Daily Lesson Plan 2
Grade 9
Mathematics
I. Title: AbsoluteValue Equations and Inequalities
II. Organization: Whole group / small groups
III.
Objectives:
·
The students will be able to solve absolutevalue equations.
·
The students will be able to solve absolutevalue inequalities.
·
The students will be able to express solutions as a range of
values on a number line.
IV.
Standards Covered:
· 2.1: Numbers and Operations
· 2.2: Computation and Estimation
· 2.3: Measurement and Estimation
· 2.4: Reasoning and Connections
· 2.5: Problem Solving and Communication
· 2.8: Patterns, Functions, and Algebra
V. Materials:
· Pencil
· Paper
· Textbook: Holt Algebra I
· Calculator
· 2 Worksheets: Absolute Value Equations and Absolute Value Ineq.
VI.
Procedure:
Intro / Motivation:
· Assign Preclass/Warmup, walk around room, take roll, and monitor students.
· Review questions from homework for the previous night.
· Give 3 or 4 absolute value practice questions for review.
o  13 – 24  = 11
o  1 – 27  = 26
o  5 – 10  = 5
o  11 – 3  = 8
Developmental Activities: The problems that we just solved involve using
absolute values. Today we will be looking at absolutevalue equations and absolutevalue
inequalities. Before we begin, let’s recall the definition of the absolute
value of a number. In section 6.4 we learned that:
For a ³ 0,  x  = a is equivalent to x = a or x = a.
When solving absolutevalue
equations, you must consider two cases. Let’s look at an example.
 3x – 2  = 10
Case 1:  3x – 2  = 10
Case 2:  3x – 2  = 10
3x – 2 = 10
3x – 2 = 10
3x = 12
3x = 8
x = 4
x = 8/3
Let’s check our work
to see if this is correct!
Case 1:  3(4) – 2  = 10
Case 2:  3(8/3) – 2  = 10
 12 – 2  = 10
8 – 2  = 10
 10  = 10
 10  = 10
TRUE!
TRUE!
Let’s look at another
example:
 2x – 4  = 8
Case 1:  2x – 4  = 8
Case 2:  2x – 4  = 8
2x – 4 = 8
2x – 4 = 8
2x = 12
2x = 4
x = 6
x = 2
Let’s check our work
to see if this is correct!
Case 1:  2(6) – 4  = 8
Case 2:  2(2) – 4  = 8
 12 – 4  = 8
 4 – 4  = 8
 8  = 8
 8  = 8
TRUE!
TRUE!
Next we will take a look at
some absolutevalue inequality problems. Solving absolutevalue inequalities
is very similar to solving absolutevalue equations. When working with absolutevalue
inequalities we must consider two cases once again. Let’s look at an example:
 x – (5)  £ 2
Solve
–2 £ x – (5) £ 2 by solving each part of the inequality separately.
Case 1: 2 £ x – (5)
Case 2: x – (5) £ 2
2 £ x + 5
x + 5 £ 2
7 £ x
x £ 3
Then combine the solutions
in the form of a conjunction:
7 £ x £ 3
The graph of this solution
will look like:
Let’s look at another
example:
 x  6  > 2
Case 1: The quantity x  6
is positive
 x – 6  > 2
x – 6 >
2
x >
8
Case 2: The quantity x – 6 is negative
 x – 6  > 2
 (x – 6) > 2

x + 6 >
2

x >
4
x < 4
We must change the direction of the
sign because we divided by a negative sign.
The graph for this solution looks
like:
After going over these example problems,
the students will be given two worksheets to work on in small groups. One worksheet
contains problems on absolutevalue equations and the other worksheet contains problems on absolutevalue inequalities. The students will be given the remainder of the period to work in small groups on
these worksheets.
Closure: Students who have not completed the assigned worksheets in class
will be expected to finish them for homework.
VII.
Adaptations:
· For ESL students, allow them to work on problems with a buddy who can help with the language.
· For special needs students, repeat key concepts.
VIII.
Evaluation:
Student: The students will be evaluated using the worksheets given
in class.
Teacher:
· Was the class enthusiastic about the lesson? Was
there a lot of participation?
· Did the students understand the difference between absolutevalue equations and absolutevalue
inequalities?
· Did the students understand how to graph absolutevalue inequalities?
IX.
Follow Up:
·
Have students complete the worksheets given in class for homework. Review these worksheets tomorrow at the beginning of class.

ABSOLUTE VALUE EQUATIONS
Worksheet 1

Quick
Review:
The expression within the absolute
value symbols can be negative or positive.
Solve
the equations below.
1.  x
 = 7
x = ______________
2.  2x  = 6 x = ______________
3.  x – 3  = 1 x = _____________
4.  x + 1  = 3 x = _____________
5.  x + 2  = 3 x = _____________
6.  x – 2  = 1 x = _____________
7.  x + 2  = 2 x = _____________
8.  x – 3  + 1 = 5 x = ____________
9.  4x – 2  = 10 x = ____________
10.  3x + 6  = 12 x = _____________
11.  2x – 3  = 4 x = ____________
12.  x – ½  = 9/2 x
= ____________
13.  3x – 1  = 8 x = ____________
14.  6x + 2   3 = 5 x = ___________
15.  5x – 7  = 3 x = _____________
16.  2x + 1  = 3 x = ______________

ABSOLUTE VALUE INEQUALITIES
Worksheet 2

Quick Review:
 The
expression within the absolute value symbols can be negative or positive.
 Reverse
the direction of the inequality for the negative case.
Solve
each inequality. Sketch the graph next to each problem.
1.  x + 1  > 1 x < _______ or x > ________
2.  x + 1  < 1 ________ < x < __________
3.  x – 1  ≥ 2
x ≤ ________ or x ≥ ________
4.  x + 1  ≤ 2 ________
≤ x ≤ ___________
5.  x – 2  ≥ 2
x ≤ ________ or x ≥ ________
6.  x + 2  ≤ 3 ________
≤ x ≤ ___________
7.  x – 4  < 1 __________ < x < __________
8.  2x – 1  ≥ 3 x
≤ ________ or x ≥ __________
9.  3x + 2  > 4 x < _______ or x > __________
10.  4x –
5  < 3 ________ < x < __________
Daily Lesson Plan 3
Grade 9
Mathematics
I. Title: Systems of Equations and Inequalities: The Substitution
Method
II. Organization: Whole group / small groups
III.
Objectives:
·
The students will be able to find an exact solution to a system
of linear equations by using the substitution method.
IV.
Standards Covered:
· 2.1: Numbers and Operations
· 2.2: Computation and Estimation
· 2.5: Problem Solving and Communication
· 2.6: Statistics and Data Analysis
· 2.8: Patterns, Functions, and Algebra
· 2.9: Geometry
V. Materials:
· Pencil
· Paper
· Textbook: Holt Algebra I
· Calculator
· Worksheet: Solving Linear Systems by Substitution
VI.
Procedure:
Intro / Motivation:
· Assign Preclass/Warmup, walk around room, take roll, and monitor students.
· Review questions from homework for the previous night.
· Give 3 or 4 substitution practice problems for review.
Solve each equation for y:
· 5x + 3y = 8
x = 2
· 2x + 4y = 16
x = 4
Developmental Activities: Today we are going to be solving systems of
equations by using the substitution method. If we know the value of one variable
in a system of equations, we can find the solution for the system by substituting the known value of the variable into one
of the equations. This method is called the substitution method.
Let’s look at an example.
Solve by using substitution:
8x + 2y = 19
x = 3
Since x = 3, we can substitute
the value of x into the first equation.
8
(3) + 2y = 19
Solve the resulting equation
for y.
24
+ 2y = 19
2y
= 5
y
= 2.5
The solution for this system
of equations is ( 3 , 2.5 ).
Let’s check our work
to see if this is correct!
We can check the solution by
substituting the values for x and y into the first equation.
8
(3) + 2 (2.5) = 19
24
+ (5) = 19
19
= 19
TRUE!
Let’s look at another
example:
Solve by using substitution:
15x – 5y = 30
y = 2x + 3
Substitute 2x + 3 for y into
the first equation, and solve for x.
15x
– 5 (2x + 3) = 30
15x
– 10x – 15 = 30
5x
– 15 = 30
5x
= 45
x
= 9
Substitute 9 for x into the
equation y = 2x + 3, and solve for y.
y
= 2 (9) + 3
y
= 18 + 3
y
= 21
The solution for this system
of equations is ( 9 , 21 ).
Let’s check our work
to see if this is correct!
We can check the solution by
substituting the values for x and y into the original equations.
15
(9) – 5 (21) = 30
135
– 105 = 30
30
= 30
TRUE!
OR,
21
= 2 (9) + 3
21
= 18 + 3
21
= 21
TRUE!
Now let’s take a look
at a trickier example!
Solve by using substitution:
3x + y = 4
5x – 7y = 11
To
use substitution for this example, we must solve one equation for y. Which equation
do you think would be easier to solve for y? The first equation is right! To solve the first equation for y, all we have to do is subtract the 3x from both
sides. To solve the second equation for y, we would have to subtract 5x from
both sides and then divide both sides by 7. So lets solve the first equation
for y.
3x + y = 4
y = 4 – 3x
Now
that we have found the value for y, we can substitute 4 – 3x for y in the second equation and solve for x.
5x – 7y = 11
5x – 7 (4 – 3x) = 11
5x – 28 + 21x = 11
26x – 28 = 11
26x = 39
x = 1.5
Now
that we have found the value of x, substitute 1.5 for x in the first equation and solve for y.
3 (1.5) + y = 4
4.5 + y = 4
y = 0.5
The
solution for this system of equations is ( 1.5 , 0.5 ).
We
can check the solution by substituting the values for x and y into the original equations.

** If there is extra time,
do another example: **
6x – 2y = 11
x + 3y = 4

After going over these example problems,
the students will be given a worksheet to work on individually or in small groups. The
worksheet contains problems on solving linear systems by using the substitution method.
The students will be given the remainder of the period to work in small groups on this worksheet.
Closure: Students who have not completed the assigned worksheet in class
will be expected to finish it for homework.
VII.
Adaptations:
· For ESL students, allow them to work on problems with a buddy who can help with the language.
· For special needs students, repeat key concepts.
· Circulate around the room and answer any questions, or help students who are having trouble
with the lesson.
· For students who finish the worksheet early, give a second worksheet for them to work
on.
VIII.
Evaluation:
Student: The students will be evaluated using the worksheet given
in class.
Teacher:
· Was the class enthusiastic about the lesson? Was
there a lot of participation?
· Did the students understand the concept of substitution?
· Could the students identify which equation was easiest to solve for?
IX.
Follow Up:
·
Have students complete the worksheet given in class for homework. Review this worksheet tomorrow at the beginning of class.