. 3x + 2 = 2x ─ 2
xxx + = xx ─
+ ─
Solution:
• Be sure you understand the visual model above. Remember that 2x – 2 is the same as 2x + (─2)
• Step 1: Remove 2 x’s from each side. Keep in mind what you do to one side, you must do to other. The
equation now shows:
+─
=
x +
─
• Step 2: Move the integers to one side. Add 2 (─1) tiles to each side. This will cancel out the left side. The
right side now has 4 (─1) tiles. +─ ──
x+─ =
──
• Step 3: Solve x=-4
Answer: x= ─4
2. ─4 – 6x = 2 – 5x
Solution:
• Step 1: Move the variables first. Choose to move the smaller variable, 5k, by adding it to both sides. The new
equation is ─4 ─ x =2
• Step 2: Move the integers to the other side. Add 4 to each side. The new equation is: ─x = 6
• Since the x is negative, divide each side by (─1) to make it positive.
Answer: x = ─6
Copyright © Swun Math Grade 8 Unit 4 Lesson 7 C TE 346
Structured Guided Practice
(A/B Partners Practice)
Directions: Solve for the variable. Use algebra tiles or drawings as needed.
1. 3x + 5 = ─ 10 ─ 2x
++ = ─ ── ─ ─ -x -x
xxx + + ─ ── ─ ─
+
Solution:
x= ─3
2. 1 + 7x = 10 + 8x
Solution:
x= ─9
347 Copyright © Swun Math Grade 8 Unit 4 Lesson 7 C TE
Final Check for Understanding
(Teacher Checks Work)
Directions: Solve for the variable. Use algebra tiles or drawings as needed.
1. 3x – 1 = x + 7
x xx ─ = + ++
x+
+ ++
Solution:
x=4
2. ─2 – 3x = ─2x – 9
Solution:
x= 7
Closure
Recap today’s lesson with one or more of the following questions:
MP7: Why is it helpful to understand the properties of equality?
MP8: Explain how this strategy made it easier to identify the solution.
Copyright © Swun Math Grade 8 Unit 4 Lesson 7 C TE 348
Homework Name: ___________________________
Date: ___________________________
Unit 4 · Lesson 7: Equations with Variables on
Both Sides
Objective: I will learn to solve equations with variables on both sides.
Vocabulary Steps:
1. Collect like terms.
2. Move variables to one side (additive inverse).
3. Move integers to one side (additive inverse).
4. Isolate the variable (multiplicative inverse).
5. Solve and check.
Example # 1 Example # 2
Directions: Solve for the variable. Use algebra tiles as needed.
3x + 2 = 2x ─ 2 ─4 – 6x = 2 – 5x
xxx + = ─
+ x x─
Solution: Solution:
• Be sure you understand the visual model above. • Step 1: Move the variables first. Choose to move the
Remember that 2x – 2 is the same as 2x + (─2) smaller variable, 5k, by adding it to both sides. The new
equation is ─4 ─ x =2
• Step 1: Remove 2 x’s from each side. Keep in mind what
• Step 2: Move the integers to the other side. Add 4 to
you do to one side, you must do to other. The equation each side. The new equation is: ─x = 6
now shows: + =─ • Since the x is negative, divide each side by (─1) to make
x +─ it positive.
Answer: x = ─6
• Step 2: Move the integers to one side. Add 2 (─1) tiles to
each side. This will cancel out the left side. The right
side now has 4 (─1) tiles. +─ ─ ─
x + ─ = ─
─
• Step 3: Solve x=-4
Answer: x= ─4
349 Copyright © Swun Math Grade 8 Unit 4 Lesson 7 C TE
Homework
Unit 4 · Lesson 7: Equations with Variables on
Both Sides
Directions: Solve for the variable. Use algebra tiles or drawings as needed.
1. ─2a = ─8 + 2a 2. ─7 ─4b = 5b
3. ─2c – 2 = ─3c – 8 4. ─6d – 10 = 10 – 4d
5. ─7g + 5 = 2 – g – 7g 6. 9k + 1 = ─9 + 7k
Copyright © Swun Math Grade 8 Unit 4 Lesson 7 C TE 350
Answer Key 2. b=7
4. d= ─10
Homework 6. k= ─5
1. a=2
3. c= ─6
5. g= ─3
351 Copyright © Swun Math Grade 8 Unit 4 Lesson 7 C TE
Notes
Copyright © Swun Math Grade 8 Unit 4 Lesson 7 C TE 352
Equations with Rational MPs Applied MP
Numbers
* Embedded MP
Procedural Lesson
Grade 8 · Unit 4 · Lesson 8 12345678
MC: 8.EE.7b * * * * *
Problem of the Day Student Journal Pages
192-197
Objective: I will solve equations with rational number coefficients and constants.
Vocabulary Teacher Resources
Fractions: Considerations:
1. Determine the LCD (least common Tell students to simplify the equation to make
denominator). solving as efficient as possible. Practice clearing
2. Multiply both sides by the LCD. equations of fractions and decimals to make
3. Use inverse operations to solve. equations easier to work with. If necessary, work
with equations that have one fraction. A review
Decimals: of LCD and using the distributive property to
1. Determine the power of 10 that will clear clear fractions and decimals may be necessary.
the decimal.
2. Multiply both sides by the power of 10. Steps for Integers:
3. Use inverse operations to solve. 1. Collect like terms.
2. Isolate the variable.
3. Solve the equation.
4. Check by inserting the answer into the
equation.
Application of MPs:
MP7: What observations do you make based on past
experience with similar problems?
When I decided to use the distributive .
property, I was considering
MP8: What happens to an equation containing
decimals when you multiply both sides by a
common power of ten?
When I multiply both sides by a common .
power of ten, the equation
353 Copyright © Swun Math Grade 8 Unit 4 Lesson 8 P TE
Input/Model
(Teacher Presents)
Directions: Solve.
1.
�23 + 1 � = �65 + 2 �
2
Solution:
Distribute the number 6 (LCD) to the equation to clear the fractions.
6 �32 + 1 � = 6 �65 + 2 �
2
12 + 6 = 30 + 12
3 2 6
4 + 3 = 5 + 12
7 = 5 + 12
−7 − 7
0 = 5 + 5
−5 − 5
−5 = 5
= −1
2.
4.6 + 6.8 = 7.4 + 1.2
Solution:
Use the number 10 to remove the decimals.
10(4.6 + 6.8) = 10(7.4 + 1.2)
46 + 68 = 74 + 12
−46 − 46
68 = 28 + 12
68 = 28 + 12
−12 − 12
56 = 28
56 28
28 = 28
2 =
Copyright © Swun Math Grade 8 Unit 4 Lesson 8 P TE 354
Structured Guided Practice
(A/B Partners Practice)
Directions: Solve.
1.
2 + 2 + 1 4 + 1
3 2= 2
Solution:
r=9
2.
2.43 − 4.6 = 1.03 − 21.4
Solution:
n= ─12
355 Copyright © Swun Math Grade 8 Unit 4 Lesson 8 P TE
Final Check for Understanding
(Teacher Checks Work)
Directions: Solve.
1.
1 + 2 = 6 + 1 + 3
2 3 2 6
Solution:
x= 6
2.
(7.8 + 3.4 ) = (−1.6 − 7.2)
Solution: Grade 8 Unit 4 Lesson 8 P TE 356
n= ─3
Copyright © Swun Math
Student Practice Name: ___________________________
Date: ___________________________
Unit 4 · Lesson 8: Equations with Rational Numbers
Directions: Solve.
1. 1 ─ 3 = 2 ─ 3 2. 1 (6 + 24) − 20 = − 1 (12 − 72)
2 4 3 4
Solution: Solution:
p=4 x=6
3. 2 + 1 = 7 4. 2.3 − 7.4 = 4.5 + 2.8
3 2
Solution: Solution:
n=6 x= ─23.8
5. 1 + 1 = 1 − 1 6. 1.2 − 11.2 = 2.8 + .8
3 5 5
Solution: Solution:
x= ─9 x=35
357 Copyright © Swun Math Grade 8 Unit 4 Lesson 8 P TE
Challenge Problems
Directions: Solve.
10 [6.4 + 2.1( − 2)] = 10(8.5 + 3.4 )
Solution:
10 [6.4 + 2.1( − 2)] = 10 (8.5 + 3.4 )
64 + 21( − 2) = 85 + 34
64 + 21 − 42 = 85 + 34
21 + 22 = 85 + 34
−21 − 21
22 = 85 + 13
22 = 85 + 13
−85 – 85
−63 = +13
−63 = +13
+13 +13
= −63 − 4 11
13 13
Extension Activity
* MP1: Make sense of the problem and persevere in solving it.
* MP4: Apply mathematics in everyday life.
• When clearing an equation of fractions or decimals, explain what is happening to the
numbers and variables on each side of the equation.
• Why does it work? Include what you know about properties of equality and how an
equation is like a scale.
• Could you ever divide all terms by a common factor and create an equivalent
equation?
Copyright © Swun Math Grade 8 Unit 4 Lesson 8 P TE 358
Closure
Reaching Consensus
*MP3: Do you agree or disagree with your classmate? Why or why not?
Student Presentations
*MP1: What steps in the process are you most confident about?
*MP6: Explain how you might show that your solution answers the problem.
Closure
Recap today’s lesson with one or more of the following questions:
MP7: What observations do you make based on past experience with similar
problems?
MP8: What happens to an equation containing decimals when you multiply both
sides by a common power of ten?
359 Copyright © Swun Math Grade 8 Unit 4 Lesson 8 P TE
Homework Name: ___________________________
Date: ___________________________
Unit 4 · Lesson 8: Equations with Rational Numbers
Objective: I will solve equations with rational number coefficients and constants.
Vocabulary Steps:
Expression Equation Variable 1. Collect like terms.
2. Isolate the variable.
− + = + 3. Solve the equation.
4. Check by inserting the answer into the
Term Coefficient Constant
equation.
Fractions:
1. Determine the LCD (least common
denominator).
2. Multiply both sides by the LCD.
3. Use inverse operations to solve.
Decimals:
1. Determine the power of 10 that will clear the
decimal.
2. Multiply both sides by the power of 10.
3. Use inverse operations to solve.
Example # 1 Example # 2
Directions Solve for the given variable. 4.6 + 6 = 7.2 + 1.2
�23 + 1 � = �65 + 2 �
2
Solution: Solution:
I will distribute the number 6 to the equation to clear the I used the number 10 to remove the decimals.
10(4.6 + 6.8) = 10(7.4 + 1.2)
fractions.
46 + 68 = 74 + 12
6 �32 + 1 � = 6 �65 + 2 � −46 − 46
2
68 = 28 + 12
4 + 3 = 5 + 12 68 = 28 + 12
−12 − 12
7 = 5 + 12 56 = 28
−7 − 7 56 28
28 = 28
0 = 5 + 5
2 =
0 = 5 + 5
−5 − 5
−5 = 5
= −1
Copyright © Swun Math Grade 8 Unit 4 Lesson 8 P TE 360
Homework
Unit 4 · Lesson 8: Equations with Rational Numbers
Directions: Solve.
1. 3 − 5 = 9 − 1 2. 5.7 − 2.3 = 4.2 − 20.3
5 10 2
3. 2 − 1 = 1 + 5 1. 0.55 + 0.05 = − 0.70
3 6 2 6
2. 1 − 3 = 3 − 4 3. 2.25 − 0.57 = 8.655
2 5 10
Explain the steps you used to solve problem number _______.
______________________________________________________________________________________________________________________
______________________________________________________________________________________________________________________
______________________________________________________________________________________________________________________
______________________________________________________________________________________________________________________
361 Copyright © Swun Math Grade 8 Unit 4 Lesson 8 P TE
Answer Key
Extension Activity
Answers will vary. Sample response:
When you multiply both sides of an equation by a common denominator or power of ten, every term is being
increased by the number being distributed. For example, when you multiply an equation by 10, you are making
all terms ten times bigger. You can multiply by any number as long as all terms are increased by the same
amount. This preserves the balance in the equation and keeps both sides equal.
Homework 2. n= 4
4. m=1.6
1. n = ─15 6. x=4.1
3. p= 6
5. x= ─17
Copyright © Swun Math Grade 8 Unit 4 Lesson 8 P TE 362
Equations: No Solution or Many MPs Applied MP 8
* Embedded MP
Conceptual Lesson 123 *
Grade 8 · Unit 4 · Lesson 9 4567
*
MC: 8.EE.7a
Problem of the Day Student Journal Pages
198-201
Objective: I will explore equations that have no solutions or many solutions.
Vocabulary Teacher Resources
No solution: If an equation yields a statement Considerations:
that is not true, such as 2=3, then the equation
has no solution. Review solving equations. The use of algebra tiles is
only modeled in the input/model for the students, but
Infinite solutions: If an equation yields a they should feel free to use tiles to help them solve in
statement that is always true, such as 2=2, then all the problems if needed. Some students feel uneasy
the equation has infinite solutions. about having “no solution;” they tend to think they
have errors. The use of algebra tiles will help them
“see”. If working procedurally, make sure they solve
slowly and show all steps. Remind students to keep
numbers in line.
Steps:
1. If using tiles, set up the problem.
2. Remove the same amount from each side of
the equation.
3. Continue to simplify.
4. Decide if your answer has no solution or
infinite solutions.
Application of MPs:
MP2: Which numbers are easiest for you to insert
into equations with multiple variables?
The easiest numbers for me to insert into .
equations with multiple variables are
MP8: What do you notice about equations that have
multiple answers? What do you notice about
equations that have no answers?
I notice that when equations have multiple
answers, and when an equation has no
answers, .
363 Copyright © Swun Math Grade 8 Unit 4 Lesson 9 C TE
Input/Model
(Teacher Presents)
Directions: Simplify. Decide if the equation has many solutions or none.
1. 2b = ─7b + 9b 2. ─4a = 8 ─ 4a
x x = -x -x -x x x x x x -x -x -x -x =
-x -x -x -x x x x x
Solution: Solution:
• If needed, use the visual models to simplify both • Visually simplify:
sides of the equation.
• Or, simplify both sides of the equation • Notice that you are left with nothing on left, this is
procedurally: 0. You have 8 positives. So, the equation looks
─7 + 9 = 2, so 2b = 2b like this: 0=8; which is not true.
• Both sides are the equal sign are the same, so the • Procedurally, simplify both sides of the equation
by using inverse operations.
statement is true.
• Add 4a to both sides: (4a) – 4a = 8 – 4a +(4a)
• Also, since any value for b will work to create a • Simplify: 0 = 8
true statement, there are infinite numbers of • Since I know this to be untrue, there is no solution
possibilities.
to this problem.
Answer: No solution
Answer: Infinite solutions
Copyright © Swun Math 364Grade 8 Unit 4 Lesson 9 C TE
Structured Guided Practice
(A/B Partners Practice)
Directions: Simplify. Decide if the equation has many solutions or none.
1. ─3 + p = p + 9 2. 2c + 8c = 10c
x =x
=
Solution: Solution:
─3 = 9 is false; No solution 0=0; Infinite solutions
*Note: When using the algebra tiles, the written
number 9 or ─3 can be used instead of the positive &
negative cubes.
365 Copyright © Swun Math Grade 8 Unit 4 Lesson 9 C TE
Final Check for Understanding
(Teacher Checks Work)
Directions: Simplify. Decide if the equation has many solutions or none.
1. ─k = ─8k + 7k 2. 6s + 3s + 7 = 4 + 9s +3
Solution: Solution:
0=0; Infinite solutions 0=0; Infinite solutions
Closure
Recap today’s lesson with one or more of the following questions:
MP2: Which numbers are easiest for you to insert into equations with multiple
variables?
MP8: What do you notice about equations that have multiple answers? What do you
notice about equations that have no answers?
Copyright © Swun Math 366Grade 8 Unit 4 Lesson 9 C TE
Homework Name: ___________________________
Date: ___________________________
Unit 4 · Lesson 9: Equations: No Solution or Many
Objective: I will explore equations that have no solutions or many solutions.
Vocabulary Steps
No solution: If an equation yields a statement 1. If using tiles, set up the problem.
that is not true, such as 2=3, then the equation
has no solution. 2. Remove the same amount from each side of
the equation.
Infinite solutions: If an equation yields a
statement that is always true, such as 2=2, then 3. Continue to simplify.
the equation has infinite solutions. 4. Decide if your answer has no solution or
infinite solutions.
Example # 1 Example # 2
Directions: Simplify. Decide if the equation has many solutions or none.
2b = ─7b + 9b ─4a = 8 ─ 4a
Solution: Solution:
• Simplify both sides of the equation: • Simplify both sides of the equation by using inverse
─7 + 9 = 2, so 2b = 2b operations.
• Both sides are the equal sign are the same, so • Add 4a to both sides: (4a) – 4a = 8 – 4a +(4a)
• Simplify: 0 = 8
the statement is true. • Since I know this to be untrue, there is no solution to
• Also, since any value for b will work to create this problem.
a true statement, there are infinite numbers
of possibilities. Answer: No solution
Answer: Infinite solutions
367 Copyright © Swun Math Grade 8 Unit 4 Lesson 9 C TE
Homework Name: ___________________________
Date: ___________________________
Unit 4 · Lesson 9: Equations: No Solution or Many
Directions: Simplify. Decide if the equation has many solutions or none.
1. 3 + 3m = 3m + 9 2. ─8r + 2r = ─6r
3. ─1 + 9k = 9k ─ 1 4. 4p = 4p + 7
5. 0 = ─5r + 5r 6. 7d─4 = ─2 + 7d
Copyright © Swun Math 368Grade 8 Unit 4 Lesson 9 C TE
Answer Key 2. ─6r=─6r; Infinite solutions
4. 0=7; No solution
Homework 6. ─4=─2; No Solution
1. 3=9; No solution
3. ─1=─1; Infinite solutions
5. 0=0; Infinite solutions
369 Copyright © Swun Math Grade 8 Unit 4 Lesson 9 C TE
Notes
Copyright © Swun Math 370Grade 8 Unit 4 Lesson 9 C TE
Equations: No Solution or Many MPs Applied MP
Procedural Lesson * Embedded MP
Grade 8 · Unit 4 · Lesson 10
1234567 8
MC: 8.EE.7a
* * * * *
Problem of the Day Student Journal Pages
202-207
Objective: I will identify equations with many or no solutions.
Vocabulary Teacher Resources
No solution: If an equation yields a statement Considerations:
that is not true, such as 2=3, then the equation
has no solution. Review the order of operations and the
distributive property and explain how it makes
Infinite solutions: If an equation yields a the work more efficient. If needed, practice
statement that is always true, such as 2=2, then simplifying equations with small numbers and
the equation has infinite solutions. many like terms. The use of algebra tiles may be
needed to model some of the smaller terms, but
with larger terms it may get too messy.
371 Copyright © Swun Math Steps:
1. Analyze the problem.
2. Simplify by collecting like terms.
If result is a true statement, then there are
infinite solutions.
If variables cancel out, unequal numbers
remain, or you must divide by zero, there is
no solution.
Application of MPs:
MP1: When solving equations, what is the first thing
you look for?
When solving equations, the first thing I look
for is_________________________________.
MP7: Why are we not able to simplify the same
variable that is raised to a different power?
Expressions containing the same variable
raised to different powers cannot be simplified
because ____________________.
Grade 8 Unit 4 Lesson 10 P TE
Input/Model
(Teacher Presents)
Directions: Simplify. Determine how many solutions.
1.
18a + 14b – 12 = 11a + 5 – 4b + 7a + 18b + 7
Solution:
• Step 1: Analyze the problem
• Step 2: Simplify the equation using inverse operations and collecting like terms to eliminate terms.
18 + 14 – 12 = 11 + 5 – 4 + 7 + 18 + 7
18 + 14 – 12 = 18 + 14 + 12
− 12 = 12
• Step 3: Since the variables cancel out and unequal numbers remain, there is no solution.
2.
3b + 5 + 2b = 5 + 1b + 5b ─ b
Solution:
• Step 1: Analyze the problem
• Step 2: Simplify the equation using inverse operations and collecting like terms to eliminate terms.
5b + 5 = 5 + 5b
• Step 3: Since the result is equivalent equations and will result in 1 = 1, then there are infinite solutions. I
can see that both sides have the same exact terms; therefore, there are infinite solutions.
Copyright © Swun Math 372Grade 8 Unit 4 Lesson 10 P TE
Structured Guided Practice
(A/B Partners Practice)
Directions: Simplify. Determine the number of solutions.
1.
2(b ─ 2) = 2b ─ 4
Solution:
─4= ─4; Infinite solutions.
2.
4x – 19 = 4x ─ 3
Solution:
No solution
373 Copyright © Swun Math Grade 8 Unit 4 Lesson 10 P TE
Final Check for Understanding
(Teacher Checks Work)
Directions: Simplify. Determine the number of solutions.
1.
─5x ─3 = 12 – 4x — x
Solution:
No solution
2.
3(2x + 4) = 8x + 12 – 2x
Solution: 374Grade 8 Unit 4 Lesson 10 P TE
Infinite solutions.
Copyright © Swun Math
Student Practice Name: ___________________________
Date: ___________________________
Unit 4 · Lesson 10: Equations: No Solution or Many
Directions: Simplify. Determine the number of solutions.
1. 12 + 4n = 4(n + 3) 2. 6(3 + c) = 6c +12 + 6
Solution: Solution:
Infinite solutions Infinite solutions
3. 19k +27 = 2(8k – 13) + 3k + 1 4. 42m + 16 – 18m + 12 = 24(m + 1) + 4
Solution: Solution:
No solution Infinite solutions
5. 3(x + 4) = 3x + 11 6. 4(8m – 1) = 32m + 19
Solution: Solution:
No solution No solution
375 Copyright © Swun Math Grade 8 Unit 4 Lesson 10 P TE
Challenge Problems
Directions: Simplify. Determine the number of solutions.
1. 2. A rectangular prism has a length of 9, a
width of 7, and a height of ℎ. The surface
�3 �43 − 2 � ÷ (1.75 ÷ .875 )� = (20 − 15 ) area is 126 + 32ℎ. What must be the
5 value of ℎ?
Solution:
�3 �34 − 2 � ÷ (1.75 ÷ .875 )� (20 − 15 ) Solution:
=5 There are infinite solutions.
(4 − 6 ) ÷ 2 = 4 − 3
2 − 3 = 4 − 3
2 = 4
In this case, c = 0, but d can have infinite answers.
Extension Activity
* MP1: Make sense of the problem and persevere in solving it.
* MP4: Apply mathematics in everyday life.
Corrine disagreed with Felicity about the difference between “no solution” and “infinite
solutions.” Corrine said that “no solution” means it was impossible to find an answer
because it was too hard. Felicity said if there’s no solution, there is no answer, and the fact
that there is no solution is the answer itself.
This continued into a difference of opinion about what “infinite solutions” meant. Felicity
says infinite solutions means any number can work, positive or negative; Corrine said that
sometimes it does matter, and infinite solutions might mean only a certain type of number,
positive or negative.
Considering both arguments, who was correct? Why?
Copyright © Swun Math 376Grade 8 Unit 4 Lesson 10 P TE
Closure
Reaching Consensus
*MP3: Do you agree or disagree with your classmate? Why or why not?
Student Presentations
*MP1: What steps in the process are you most confident about?
*MP6: Explain how you might show that your solution answers the problem.
Closure
Recap today’s lesson with one or more of the following questions:
MP1: When solving equations, what is the first thing you look for?
MP7: Why are we not able to simplify the same variable that is raised to a different
power?
377 Copyright © Swun Math Grade 8 Unit 4 Lesson 10 P TE
Homework Name: ___________________________
Date: ___________________________
Unit 4 · Lesson 10: Equations: No Solution or Many
Objective: I will identify equations with many or no solutions.
Vocabulary Steps
No solution: If an equation yields a statement 1. Analyze the problem.
that is not true, such as 2=3, then the equation 2. Simplify: use inverse operations and combine
has no solution.
like terms.
Infinite solutions: If an equation yields a If result is a true statement, then there
statement that is always true, such as 2=2, then
the equation has infinite solutions. are infinite solutions.
If variables cancel out, unequal
numbers remain, or you must divide by
zero, there is no solution.
Example # 1 Example # 2
Directions: Simplify. Determine the number of solutions.
18a + 14b – 12 = 11a + 5 – 4b + 7a + 18b + 7 3b + 5 + 2b = 5 + 1b + 6b ─ b
Solution: Solution:
• Step 1: Analyze the problem • Step 1: Analyze the problem
• Step 2: Simplify the equation using inverse operations • Step 2: Simplify the equation using inverse operations
and collecting like terms to eliminate terms. and collecting like terms to eliminate terms.
18 + 14 – 12 = 11 + 5 – 4 + 7 + 18 + 7 5b + 5 = 5 + 5b
• Step 3: Since the result is equivalent equations and will
18 + 14 – 12 = 18 + 14 + 12 result in 1 = 1, then there are infinite solutions. I can see
that both sides have the same exact terms; therefore,
− 12 = 12 there are infinite solutions.
• Step 3: Since the variables cancel out and unequal
numbers remain, there is no solution.
Copyright © Swun Math 378Grade 8 Unit 4 Lesson 10 P TE
Homework
Unit 4 · Lesson 10: Equations: No Solution or Many
Directions: Simplify. Determine the number of solutions.
1. 8(5b + 3) = 10(4b + 2) 2. 6s + 3 – 6s = 3
3. 16(2w + 11) = 8(4w + 22) 4. 6 – 4r + 4r = 8
5. ─11 + b = ─7b – 8(─b +1) 6. ─7(3x + 4) + 21x = ─28
Explain the steps you used to solve problem number _______.
______________________________________________________________________________________________________________________
______________________________________________________________________________________________________________________
______________________________________________________________________________________________________________________
______________________________________________________________________________________________________________________
______________________________________________________________________________________________________________________
379 Copyright © Swun Math Grade 8 Unit 4 Lesson 10 P TE
Answer Key
Extension Activity
Answers will vary. Sample response:
Felicity is correct in the first argument. Having no solution simply means that, mathematically, there is no
possible number that would satisfy the equation. This is not because it’s too hard, but because the situation is
mathematically impossible.
Felicity is correct in the second equation. While sometimes an infinite number of solutions might mean all
numbers, positive or negative, there are cases where infinite solutions can only be positive or negative. For
example, with area, a length cannot be a negative number, and therefore, the infinite number of possible
lengths can only include positives.
Homework 2. Infinite solutions
4. No solution
1. No solution 6. Infinite solutions
3. Infinite solutions
5. No solution
Copyright © Swun Math 380Grade 8 Unit 4 Lesson 10 P TE
Write Linear Equations MPs Applied MP 8
Procedural Lesson * Embedded MP *
Grade 8 · Unit 4 · Lesson 11
1234567
MC: 8.EE.7
* * * *
Problem of the Day Student Journal Pages
208-213
Objective: I will write and solve linear equations.
Vocabulary Teacher Resources
Read for Understanding: Considerations:
Read problems slowly. Stop at each punctuation
mark to identify key words that indicate When setting up word problems as equations,
operations. encourage students to read the problem slowly and
more than once. Students should focus on the verbs
Define a Variable: or action in the sentences to help them determine the
Identify the missing quantity in an equation. operation. Some problems will require students to
Before writing and solving the equation, choose a solve for a variable, but that is not the final answer.
variable and state what the variable represents. There may be multiple answers to an equation and all
Be sure to include labels or units of measure. answers must be given to have a solution. Request
that students use substitution to plug answers back
Solve and Check: into the equation to check work. Encourage pictures
Plug the solution back into the equation and or diagrams to help organize information.
check the accuracy. Does the answer make sense
with the question? Is the answer reasonable? *Reminder: All forms of the word “is” indicate the
equal sign in a word problem (is, was, will be).
Steps:
1. Read and pause at the punctuation to
record thinking.
2. Circle keywords that indicate the
operation to use.
3. Underline the question. This is the
missing variable.
4. Write the equation and solve.
5. Plug the solution into the equation.
6. Check for reasonableness.
Application of MPs:
MP4: How would it help to visually organize the
information given in a word problem?
Visually organizing the information will help
to ______________.
MP6: What words, symbols or mathematical
notations are important in this problem?
The word, symbols or mathematical notations
that are important in this problem are .
381 Copyright © Swun Math Grade 8 Unit 4 Lesson 11 P TE
Input/Model
(Teacher Presents)
Directions: Write an equation and solve. Label the answer.
1. Two more than five times a number is 57.
x = a number
Solution:
• Write an equation to match the problem: 2 + 5x = 57
• Solve. Subtract 2 from both sides. 5x = 55
• Solve for x by dividing both sides by 5. 5 = 55
5 5
Answer: x =11
2. Hank scored 32 points in a basketball game. This is 4 more than twice as many points as
he scored the week before. How many points did Hank score the week before?
x = Hank’s points in previous game
Solution:
• Write an equation to match the problem: 32 = 4 + 2x (32 is 4 more than twice the points in previous games)
• Solve. Subtract 4 from both sides. 28 = 2x
28 2
• Solve for x by dividing both sides by 2. 2 = 2
Answer: x = 14
Copyright © Swun Math Grade 8 Unit 4 Lesson 11 P TE 382
Structured Guided Practice
(A/B Partners Practice)
Directions: Write an equation and solve. Label the answer.
1. The quotient of a number and seven, decreased by six, is 8.
x = a number
S7 o−lu6tio=n:8
+ 6 + 6
7 = 14
(7) = 14(7)
7
= 98
2. A repairperson charges $55 for a service call to fix a washing machine plus $62 per hour
to fix the machine. If the repairperson charged the Smith family three hundred ninety-
six dollars to fix their washer, how many hours did the repair take?
x = number of hours for job
Solution:
55 + 62 = 396
−55 − 55
62 = 341
62 341
62 = 62
= 5.5
5.5 hours
383 Copyright © Swun Math Grade 8 Unit 4 Lesson 11 P TE
Final Check for Understanding
(Teacher Checks Work)
Directions: Write an equation and solve. Label the answer.
1. The cost of 6 boxes of cereal and blueberries for $4.98 totaled $26.46. How much was each
box of cereal?
x = cost of a box of cereal
Solution:
6 + 4.98 = 26.46
− 4.98 − 4.98
6 = 21.48
6 21.48
6= 6
= $3.58
Each box of cereal costs $3.58.
2. Together, Susan and Laura earned $95 babysitting during the month of September. Susan
earned twelve dollars more than Laura. How much did each girl earn during September?
x = Laura’s earnings
Solution:
= Laura
+ 12 = Susan
+ + 12 = 95
2 + 12 = 95
−12 – 12
2 = 83
2 83
2=2
= $41.50
Laura: = $41.50
Susan: + 12 = $53.50
Copyright © Swun Math Grade 8 Unit 4 Lesson 11 P TE 384
Student Practice Name: ___________________________
Date: ___________________________
Unit 4 · Lesson 11: Linear Equations: Word Problems
Directions: Write an equation and solve. Label your answer.
1. On a recent shopping trip, Max bought a 2. Five times the sum of a number y and
watch that costs $12.75 and six bracelets. eleven is three less than seventeen. Find
Each bracelet was the same price. The y.
total, before tax, was $38.25. How much
did each bracelet cost? y = a number
n = cost of a bracelet
Solution: Solution:
$4.25 per bracelet
= −8 1
5
3. A delivery truck is carrying 8,742 pounds 4. Sydney bought some plants at $3.79 each.
of dog food as its only cargo. At a weigh After deducting a $10 coupon, the cost of
station, the truck weighed 27,642 pounds. the plants was thirty-five dollars and
What is the weight of the empty truck? forty-eight cents. How many plants did
Sydney buy?
x = weight of truck
x = number of plants
Solution: Solution:
18,900 pounds 12 plants
5. Jackson scored 27 points in a game. This 6. Paul and Mary scored 114 points together
is 7 more than twice as many points as in the race. Mary scored 36 points more
Peter scored. How many points did they than Paul. How much did each player
score together? score during the race?
x = Peter’s score
x = Paul’s score
Solution: Solution:
Peter scored 10 points. Peter and Jackson scored 37 Paul scored 39 points and Mary scored 75 points.
points together.
385 Copyright © Swun Math Grade 8 Unit 4 Lesson 11 P TE
Challenge Problems
Directions: Write an equation and solve. Label the answer.
Julianne made 512 grams of trail mix. She used twice as many grams of peanuts as raisins.
She used twelve more grams of dried pumpkin seeds than raisins. How much of each
ingredient did Julianne use?
Solution:
= raisins
2 = peanuts
+ 12 = pumpkin seeds
+ 2 + + 12 = 512
4 + 12 = 512
−12 − 12
4 = 500
4 = 500
4 4
= 125
= 125 grams of raisins
2 = 250 grams of peanuts
+ 12 = 137 grams of pumpkin seeds
Extension Activity
* MP1: Make sense of the problem and persevere in solving it.
* MP4: Apply mathematics in everyday life.
Gina has some dimes and some nickels with a total of $4.60. The number of dimes is four
more than three times the number of nickels she has. How many nickels does she have?
Which quantity depends on the other? Remember to distinguish the difference in value
between nickels and dimes when writing your equation. You can only use one variable to
write your equation, but you must use it more than once. Define the expression for nickels
and dimes. Write an equation and solve.
Copyright © Swun Math Grade 8 Unit 4 Lesson 11 P TE 386
Reaching Consensus
Reaching Consensus
*MP3: Do you agree or disagree with your classmate? Why or why not?
Student Presentations
*MP1: What steps in the process are you most confident about?
*MP6: Explain how you might show that your solution answers the problem.
Closure
Recap today’s lesson with one or more of the following questions:
MP4: How would it help to visually organize the information given in a word
problem?
MP6: What words, symbols or mathematical notations are important in this
problem?
387 Copyright © Swun Math Grade 8 Unit 4 Lesson 11 P TE
Homework Name: ___________________________
Date: ___________________________
Unit 4 · Lesson 11: Write Linear Equations
Objective: I will write and solve linear equations.
Vocabulary Steps:
Read for Understanding: 1. Read and pause at the punctuation
Read problems slowly. Stop for punctuation and identify to record thinking.
key words that indicate operations.
2. Circle keywords that indicate the
Define a Variable: operation to use.
Identify the missing quantity in an equation. Before
writing and solving the equation, choose a variable and 3. Underline the question. This is the
state what the variable represents. Be sure to include missing variable.
labels or units of measure.
4. Write the equation and solve.
5. Plug the solution into the equation.
6. Check for reasonableness.
Solve and Check:
Plug the solution back into the equation and check the
accuracy. Does the answer make sense with the question?
Is the answer reasonable?
*Reminder: Is, was, will be all indicate the = sign in a word
problem.
Example # 1 Example # 2
Directions: Write an equation and solve. Label your answer.
Two more than five times a number is 57. Hank scored 32 points in a basketball game. This is 4
x = a number more than twice as many points as he scored the
week before. How many points did Hank score the
week before?
x = Hank’s points in previous game
Solution: Solution:
5 + 2 = 57 32 = 2 + 4
−4 − 4
−2 − 2 28 = 2
5 = 55
28 2
5 = 55 2=2
5 5 14 points =
= 11
Copyright © Swun Math Grade 8 Unit 4 Lesson 11 P TE 388
Homework
Unit 4 · Lesson 11: Write Linear Equations
Directions: Write an equation and solve. Label the answer.
1. The sum of 4 and seven times a number is 2. The sum of 2 times p and 2 is 48. What is
130. What is the number? p?
3. Jim bought 2 books yesterday. The price 4. In a survey tasting lemonade, one fifth of
of one book was 4 times the cost of the the people surveyed were under 25 years
other. If both books cost $26.40 old. If 65 people were under 25, how
altogether, how much was each book? many people were surveyed?
389 Copyright © Swun Math Grade 8 Unit 4 Lesson 11 P TE
Homework
Unit 4 · Lesson 11: Write Linear Equations
5. “My new phone cost a fortune,” said Amy. 6. The Gophers won 2/3 of their games. The
“Mine costs twice as much,” said Jill. If Rattlers won half of their games.
together, the two phones cost $575.25, how Together, the teams won 56 games. Both
much was each girl’s phone? teams played the same number of games.
How many games did each team win?
Explain the steps you used to solve problem number _______.
______________________________________________________________________________________________________________________
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______________________________________________________________________________________________________________________
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______________________________________________________________________________________________________________________
______________________________________________________________________________________________________________________
Copyright © Swun Math Grade 8 Unit 4 Lesson 11 P TE 390
Answer Key
Extension Activity
x= nickels and because nickels are worth five cents, the value of nickels is 5x. Dimes depend on nickels, so
dimes will equal 4 + 3x, but because dimes are worth ten cents, they will be expressed as 10(4 + 3x). The
equation for how many nickels and dimes is:
5 + 10(4 + 3 ) = 460
5 + 40 + 30 = 460 or 35 + 40 = 460
35 = 420
x= 12 nickels
Homework 2. (2 x p) + 2 = 48; p = 23
1. 4 + 7x = 130; x = 18 4. 1 = 65; x = 325 people surveyed
5
3. x + 4x = 26.40; x (first book) = $5.28; 4x (second
book) = $21.12
5. x + 2x + $575.25; x = $191.75 (Amy’s phone); 2x = 5. 2 + 1 = 56 ; x = 48 games; 32 gopher wins; 24
$383.50 (Jill’s phone) 3 2
rattler wins
391 Copyright © Swun Math Grade 8 Unit 4 Lesson 11 P TE
Notes
Copyright © Swun Math Grade 8 Unit 4 Lesson 11 P TE 392
Linear Equations MPs Applied MP
Math Task * Embedded MP
Grade 8 · Unit 4 · Lesson 12 12345678
MC: 8.EE.7 * ** * *
Student Journal Pages
Objective: I will demonstrate my understanding of solving linear equations. 214-216
Student Practice
• Explain the problem and ask students to solve independently.
• Explain the problem and ask students to independently solve the problem on
their recording sheet.
independently.
• While students are working, look for 3 - 4 student examples to review during
presentations.
Reaching Consensus
• Students listen to other members explain how they solved the task.
• Students reach consensus on the answer, not on the solution process.
Student Presentations
• When selecting students to give presentations, choose those who will provide
the greatest teaching opportunity.
• Students who completed the Student Practice incorrectly and received help
during Reaching Consensus should present their mistake and how they
corrected it.
393 Copyright © Swun Math Grade 8 Unit 4 Lesson 12 T TE
Final Check for Understanding
• Are students able to articulate metacognition?
• Are students able to articulate what they would do differently to solve the problem?
• How did you solve the problem?
• Are the students’ answers reasonable?
• How would you describe the problem in your own words?
• In what way does this problem connect to other mathematical concepts?
• What approach are you considering trying first?
Reflective Closure
• Were students able to demonstrate their knowledge of solving linear equations?
• Were students able to verbalize and use mathematical language?
Answer Key
1. Draw each side separately. Remember there are 2 of each side, so the area of each side must be doubled.
Drawing the sides separately is an excellent resource for understanding surface area.
Front & back Top & bottom 8 sides
12 8
12
2. Volume = l·w·h
6(4.5)x = 229.5 cubic inches
27x = 229.5
27 27
x = 8.5 inches; The height is 8.5 in.
3. A=2(12)(16) 16 in
12 in
A=24(16)
A=384 square inches
4. A=(2 +12)16
384=32x+192
=6 in flap
Copyright © Swun Math 394Grade 8 Unit 4 Lesson 12 T TE
Math Task Name: ___________________________
Date: ___________________________
Unit 4 · Lesson 12: Linear Equations
Directions: Solve and interpret.
1. You need 80 inches of ribbon for each package like the one below. The bow takes 34
inches of ribbon. The width of the box is 16 inches. The ribbon will be wrapped once
around the width of the package and then tied into a bow. Write an equation to find the
height of the box.
16
Solution:
= height of box
2 + 2(16) + 34 = 80
2 + 32 + 34 = 80
2 + 66 = 80
2 = 14
= 7 inches for height of box
2. Using the information in Problem #1, what would be the total amount of ribbon needed to
wrap 32 boxes like the one above? What would be the total amount of ribbon for 20
smaller boxes that use half as much ribbon? How many yards is that? Explain how
much you would buy.
Solution:
= total ribbon needed
= 32(80) + 20(40)
= 2560 + 800
= 3360 inches of ribbon
= 93.33 yards or about 93 yards
395 Copyright © Swun Math Grade 8 Unit 4 Lesson 12 T TE