Writing Equations(Week 2 Day 1)

Kristy ChimeraCasie Croff

Nichole Oswald

Writing EquationsMath A (Grade 9)

Materials: Calculator, worksheets, pencils, guided notes, chalk, chalkboard ***All worksheets and notes will be provided for all groups.Lesson Overview: Students will develop a clear understanding of how to create algebraic equations from verbal sentences and also how to create sentences from algebraic equations. The students will do this by working with a partner on worksheets, after they are taught the material through the use of guided notes.

Lesson Objectives: Students will be able to execute the Four-Step Problem Solving Plan when

solving word problems. (Application Level) Students will be able to translate sentences into algebraic equations. (Analysis

Level) Students will be able to translate equations into sentences. (Analysis Level)

NYS Standards: Key Idea 3: Operations

~Use addition, subtraction, multiplication, division, and exponentiation with real numbers and algebraic expressions. (3A)

Key Idea 4: Modeling/Multiple Representation ~Represent problem situations symbolically by using algebraic

expressions. (4A)

Anticipatory Set: (time 5 minutes)Pass out the Let’s Get Started worksheet out to the students. The students will be required to write an algebraic expression based on a sentence, and also a sentence based on an algebraic expression. (This was introduced in the very first lesson taught.)

5 x2 -3 7

Answer: 37 less than 5 times the square of a number

You are having a birthday party; therefore you need to mail out invitations. Each invitation costs x dollars at the store. What would the algebraic expression be to show the cost of the greeting cards if you were to invite 17 people?

Answer: 17xGo over the answers as a class.Developmental Activity: (20 minutes)

1.) Pass out Guided Notes for Writing Equations worksheet.2.) Fill in the blanks with the students.3.) Work through the examples as a class as they appear in the notes on the board.

4.) If the students seem to be struggling see attached sheet called Extra Examples. This sheet contains 2 more four-step problem-solving plan questions. That is what the students will most likely struggle with.

Closure: (14 minutes)Students will work in pairs on Write it Out worksheet. I will circulate around as the students are working to make sure that they are on task and know how to complete the questions. Once the students are finished I will go over the answers on the chalk board with the students. The students will tell me the steps that they took to obtain their answers. If students are struggling or still have questions go back and review problem areas.

Assessment: (pass out right before students leave 1 min)Students will be given a homework worksheet that will be collected next class. During the next class students’ questions that they had on the homework should be addressed.

Name: Date:

Let’s Get Started!!

1 .) 5 x2 -3 7

2.) You are having a birthday party; therefore you need to mail out invitations. Each invitation costs x dollars at the store. What would the algebraic expression be to show the cost of the greeting cards if you were to invite 17 people?

Name: Teacher Copy Date:

Guided Notes for Writing Equations

An equation is an algebraic expression that is set equal to something.

Examples of other expressions that imply the equals sign:is, is equal to, is as much as, equals, is the same as, is identical to

Recall, variables are used to represent the unspecified values in a sentence or problem.

Translate a sentence into an equation, example:Two times a number n decreased by eight is equal to seventy.

2 n -8=70

Example: Five times the sum of x and y is the same as seven times x.5 ( x+y)=7x

Now let us try it the other way!Translate and equation into a sentence:

3(a+b)=3aThree times the sum of a and b is the same as three times a.

Example: 20(d-f)=18dTwenty times the difference between d and f equals eighteen times d.

The four-step problem-solving plan can help you solve any word problem!!

Step 1: Explore the problem~Identify what info is given and what you are suppose to find

Step 2: Plan the solution~Write an equation

Step 3: Solve the problemStep 4: Examine the solution

~Does the answer make sense?

Example: Use the Four-Step Plan. The population of the United States in 2001 was about 284,000,000 and the land area of the United States is about 3,500,000 square miles. Find the average number of people per square mile in the United States.

Explore: You know there are 284,000,000 people. You want to find out the number of people per square mile.

Plan: Write equation-let p be the number of people per square mile3,500,000 x p=284,000,000

Solve: p is approximately 81.14. Therefore there are about 81 people per square mile.

Examine: 81 x 3,500,000=283,500,000 which is about 284,000,000

Name: Date:

Guided Notes for Writing Equations

An _____________ is an algebraic expression that is set ______________to something.

Examples of other expressions that imply the equals sign:______, ______________, ____________, __________, ___________,

_______________

Recall, ___________ are used to represent the unspecified values in a sentence or problem.

Translate a sentence into an equation, example:Two times a number n decreased by eight is equal to seventy.

Example: Five times the sum of x and y is the same as seven times x.

Now let us try it the other way!Translate and equation into a sentence:

3(a+b)=3a

Example: 20(d-f)=18d

The __________________________________ can help you solve any word problem!!Step 1: ___________ the problem

~Identify what info is given and what you are suppose to _________.Step 2: Plan the _____________

~Write an _____________Step 3: __________ the problemStep 4: Examine the _____________

~Does the answer make sense?

Example: Use the Four-Step Plan. The population of the United States in 2001 was about 284,000,000 and the land area of the United States is about 3,500,000 square miles. Find the average number of people per square mile in the United States.

Explore:

Plan:

Solve:

Examine:

Extra Examples(use if students are struggling with the four-step problem-solving plan)

1.) A rectangular garden has a width that is 8 feet less than twice the length.  Find the dimensions if the perimeter is 20 feet.

Explore: We are looking for the length and width of the rectangle.  Since width can be written in terms of length, we will let

L = lengthWidth is 8 feet less than twice the length:

2L - 8 = widthPlan: Perimeter = 2 (length) + 2 (width)20 = 2L + 2 ( 2L - 8)

Solve:20 = 2L +2 ( 2L -8)20 = 2L +4L -1620 = 6L -1636 = 6L6 = L

Examine:If length is 6, then width, which is 8 feet less than twice the length, would have to be 4.  The perimeter of a rectangle with width of 4 feet and length of 6 feet is 20 feet.

2.) An investor with $70,000 decides to place part of her money in corporate bonds paying 12% per year and the rest in a certificate of deposit paying 8% per year.  If she wishes to obtain an overall return of $6300 per year, how much should she place in each investment?

Explore: We are looking for how much she invested in EACH account. x = amount invested in 12% Since the two accounts together need to be $70000, then we can take the total (70000) and subtract from it the “given” number in the 12% account (x): 70,000 - x = amount invested in 8%

Plan: .12x + .8 (70,000 - x) = 6300

Solve: .12x + .08 (70,000 - x) = 6,300.12x + 5,600 - .08x = 6,300.04x = 700X = 17,500

Examine: If she invested $17500 in corporate bonds, then she would have to invest $70000 - $17500 = $52000 in the certificate of deposit. If you take 12% of $17500 and add it to 8% of $52500 you do get $6300

Problems taken from: http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut17_prob2.htm

Name: Date:

Write it Out

Translate the sentence into an equation:

One-third of the sum of m and n is 72.

Translate the equation into a sentence:

2 x2 + y2 = 1 8

Use four-step problem-solving plan:The first ice cream plant was established in 1851 by Jacob Fussell. Today, 2,000,000 gallons of ice cream are produced in the United States each day. Use this information to find out: in how many days can 40,000,000 gallons of ice cream can be produced in the United States?

Explore:

Plan:

Solve:

Examine:

Name: Date:

Homework

Show All Work!!!

Mrs. Smith is planning to place a fence around her garden. The fencing costs $2.25 per yard. She buys f yards of fencing and pays $5.55 in tax. If the total cost of the fencing is $52.80, write an equation to represent the situation.

Translate into a sentence:

Use the following information to answer the next two questions:

Matthew is training for to prepare for a wrestling tournament. He weighs 170 pounds now. He wants to lose weight so that he can start the tournament at 160 pounds.

If t represents the number of pounds he wants to lose, write an equation to represent the situation.

How many pounds does Matthew need to lose to reach his goal?Name: Answer Key Date:

Write it Out

Translate the sentence into an equation:

One-third of the sum of m and n is 72.

1/3(m + n)=72

Translate the equation into a sentence:

2 x2 + y2 = 1 8

Y squared added to two times x squared is equal to 18

Use four-step problem-solving plan:The first ice cream plant was established in 1851 by Jacob Fussell. Today, 2,000,000 gallons of ice cream are produced in the United States each day. Use this information to find out: in how many days can 40,000,000 gallons of ice cream can be produced in the United States?

Explore:

You know that 2,000,000 gallons of ice cream are produced in the US each day. You want to know how many days it will take to produce 40,000,000 gallons of ice cream.

Plan:Write an equation to represent the situation. Let d=number of days needed to produce the ice cream.2,000,000 x d = 40,000,000

Solve:D = 20 it will take 20 days

Examine: 2,000,000 x 20 = 40,000,000 the answer makes sense.

Name: Answer Key Date:

HomeworkShow All Work!!!

Mrs. Smith is planning to place a fence around her garden. The fencing costs $2.25 per yard. She buys f yards of fencing and pays $5.55 in tax. If the total cost of the fencing is $52.80, write an equation to represent the situation.

$2.25 x f + $5.55 = $52.80

Translate into a sentence:

6 added to 2 times h is equal to 19

2 times j equals k subtracted from 7 times j

Use the following information to answer the next two questions:

Matthew is training for to prepare for a wrestling tournament. He weighs 170 pounds now. He wants to lose weight so that he can start the tournament at 160 pounds.

If t represents the number of pounds he wants to lose, write an equation to represent the situation.170 - t = 160

How many pounds does Matthew need to lose to reach his goal? 10 pounds

Solving Multi-Step Equations(Week 2 Day 2)

Kristy ChimeraCasie Croff

Nichole Oswald

Title: Solving Multi-step EquationsGrade 10

Materials:1. Chalkboard (or whiteboard)2. Chalk (or whiteboard markers)3. Index cards (I will have enough for everyone)4. Worksheets (I will also have all worksheets and answer keys)

Lesson Overview:Students will learn, through the index card activity, group work, and direct instruction, how to work backwards in a word problem, and how to solve multi-step equations.

Objectives:1. The student will be able to work backwards when solving a multi-step equation

(application level of Bloom’s taxonomy).2. The student will be able to solve multi-step equations for a specified variable

(application level of Bloom’s taxonomy).3. The student will be able to isolate the variable on one side of an equation (application

level of Bloom’s taxonomy).4. The student will be able to identify all words associated with addition, subtraction,

multiplication, and division (comprehension level of Bloom’s taxonomy).5. The student will be able to explain, using mathematical language, how they reached

their solution to the given problems (application level of Bloom’s taxonomy).6. The student will be able to summarize the steps involved in working backwards, and

solving multi-step equations (synthesis level of Bloom’s taxonomy).7. The student will be able to analyze word problems to determine the steps that are

needed to work backwards in a word problem and to solve multi-step equations (analysis level of Bloom’s taxonomy).

NYS Standards:Key Idea 1 – Mathematical Reasoning1A: Construct valid arguments.1B: Follow and judge the validity of arguments.Key Idea 3 - Operations3A: Use addition, subtraction, multiplication, division, and exponentiation with real numbers and algebraic expressions.Key Idea 4 – Modeling and Multiple Representations

4A: Represent problem situations symbolically by using algebraic expressions, sequences, tree diagrams, geometric figures, and graphs.4E: Model real-world problems with systems of equations and inequalities.

Anticipatory Set: (6 minutes)As the students take their seats in their groups, give the students an index card with a vocabulary word on it (the teachers can play too). The students will have to group themselves together with other students whose word means the same as theirs (for example add and plus). Leftover cards will be discussed and placed in the appropriate group as a class. The groups should be as follows:

Group 1: add, sum, increased, plus, more thanGroup 2: subtract, difference, decreased, minus, less thanGroup 3: multiply, product, timesGroup 4: divide, quotient, per

Developmental Activity: (15 minutes)1. Hand out the worksheet titled: Review Sheet: Operations.2. Tell the students that the groups they just made with the index cards are listed here.

They should know and understand all of those terms and their meanings.3. Ask the students what goes in each blank on the second half of the page. You should

try to get multiple students to participate. Then complete each example that goes with each blank.

4. Next, hand out the worksheet titled: “Undoing Operations Practice”.5. Work through problems 1-3 in section 1, and problems 1-3 in section 2 as a class. Try

to get students to explain how you should complete the problem. Give them guidance if needed.

Guided Practice: (15 minutes)1. Have the students work in groups of 2 to complete numbers 3-6 in sections 1 and

numbers 3-6 in section 2 of their worksheets.2. Have every student put an answer to one of the problems that they completed on the

board, and have them explain to the class how they reached their answer, using mathematical language.

Closure: (4 minutes)Hand out the worksheet titled: Self evaluation. Have the students fill out the

worksheet. Collect the worksheet. Tell the students that this is NOT for a grade, but they should be able to fill out the worksheet based on the material covered in class today. If they cannot complete the worksheet, then they should review what was done today so they can complete the worksheet and understand the material.

Independent Practice:For homework, finish number 7 in section 1 and numbers 7-10 in section 2. Also complete the extra credit problems if desired. Tell the students that we will choose a couple of the problems to correct for a grade (but don’t tell them which ones). (We will grade number 7 in section 1, and numbers 7 and 9 in section 2. The extra credit problems

will be worth ½ of a point and the problems in section 1 and 2 that we correct will be worth 2 points each (1 point for the correct answer and 1 point for the work shown, for a total of 6 possible points (or with extra credit 7.5 points)Name: ________________________________________ Date: _________________

Review Sheet: OperationsAddition Subtraction Multiplication Division

Add Subtract Multiply DivideSum Difference Product QuotientPlus Minus Times PerIncreased Decreased More Than Less Than

Make sure you know and understand that these terms are interchangeable (they mean the same thing)_____________________________________________________________

“Undoing Operations”Directions: As a class, we will fill in the blanks and complete the examples.

1.) To “undo” addition, we use ____________________ Ex.) X + 5 = 8 Check:

2.) To “undo” multiplication, we use _________________ Ex.) 6X = 24 Check:

3.) To “undo” division, we use ____________________ Ex.) X/10 = 5 Check:

4.) To “undo” subtraction, we use ____________________ Ex.) X - 50 = 50 Check:Name: ________________________________________ Date: __________________

“Undoing Operations” PracticeSection 1: Working Backwards

Directions: Solve the following problems by working backwards. Show ALL work to receive full credit. Explain, using mathematical language, how you reached your solution.

1. A number is divided by 3, and then 4 is added to the quotient. The result is 8. Find the number.Explanation: _____________________________________________________________

_______________________________________________________________________

2. Three times a number plus 3 is 24. Find the number.

Explanation: ____________________________________________________________________________________________________________________________________

3. Car Rental: Angela rented a car for $29.99 a day plus a one-time insurance cost of $5.00. Her bill was $124.96. For how many days did she rent the car?

Explanation: ____________________________________________________________________________________________________________________________________

4. A number is multiplied by 5, and then 3 is subtracted from the product. The result is 12. Find the number.

Explanation: ____________________________________________________________________________________________________________________________________5. Money: Mike withdrew an amount of money from his bank account. He spent one fourth for gasoline and had $90 left. How much money did he withdraw?Explanation: ____________________________________________________________________________________________________________________________________

6. Eight is subtracted from a number, and then the difference is multiplied by 2. The result is 24. Find the number.

Explanation: ____________________________________________________________________________________________________________________________________

7. Math A problem from the June 2004 exam:At the beginning of her mathematics class, Mrs. Reno gives a warm-up problem.

She says, “I am thinking of a number such that 6 less than the product of 7 and this number is 85”. Which number is she thinking of?

Explanation: ___________________________________________________________________________________________________________________________________________________________________________________________________________

Section 2: Solving Multi-step Equations

Directions: Solve each question for the variable. You MUST show ALL work to receive full credit! Check you answers!

1. 5X + 2 = 27 Check: 2. 0.2X - 8 = -2 Check:

3. 284

−+b

= 10 Check: 4. 6X + 9 = 27 Check:

5. 14n - 8 = 34 Check: 6. 16 = 1412−d

Check:

7. 0.6X - 1.5 = 1.8 Check: 8. (7/8)p - 4 = 10 Check:

9. 0 = 10y - 40 Check: 10. 3.2y - 1.8 = 3 Check:Extra Credit Problems:

Directions: Write an equation and solve each problem.

1. Find 3 consecutive integers whose sum is 96.

2. Find 2 consecutive odd integers whose sum is 176.

3. Find 3 consecutive integers whose sum is -93.

Name: _______________________________________ Date_______________

Self Evaluation(This is NOT for a grade, but it will be collect for completeness)

Directions: Match the term on the left with the corresponding term on the right.

1. _______ Divide a. Add

2. _______ Product b. Quotient

3. _______ Subtract c. Difference

4. _______ Sum d. TimesDirections: Math the term on the left with the term on the right that is the “undo” operation.

1. _______ Add a. Divide

2. _______ Multiply b. Subtract

3. _______ Subtract c. Multiply

4. _______ Divide d. AddNow summarize in your own words, using complete sentences, the steps involved in

working backwards in a problem, and solving multi-step equations and why these steps work (use the back of this paper if necessary).________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________Name: ______Answer Key________________________ Date: _________________

Review Sheet: OperationsAddition Subtraction Multiplication Division

Add Subtract Multiply DivideSum Difference Product QuotientPlus Minus Times Increased Decreased

Make sure you know and understand that these terms are interchangeable (they mean the same thing)_____________________________________________________________

“Undoing Operations”Directions: As a class, we will fill in the blanks and complete the examples.

1.) To “undo” addition, we use Subtraction_________ Ex.) X + 5 = 8 Check: - 5 - 5 3 + 5 = 8X = 3 8 = 8

2.) To “undo” multiplication, we use __Division__________ Ex.) 6X = 24 Check: 6 6 6*4=24 X = 4 24 = 24

3.) To “undo” division, we use _Multiplication_____ Ex.) X/10 = 5 Check: *10 *10 50/10 = 5 X = 50 5 = 5

4.) To “undo” subtraction, we use ___Addition_____Ex.) X - 50 = 50 Check: + 50 + 50 100-50=50 X = 100 50 = 50

Name: _______Answer Key_______________________ Date: ___________________

“Undoing Operations” Practice

Section 1: Working Backwards

Directions: Solve the following problems by working backwards. Show ALL work to receive full credit. Explain, using mathematical language, how you reached your solution.

1. A number is divided by 3, and then 4 is added to the quotient. The result is 8. Find the number.

8 - 4 = 4 4 * 3 = 12. The original number is 12. Check: 12/3 = 4. 4 + 4 = 8.

Explanation: ____I started out with 8. To undo the addition, I subtracted. Then, to undo the division, I multiplied. The original number is 12.

2. Three times a number plus 3 is 24. Find the number.

24 - 3 = 21. 21/3 = 7. The original number is 7. Check: 7 * 3 = 21. 21 + 3 = 24.

Explanation: ____I started with 24. To “undo“ the plus 3, I subtracted 3. Then to undo the three times, I divided by 3. The original number is 7.

3. Car Rental: Angela rented a car for $29.99 a day plus a one-time insurance cost of $5.00. Her bill was $124.96. For how many days did she rent the car?

124.96 - 5.00 = 119.96 119.96/29.99 = 4. She rented the car for 4 days.Check: 4 * 29.99 = 119.96 119.96 + 5.00 = 124.96

Explanation: __I started with 124.96. Then I subtracted the one-time insurance cost of 5.00. Then I divided 119.96 by 29.99 because it cost 29.99 per day. My answer is 4. 4. A number is multiplied by 5, and then 3 is subtracted from the product. The result is 12. Find the number.

12 + 3 = 15. 15/5 = 3. The original number is 3. Check: 3 * 5 = 15 15-3 = 12.

Explanation: ___I started with 12. To undo the subtraction, I used addition. Then, to undo the multiplication, I used division. The original number is 3.

5. Money: Mike withdrew an amount of money from his bank account. He spent one fourth for gasoline and had $90 left. How much money did he withdraw?

90/(3/4) = 120. That is the total money that he withdrew. So Mike spent (1/4) of 120 on

gas. So he spent 120*(1/4) which is 30.

Explanation: __I started with 90. I divided by 3/4 because 90 is ¾ of the money that Mike has left. Dividing 90 by ¾ will give you the total amount he started with. I get 120. Then to take (1/4) of 120, I multiply (1/4)*120 and get 30, which is the amount of money Mike spent on gas.

6. Eight is subtracted from a number, and then the difference is multiplied by 2. The result is 24. Find the number.

24/2 = 12. 12 + 8 = 20. The original number is 20. Check: 20 - 8 = 12. 12 * 2 = 24.

Explanation: __I started with 24. I divided by 2 to undo the multiplication. Then I added 8 to undo the subtraction. The original number is 20.

7. Math A problem from the June 2004 exam:At the beginning of her mathematics class, Mrs. Reno gives a warm-up problem.

She says, “I am thinking of a number such that 6 less than the product of 7 and this number is 85”. Which number is she thinking of?

85 + 6 = 91. 91/7 = 13. The original number is 13. Checking the answer, we get 13*7 = 91. Then 91 – 6 = 85.

Explanation: __I started with 85. I wanted to undo subtraction, so I added 6. Then I wanted to undo multiplication, so I divided by 7. S ection 2: Solving Multi-step Equations

Directions: Solve each question for the variable. You MUST show ALL work to receive full credit!

1. 5X + 2 = 27 2. 0.2X - 8 = -2 - 2 - 2 + 8 + 8 5X = 25 0.2 X = 6 5 5 0.2 0.2 X = 5 X = 30

3. 284

−+b

= 10 4. 6X + 9 = 27

*(-2) *(-2) - 9 - 9 4b+8 = -20 6X = 18 -8 -8 6 6 4b = -28 X = 3 4 4 b = -7

5. 14n - 8 = 34 6. 16 = 1412−d

+ 8 + 8 *14 *14 14n = 42 224 = d-12 14 14 +12 +12 n = 3 236 = d

7. 0.6X - 1.5 = 1.8 8. (7/8)p - 4 = 10 +1.5 +1.5 +4 +4 0.6X = 3.3 (7/8)p = 14 (The other 0.6 0.6 *8 *8 way is to use X = 5.5 7p = 112 *8/7) Either 7 7 way is fine. p = 16

9. 0 = 10y - 40 10. 3.2y - 1.8 = 3 +40 +40 +1.8 +1.8 40 = 10y 3.2y = 4.8 10 10 3.2 3.2 4 = y y = 1.5Extra Credit Problems:

Directions: Write an equation and solve each problem.

1. Find 3 consecutive integers whose sum is 96.

n + n + 1 + n + 2 = 963n + 3 = 96 -3 -3 So the three integers are n = 31, n + 1 = 32 and n + 2 = 333n = 93 3 3 n = 31

2. Find 2 consecutive odd integers whose sum is 176.

(2n + 1) + (2n + 3) = 176 4n + 4 = 176 So the two consecutive odd integers are (2n+1) = 87 and (2n+3) = 89 -4 -4 4n = 172 4 4 n = 43

3. Find 3 consecutive integers whose sum is -93.

n + (n+1) + (n+2) = -93 We know that our n’s must be negative because the final

-n +-( n + 1)+-( n + 2) = -93 sum is negative. -3n - 3 = -93 So the three consecutive integers are -n = 30, -(n+1) = 31, and –(n+2) =

+3 +3 32. So plugging n back into my equation, I get –n = -30, -(n+1) = 3n = -90 -31, and –(n+2) = -32 3 3 n = 30

Name: ____Answer Key _________________________ Date_______________

Self EvaluationDirections: Match the term on the left with the corresponding term on the right.

1. ___b___ Divide a. Add

2. ___d___ Product b. Quotient

3. ___c___ Subtract c. Difference

4. ___a___ Sum d. Times

Directions: Math the term on the left with the term on the right that is the “undo” operation.

1. ___b___ Add a. Divide

2. ___a___ Multiply b. Subtract

3. ___d___ Subtract c. Multiply

4. ___c___ Divide d. Add

Now summarize in your own words, the steps involved in working backwards in a problem, and solving multi-step equations and why these steps work.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Solving Equations with Variables on Each Side(Week 2 Day 3)

Kristy ChimeraCasie Croff

Nichole OswaldLesson 3.5 – Day 3

Solving Equations with the Variable on Each SideAlgebra 1Topic: Solving equations with variables on each side

Materials: Students: pen/pencil notes (provided) homework / practice worksheet (provided)

Teacher:  notes for class homework / practice worksheet for class

post – it notes with solutions for Algebra Match Game (I will provide these for all

groups.)

Lesson Overview: - Students will recall knowledge of solving equations and steps involved in solving equations containing variables.  - Students will discover how and learn to solve equations with variables on each side through guided notes and examples and cooperative work. 

Lesson Objectives:- Students will propose a plan (steps needed) to solve equations with variables on each side. (Synthesis)- Given an equation with one variable on each side, students will be able to calculate and determine the value of the specified variable. (Application)

NYS Standards:  Key Idea 3 - Operations3A: Use addition, subtraction, multiplication, division, and exponentiation with real numbers and algebraic expressions. Key Idea 4 – Modeling / Multiple Representations4A: Represent problem situations symbolically by using algebraic expressions, sequences, tree diagrams, geometric figures, and graphs.

Anticipatory set:  (8 min) Algebra match game - Teacher places post-it notes on the board in random order (I will provide the post-it notes for all groups. This could be done before students arrive). - Teacher passes out an equation to each student. - Student solves equation on his/her sheet and when they have come up with an answer, they can go up to the board and take the post-it note that has that answer on it. - Teacher verifies the students' answers as they return to their desks.  If student has taken the wrong answer, have them put the post-it note back on the board and re-solve their equation.

Developmental Activity:  (25 minutes)

- Pass out the “Solving Equations with Variables on Each Side” worksheet to all students. (I will provide these for all groups.) - Teacher writes an equation (example 1) on the board that has a variable term on each side of the equation.  Teacher asks for student suggestions on how to solve such an equation. - Students lead the discussion with their suggestions on how to solve the equation. Teacher performs students’ suggested operations and solving steps on the board as they are offered. Teacher validates and guides the discussion by correcting students’ errors and misconceptions. Be sure to explain the use of the Properties of Equality for Addition, Subtraction, Multiplication, and Division while completing example 1.  Be sure to point out that the steps are similar when solving these equations and equations with variables on just one side of the equation.  (Sometimes the steps are just repeated in the solving process). - Teacher completes examples 2 and 3 on the board as the students dictate what steps to take and how to solve the equation. Be sure to discuss the results of no solution (example 2) and an identity solution (example 3).- Students complete the Try These Together problems in small groups. Students share and work out their examples for the class. Be sure to ask key questions while going over the solutions. Ex.  What property was used in this step of solving?

Closure:  (5 minutes) - Teacher has students lead discussion by having students come up with and outline the general steps used when solving equations with variables on each side. Students complete the Concept Summary in the notes as the discussion progresses.       Assessment: (1 minute)- Pass out practice / homework worksheets. (This can be done during Closure Activity.)  - Students will complete for next class.

Resources:

http://www.glencoe.com/sec/math/algebra/algebra1/algebra1_05/index.php4/ny

http://www.glencoe.com/sec/math/fybh/na/alg1_index.html

http://www.floridatechnet.org/GED/LessonPlans/Mathematics/Mathlesson46.pdf(Algebra Matching Game)

NAME __________________________________________ DATE ____________

Solving Equations with the Variable onEach Side

Example 1Solve 3(x – 2) = 4x + 5 Steps for solving:

1. _____________________2. _____________________3. _____________________4. _____________________5. _____________________

6. _____________________7. _____________________

Many equations contain variables on each side. To solve these types of equations, first use the Addition or Subtraction Properties of ___________ to write an equivalent equation that has all of the variables on only one side. Then solve.

Properties of Equality

Subtraction Property of Equality: When you subtract both sides of an equation by the same number, the two sides remain equal.

Addition Property of Equality: When you add the same number to both sides of an equation, the two sides remain equal.

Multiplication Property of Equality: When you multiply both sides of an equation by the same number, the two sides remain equal.

Division Property of Equality: When you divide both sides of an equation by the same number, the two sides remain equal.

Example 2Solve: 4(s + 2) – s = 3(s + 1)Some equations may have no solution because there is no value of the variable that will result in a true equation. For example, x+1 = x+2 has no solution; it cannot be true.

Example 3Solve: 4p+2 = 1/3 (12p+3) + 1

An equation that is true for every value of the variable is called an identity. For example, x + x = 2x is true for every value of x.

Try These Together

a. 2(2x – 5) = 6x + 4 b. x + 4 = 3[x-2(x+2)]

Concept Summary:Step 1: Use the __________________ ______________________.Step 2: ____________________ the expressions on each side.Step 3: Use the _________________ and/or _________________ Properties of Equality to get the variables on one side and the numbers without variables on the other side.Step 4: __________________ the expressions on each side of the equals sign.Step 5: Use the _________________ or ______________________ Property of Equality to solve.Name: ________________________________________ Date: ________________

Practice Solving Equations with the Variable on Each Side

Name the property of equality used

x + 57.3 = 91.2 65 = y - 156 - 57.3 -57.3 +156 +156 x = 91.2 – 57.3 65 + 156 = y

Solve each equation. Then check your solution.

18 + 2n = 4n – 9

2 n + 6 = 1 n – 33 4

3 – 4x = 8x + 8

-6(4x + 1) = 5 – 11x

Nine less than half n is equal to one plus the product of -1/8 and n. Find the value of n.

NAME ANSWER KEY DATE ____________

Solving Equations with the Variable onEach Side

Example 1Solve 3(x – 2) = 4x + 5 Steps for solving:3x – 6 = 4x + 5 1. Distributive property to remove parenthesis -3x -3x 2. Collect all terms with x on one side of the equal3x-3x-6 = x+ 5 sign by subtracting 3x from each side -6 = x+5 3. Simplify by adding like terms on each side. -5 -5 4. Subtract 5 from each side -11 = x 5. SimplifyMany equations contain variables on each side. To solve these types of equations, first use the Addition or Subtraction Properties of Equality to write an equivalent equation that has all of the variables on only one side. Then solve using the Multiplication and Division Properties of Equality.

Properties of Equality

Subtraction Property of Equality: When you subtract both sides of an equation by the same number, the two sides remain equal.

Addition Property of Equality: When you add the same number to both sides of an equation, the two sides remain equal.

Multiplication Property of Equality: When you multiply both sides of an equation by the same number, the two sides remain equal.

Division Property of Equality: When you divide both sides of an equation by the same number, the two sides remain equal.

Example 2Solve: 4(s + 2) – s = 3(s + 1)

4s + 8 – s = 3s + 34s – s + 8 = 3s + 33s + 8 = 3s + 3-3s -3s

8 = 3 This statement is false. Since 8 = 3 is a false statement, this equation has no solution.

NOTE:Some equations may have no solution because there is no value of the variable that will result in a true equation. For example, x+1 = x+2 has no solution; it cannot be true.

Example 3Solve: 4p+2 = 1/3 (12p+3) + 1

4p + 2 = 12/3 p + 3/3 +14p + 2 = 4p +1 +14p + 2 = 4p + 2 Since the expressions on each side of the equation are the same,

this equation is an identity. The statement 4p+2 = 1/3 (12p+3) + 1 is true for all values of p.

NOTE:An equation that is true for every value of the variable is called an identity. For example, x + x = 2x is true for every value of x.

Try These Together

a. 2(2x – 5) = 6x + 4 b. x + 4 = 3[x-2(x+2)]

4x – 10 = 6x + 4 x + 4 = 3[x – 2x – 4]-4x -4x x + 4 = 3x – 6x -12 -10 = 2x +4 x + 4 = -3x –

12 -4 -4 +3x +3x -14 = 2x 4x + 4 = -12 2 2 -4 = -4

4x = -16

-7 = x 4 4 x = -4

Concept Summary:Step 1: Use the DISTRIBUTIVE PROPERTY.Step 2: SIMPLIFY the expressions on each side.Step 3: Use the ADDITION and/or SUBTRACTION Properties of Equality to get the

variables on one side and the numbers without variables on the other side.Step 4: SIMPLIFY the expressions on each side of the equals sign.Step 5: Use the MULTIPLICATION or DIVISION Property of Equality to solve.

Name: ANSWER KEY Date: ________________

Practice Solving Equations with the Variable on Each Side

Name the property of equality used

x + 57.3 = 91.2 65 = y - 156 - 57.3 -57.3 +156 +156 x = 91.2 – 57.3 65 + 156 = y

SUBTRACTION PROPERTY ADDITION PROPERTYOF EQUALITY OF EQUALITY

Solve each equation. Then check your solution.

18 + 2n = 4n – 9 -2n -2n 18 = 2n - 9 +9 +9 27 = 2n 2 2 27 = n or n = 13.5 2

2 n + 6 = 1 n – 33 4 2n – 1n + 6 = -3 find lcd for 2 and 1 lcm = 12 3 4 3 4 2n - 1n = -9 2 = 8 and 1 = 3 3 4 3 12 4 12 8n – 3n = -912 12 5n = -912 5n = -108 n = 21.6 or n = -108 = -21 3

5 53 – 4x = 8x + 8 +4x +4x 3 = 12x +8 -8 = -8 -5 = 12x 12 12 -5 = x or x = -.416667 12-6(4x + 1) = 5 – 11x -24x – 6 = 5 – 11x+24x +24x -6 = 5 + 13x

-5 -5 -11 = 13x 13 13 -11 = x or x = -.846154 13Nine less than half n is equal to one plus the product of -1/8 and n. Find the value of n.

9 – 1 n = 1 + -1 n 2 89 – 4 n + 1 n = 1 8 89 – 3 n = 1 8-9 -9 -3 n = -8 8*8 *8 -3n = -64 -3 -3 n = 64 = 7 1 or n = 21.333 3 3

Solving Equations and Formulas(Week 2 Day 4)

Kristy ChimeraCasie Croff

Nichole Oswald

Lesson 3.8 – Day 4 Solving Equations and Formulas Algebra 1Topic: Solving Equations and Formulas for Given Variables

Materials: Students: pen/pencil, notes (provided), homework / practice worksheet (provided) Teacher:  notes for class (provided), homework / practice worksheet for class (provided)

Lesson Overview: - Students will solve equations in which there are multiple variables for a specified variable. - Students will use formulas to solve real-world problems.

Lesson Objectives:- Students will propose a plan to solve equations and formulas for a given variable. (Synthesis) - Using hypothetical measurements and data, students will be able to calculate the value of a given variable, including appropriate units in terms of other variables. (Application) - Students will compare and contrast the procedures used for solving equations with one variable to those used for solving equations and formulas with two or more variables. (Evaluation)

NYS Standards: Key Idea 3 - Operations3A:  Use addition, subtraction, multiplication, division, and exponentiation with real numbers and algebraic expressions.Key Idea – Modeling / Multiple Representations 4A:  Represent problem situations symbolically by using algebraic expressions, sequences, tree diagrams, geometric figures, and graphs. Key Idea - Measurement 5A:  Apply formulas to find measures such as length, area, volume, weight, time, and angle in real-world contexts.  5C:  Use dimensional analysis.

Anticipatory Set:  (5 – 7 minutes) - Teacher draws a rectangle on the board and identifies it as a table top that we need to find the measurements of so we can buy tiles to decorate it. - Teacher writes P=86 inches and identifies this as the perimeter of the table top.- Teacher tells students that the length of the table is 11 inches longer than the width of the table and asks students how to label the sides (length and width) of the table top. - Teacher asks students for the formula for perimeter and writes it on the board.- Students work through the equation solving for W.

P = L + L + W + W = 2L + 2W86 = 2(W + 11) + 2W (FILL IN WHAT YOU KNOW/ WHAT IS

GIVEN)86 = 2W + 22 + 2W (DISTRIBUTE)86 = 4W + 22 (SIMPLIFY / COMBINE LIKE TERMS)64 = 4W (GET VARIABLE TERM ALONE ON ONE SIDE)16 = W (SOLVE)

- Now ask students how to find the length.L = W + 11 inches

- Be sure to have students observe and point out that we are solving for L in terms of W as this is what we will be working on in today’s lesson).

L = 16 + 11 =27 inches

Developmental Activity:  (27 minutes) - Pass out “Solving Equations and Formulas” worksheet to all students. (I will provide these for all groups.)- Teacher writes example 1 on the board and asks students how they would possibly start solving an equation like this (with 2 variables).  Teacher informs students that the steps they've been using to solve multi-step equations are used in solving these equations too.  - Teacher leads class through example 1 outlining the properties and steps used while working through the equation.  (Be sure to point out that the variable solved for is in terms of the other variable and these are all random variables that could stand for anything.) ex. The variable t has been multiplied by r, so divide each side by r to isolate t. “Now we’ve solved for t in terms of d and r.”- Teacher goes over example 2 asking for student participation and suggestions while solving. Be sure to discuss dimensional analysis while solving example 2. Discuss that the units of measure must be carried through the equation so the solution is in the correct units. Show students how to do this as you are going over the example.- Students work in groups to complete examples 3 and 4.  Teacher has students offer solutions to the class.

Closure:  (5 minutes)

- Teacher asks students questions reviewing the lesson (while passing out the homework). Ex. Suppose you have an equation with several variables for example V = lwh. You want to solve for a particular variable, say h, but we do not know the values of V, w, or l. How does the procedure compare with that for solving the equation with just one variable, for instance if we knew the values of V, w, and l? How does the solution compare with the solution for an equation with one variable?

Assessment: (1 minute)  Practice / Homework worksheet- Teacher passes out practice / homework worksheet to students (This could be done during Closure Activity). Students should complete the worksheet and bring it back to next class. 

Resources:http://www.glencoe.com/sec/math/algebra/algebra1/algebra1_05/index.php4/ny

http://www.glencoe.com/sec/math/fybh/na/alg1_index.htmlName: ______________________________________ Date: ____________

Solving Equations and FormulasSome equations contain more than one variable. To solve an equation or formula for a specific variable, you need to get that variable by itself on one side of the equation (just like we had to do when solving for one variable).

Example 1Solve the formula d = rt for t.NOTE:When you divide by a variable in an equation, remember that division by 0 is undefined.Example 2Find the time it takes to drive 75 miles at an average rate of 35 miles per hour. Use the formula you found in example 1 where t = time, d = distance, and r = rate.NOTE: When you use a formula you made need to use dimensional analysis, which is the process of carrying units throughout a computation / problems.

Try These Together

1. Solve 4a + b = 3a for a. 2. Solve y = mx + b for b.

3. Fuel Economya. A car’s fuel economy E (miles per gallon) is given by the formula E = m ,

where m is g

the number of miles driven and g is the number of gallons of fuel used. Solve the formula for m.

b. If Claudia’s car has an average fuel economy of 30 miles per gallon and she used 9.5 gallons, how far did she drive?

Name: ____________________________________ Date: _________________

Practice Solving Equations and Formulas

Solve each equation for the variable specified.12 g + 31h = -8g, for h

v = r + at, for rThe perimeter of a square field is given by the equation P = 2L + 2W, where P represents the perimeter of the square, L represents the length of the field, and W represents the width of the field.a. Solve the formula for l.b. Find the length of a field that is 50 yards wide and has a perimeter of 220 yards.Name: ANSWER KEY Date: ____________

Solving Equations and FormulasSome equations contain more than one variable. To solve an equation or formula for a specific variable, you need to get that variable by itself on one side of the equation (just like we had to do when solving for one variable).Example 1Solve the formula d = rt for t.d = t, where r cannot be 0rNOTE:When you divide by a variable in an equation, remember that division by 0 is undefined.Example 2Find the time it takes to drive 75 miles at an average rate of 35 miles per hour. Use the formula you found in example 1 where t = time, d = distance, and r = rate.t = d rt = 75 mi 35 mi hr Use dimensional analysis: mi = mi x h = ht = 2 1 hours mi 1 mi 7 hNOTE: When you use a formula you made need to use dimensional analysis, which is the process of carrying units throughout a computation / problems.

Try These Together1. Solve 4a + b = 3a for a. 2. Solve y = mx + b for b.

-4a -4a -mx -mx b = -a y-mx = b -1 -1 -b = a

3. Fuel Economya. A car’s fuel economy E (miles per gallon) is given by the formula E = m ,

where m is g

the number of miles driven and g is the number of gallons of fuel used.

Solve the formula for m.E = m g*g *gEg = m

b. If Claudia’s car has an average fuel economy of 30 miles per gallon and she used 9.5 gallons, how far did she drive?

Eg = m30(9.5) = m285 = m She drove 285 miles.

Name: ANSWER KEY Date: _________________Practice Solving Equations and FormulasSolve each equation for the variable specified.12g + 31h = -8g, for h12g + 31h = -8g-12g -12g 31h = -20g 31 31 h = -20 g 31v = r + at, for rv = r + at-at -atv – at = r r = v - atThe perimeter of a square field is given by the equation P = 2L + 2W, where P represents the perimeter of the square, L represents the length of the field, and W represents the width of the field.a. Solve the formula for L.

P = 2L + 2WP – 2W = 2LP – 2W = L

2

b. Find the length of a field that is 50 yards wide and has a perimeter of 220 yards.

L = P – 2W 2

L = 220 – 2(50) 2L = 220 - 100

2L = 120 2L = 60 YARDS