Download Writing Equations

Writing Equations
(Week 2 Day 1)
Kristy Chimera
Casie Croff
Nichole Oswald
Writing Equations
Math 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.)
5x2 -37
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: 17x
Go 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.) 5x2 -37
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.
2n-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)=3a
Three times the sum of a and b is the same as three times a.
Example: 20(d-f)=18d
Twenty 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 problem
Step 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 mile
3,500,000 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 problem
Step 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 = length
Width is 8 feet less than twice the length:
2L - 8 = width
Plan:
Perimeter = 2 (length) + 2 (width)
20 = 2L + 2 ( 2L - 8)
Solve:
20 = 2L +2 ( 2L -8)
20 = 2L +4L -16
20 = 6L -16
36 = 6L
6=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 = 700
X = 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_pr
ob2.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:
2x2 + y2 = 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:
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:
2h + 6 = 19
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:
2x2 + y2 = 18
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:
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.
$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 Chimera
Casie Croff
Nichole Oswald
Title: Solving Multi-step Equations
Grade 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 Reasoning
1A: Construct valid arguments.
1B: Follow and judge the validity of arguments.
Key Idea 3 - Operations
3A: 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 than
Group 2: subtract, difference, decreased, minus, less than
Group 3: multiply, product, times
Group 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: Operations
Addition
Subtraction
Multiplication
Division
Add
Sum
Plus
Increased
More Than
Subtract
Difference
Minus
Decreased
Less Than
Multiply
Product
Times
Divide
Quotient
Per
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 ____________________
Check:
Ex.) X/10 = 5
4.) To “undo” subtraction, we use ____________________ Ex.) X - 50 = 50 Check:
Name: ________________________________________ 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.
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:
4b  8
3.  2 = 10
Check:
4. 6X + 9 = 27
Check:
5. 14n - 8 = 34
Check:
d  12
6. 16 = 14
Check:
7. 0.6X - 1.5 = 1.8
Check:
8. (7/8)p - 4 = 10
Check:
9. 0 = 10y - 40
Extra Credit Problems:
Check:
10. 3.2y - 1.8 = 3
Check:
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. Times
Directions: 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. Add
Now 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: Operations
Addition
Subtraction
Multiplication
Division
Add
Sum
Plus
Increased
Subtract
Difference
Minus
Decreased
Multiply
Product
Times
Divide
Quotient
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
-5 -5
X
=3
2.) To “undo” multiplication, we use __Division__________
3.) To “undo” division, we use _Multiplication_____
Check:
3+5=8
8 =8
Ex.) 6X = 24 Check:
6
6 6*4=24
X = 4 24 = 24
Ex.) X/10 = 5
*10 *10
X = 50
Check:
50/10 = 5
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.
Section 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 -2
5X = 25
5
5
X
=5
2. 0.2X - 8 = -2
+8 +8
0.2 X = 6
0.2
0.2
X
= 30
4b  8
3.  2 = 10
4. 6X + 9 = 27
*(-2) *(-2)
4b+8 = -20
-8
-8
4b
= -28
4
4
b = -7
-9 -9
6X
= 18
6
6
X
=3
d  12
6. 16 = 14
5. 14n - 8 = 34
+8 +8
14n = 42
14
14
n =3
*14 *14
224 = d-12
+12
+12
236 = d
7. 0.6X - 1.5 = 1.8
+1.5 +1.5
0.6X
= 3.3
0.6
0.6
X
= 5.5
8. (7/8)p - 4 = 10
+4 +4
(7/8)p
= 14
*8
*8
7p
= 112
7
7
p
= 16
9. 0 = 10y - 40
+40
+40
40 = 10y
10 10
4 =y
Extra Credit Problems:
10. 3.2y - 1.8 = 3
+1.8 +1.8
3.2y
= 4.8
3.2
3.2
y
= 1.5
(The other
way is to use
*8/7) Either
way is fine.
Directions: Write an equation and solve each problem.
1. Find 3 consecutive integers whose sum is 96.
n + n + 1 + n + 2 = 96
3n + 3 = 96
-3
-3
3n
= 93
3
3
n
= 31
So the three integers are n = 31, n + 1 = 32 and n + 2 = 33
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 Evaluation
Directions: 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 Chimera
Casie Croff
Nichole Oswald
Lesson 3.5 – Day 3
Solving Equations with the Variable on Each Side
Algebra 1
Topic: 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 - Operations
3A: Use addition, subtraction, multiplication, division, and exponentiation with real numbers
and algebraic expressions.
Key Idea 4 – Modeling / Multiple Representations
4A: 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 __________________________________________
Solving Equations with the Variable on
Each Side
DATE ____________
Example 1
Solve 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 2
Solve: 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 3
Solve: 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
- 57.3 -57.3
x
= 91.2 – 57.3
65 = y - 156
+156
+156
65 + 156 = y
Solve each equation. Then check your solution.
18 + 2n = 4n – 9
2n+6=1n–3
3
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 on
Each Side
Example 1
Solve 3(x – 2) = 4x + 5
3x – 6 = 4x + 5
Steps for solving:
1. Distributive property to remove parenthesis
-3x
-3x
2. Collect all terms with x on one side of the equal
3x-3x-6 = x+ 5
sign by subtracting 3x from each side
-6 = x+5
3. Simplify by adding like terms on each side.
-5
4. Subtract 5 from each side
-5
-11 = x
5. Simplify
Many 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 2
Solve: 4(s + 2) – s = 3(s + 1)
4s + 8 – s = 3s + 3
4s – s + 8 = 3s + 3
3s + 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 3
Solve: 4p+2 = 1/3 (12p+3) + 1
4p + 2 = 12/3 p + 3/3 +1
4p + 2 = 4p +1 +1
4p + 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
4x – 10 = 6x + 4
-4x
-4x
-10 = 2x +4
-4
-4
-14 = 2x
2
2
b. x + 4 = 3[x-2(x+2)]
x + 4 = 3[x – 2x – 4]
x + 4 = 3x – 6x -12
x + 4 = -3x – 12
+3x
+3x
4x + 4 = -12
-4 = -4
4x
= -16
4
4
x = -4
-7 = x
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
- 57.3 -57.3
x
= 91.2 – 57.3
SUBTRACTION PROPERTY
OF EQUALITY
65 = y - 156
+156
+156
65 + 156 = y
ADDITION PROPERTY
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
2n+6=1n–3
3
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 = -9
12 12
5n = -9
12
5n = -108
n = 21.6 or n = -108 = -21 3
5
5
3 – 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
13
Nine 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
8
9–4n+1n=1
8
8
9–3n=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 Chimera
Casie Croff
Nichole Oswald
Lesson 3.8 – Day 4
Solving Equations and Formulas
Algebra 1
Topic: 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 - Operations
3A: 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 + 2W
86 = 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 Formulas
Some 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 1
Solve the formula d = rt for t.
NOTE:
When you divide by a variable in an equation, remember that division by 0 is undefined.
Example 2
Find 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 Economy
a. 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: ____________________________________
Practice Solving Equations and Formulas
Solve each equation for the variable specified.
12 g + 31h = -8g, for h
Date: _________________
v = r + at, for r
The 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 Formulas
Some 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 1
Solve the formula d = rt for t.
d = t, where r cannot be 0
r
NOTE:
When you divide by a variable in an equation, remember that division by 0 is undefined.
Example 2
Find 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
r
t = 75 mi
35 mi
hr
Use dimensional analysis: mi = mi x h = h
t = 2 1 hours
mi 1 mi
7
h
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.
-4a
-4a
-mx -mx
b = -a
y-mx = b
-1 -1
-b = a
3. Fuel Economy
a. 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 *g
Eg = 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 = m
30(9.5) = m
285 = m
She drove 285 miles.
Name: ANSWER KEY
Date: _________________
Practice Solving Equations and Formulas
Solve each equation for the variable specified.
12g + 31h = -8g, for h
12g + 31h = -8g
-12g
-12g
31h = -20g
31
31
h = -20 g
31
v = r + at, for r
v = r + at
-at
-at
v – at = r
r = v - at
The 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 + 2W
P – 2W = 2L
P – 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)
2
L = 220 - 100
2
L = 120
2
L = 60 YARDS