DETAILED LESSON PLAN IN
MATHEMATICS 8
DETAILED LESSON PLAN IN ENGLISH
School:
Grade Level: 8
Teacher:
Learning Area: Mathematics
Time & Dates:
Quarter: 4th Quarter
I. OBJECTIVES/ LEARNING OUTCOMES
A. Content Standard
The learners demonstrate understanding of
key concepts of factors of polynomials,
rational algebraic expressions, linear
equations and inequalities in two variables
and linear functions, algebraic expressions,
the properties of real numbers as applied in
linear equations, and inequalities in one
variable.
B. Performance Standard
The learners are able to formulate real-life
problems involving factors of polynomials,
rational algebraic expressions, linear
equations and inequalities in two variables,
systems of linear functions, and solve these
problems accurately using a variety of
strategies.
C. Learning Competencies
At the end of 2 hours, the students are
expected to achieve at least 80% accuracy
level in the following behavior:
1. Solve a system of linear equations
in two variables by graphing and
substitution.
2. Identify the number of solutions of
a linear equation in two variables.
3. Participate actively in class
discussion.
II. Subject Matter
Topic:
Solving s System of Linear Equations by
Graphing and Substitution
References:
Teacher’s Book pages 96-106
Mathematics 8 textbook pages 96-106
Materials:



PowerPoint Presentation
TV
White board marker
2 hours
d. Duration:
IFVL:
III. PROCEDURE
TEACHER’S INSTRUCTION
Greetings:
Good Morning class.
LEARNER’S RESPONSE
Good morning Ma’am
Prayer:
May request, Cyrene Faith to lead us in prayer.
Before you take your sit, kindly look and pick
up piece of trash and arrange your chairs (Students do so)
properly.
All right, you may now take your sits.
(Students do so)
Checking of Attendance:
Secretary, kindly check the attendance.
B. SETTING OF STANDARDS
Our classroom rules spell MATH which
means:
M- inimize your voice
A-ll ideas count
T-ry your best
H-ave fun
.
Have any problem with our rules?
None, Ma’am
REVIEW/MOTIVATION
What was our topic last meeting?
Let’s have a recap.
It’s all about Graphing Systems of Linear
Equation
Let us graph the examples of system of linear
equations in two variables.
Example number 1.
Equation 1: x +y = 5
Equation 2: 2x + 3y = 9
First, we will solve for y and find the slope and
y-intercept then we categorize the graph.
Eq. 1:
x+y = 5
y= -x +5
m = -1, b = 5
Eq. 2:
2x + 3y = 9
3y = -2x + 9
2
y = −3𝑥 + 3
-Divide both sides by 3.
2
m = −3 , b = 3
Now, after we get the slope and y-intercept
what is the next step?
The next step is to graph using the slope and yintercept to identify the categorization of the
graph.
Yes, exactly! We have to graph the equation
using our slope and y-intercept then
categorize them if it is intersecting, parallel or
coincide.
What have you notice in our graph? Is it
parallel, intersecting or coincide?
I have notice that the graph is intersecting.
Yes, as you can see our two equations
intersect to each other.
Let’s have another example.
Eq. 1: 2x + 3y = 3
Eq. 2: 2x + 3y = -6
Who can try on the board?
Example number 2:
Eq. 1: 2x + 3y = 3
2x + 3y = 3
3y = -2x +3
-Divide both sides by 3
2
y = - 3𝑥 + 3
2
m = - 3, b = 3
Eq. 2: 2x + 3y = -6
2x + 3y = -6
3y = -2x – 6
-Divide both sides by 3
2
y = - 3x – 2
2
m = − 3, b = -2
Let’s check if it’s correct.
Eq. 1: 2x + 3y = 3
2x + 3y = 3
3y = -2x +3
-Divide both sides by 3
2
y = - 3𝑥 + 3
2
m = - 3, b = 3
Eq. 2: 2x + 3y = -6
2x + 3y = -6
3y = -2x – 6
-Divide both sides by 3
2
y = - 3x – 2
2
m = − , b = -2
3
The answer of Lovely is correct. Now, who can
graph the equation using the slope and the yintercept?
How can you describe the graph?
The graph is categorizing as parallel.
Yes, very good. The graph shows that it is
parallel.
I am happy that you fully understand our
lesson.
D. DISCUSSION
Now, let’s proceed to our new lesson this
morning is all about Solving Systems of Linear
Equations by Graphing and Substitution.
By the end of the lesson you must be able to:
1. Solve a system of linear equations by
graphing.
2. Solve a system of linear equations in
two variables by substitution.
3. Participate actively in class discussion.
Let us solve the equations through the first
method by graphing. First thing to do is to
solve the y in terms of x.
Example 1:
Eq. 1 y = -x + 4
m = -1
b=4
Eq. 2 2x – y = 2 → -y = -2x + 2 → y = 2x -2
m=2
b = -2
Now, we to graph the equation using our slope
and y-intercept.
What have you notice in the graph?
Yes, the graph shows two lines that intersect
at the point (2,2). It follows that the point
(2, 2) is the solution of the system. This
means that, x = 2, y = 2. These values of x
and y satisfy the two equations in the system.
Let us verify by substitution.
y = -x + 4 → x + y = 4
2+2=4
- substitute the x and y values
4=4√
2x – y = 2
2(2) – 2 = 2
- substitute the x and y values
4 -2 = 2
2=2√
That’s first method to solve the equations
through graphing. Now, let us solve the
equations by the second method the
substitution method.
Example 2: Solve by substitution.
x – y = -3
x + 4y = 7
Solution:
Step 1: Solve for one variable. Solve for x in
the first equation.
x – y = -3
x=y–3
Step 2: Substitute y-3 to x in the second
equation; then solve for the value of y.
x + 4y = 7
(y-3) + 4y = 7
5y -3 = 7
I have notice in the graph that they intersect to
each other.
5y = 10 →y = 2
Step 3: Substitute 2 to y in the first equation
or in the second equation; then solve for the
value x.
x-y = -3
x- 2 = -3
x + 4y = 7
or
x = -3+2
x= -1
x + 4(2) = 7
x+8=7
x=7–8
x = -1
Step 4: Check. Substitute -1 to x and 2 to y in
both equations.
x – y = -3
x + 4y = 7
-1 – 2 = -3
-1 + 4 (2) = 7
- 3 = -3
7=7
The solution is ( -1, 2) or SS = (-1, 2)
Example 3: Solve by substitution.
2x -5y = -28
Example 3:
3x + 15y = 3
2x -5y = -28
Who wants to try?
3x + 15y = 3
Step 1: Solve for one variable. Solve for x in
the second equation.
3x + 15y = 3
- Transfer y to the other side
3x = -15y + 3
-Divide both sides by 3
x = -5y + 1
Step 2: Substitute -5y + 1 to x in the first
equation; then solve for the value of y.
2x – 5y = -28
2 (-5y + 1) -5y = -28
-10y + 2 -5y = -28
-15y = -28-2
-15y = -30 -Divide both sides by -15
y=2
Step 3: Substitute 2 to y in the first equation
and the solve for x.
2x – 5y = -28
2x – 5(2) = -28
2x – 10 = -28
2x = -28 + 10
2x = -18 – Divide both sides by 2
x = -9
Step 4: Check. Substitute -9 to x and 2 to y in
both equations.
Let’s check if it is correct.
The solutions of the two equations
2x -5y = -28
3x + 15y = 3
2x – 5y = -28
3x + 15y = 3
2(-9) – 5(2) = -28
3(-9) + 15(2) = 3
-18 – 10 = -28
-27 + 30 = 3
-28 = -28
3=3
The solution is (-9, 2) or SS = (-9, 2).
is (-9, 2) or SS = (-9, 2).
Very good! Let’s have another example.
Example 4: Solve by substitution.
y = 2(x-3)
3x + 4y = -24
Solve for y in the first equation.
y = 2(x-3)
y = 2x – 6
Substitute 2x -6 to y in the second equation
3x + 4y = -24
3x + 4(2x – 6) = -24
3x + 8x – 24 = -24
11x = -24 +24
11x = 0
x=0
Substitute 0 to x in the first equation then solve
for y.
y = 2(x-3)
y = 2 (0-3)
y = 2(-3)
y = -6
The solution is (0, -6)
Let’s check and substitute (0, -6) in both
equations.
y = 2(x-3)
-6 = 2(0-3)
-6 = -6
3x + 4y = -24
3(0) + 4(-6) = -24
-24 = -24
That’s the steps in substitution method.
Any clarifications regarding to our lesson this
morning?
None, Ma’am
E. ENRICHMENT ACTIVITIES
Try this in a one half sheet of paper.
Instruction: Solve each system of linear
equations in two variables by substitution.
1. 5x -2y = 5
y = -3x + 3
2. y = 2x – 1
3y = 2x -3
3. 2x -3y = -6
x=y+3
Are you done? Let’s check your answers.
Who wants to answer number 1?
1. 5x -2y = 5
y = -3x + 3
Substitute -3x + 3 to equation 1 and solve for
x.
5x – 2y =5
5x – 2(-3x +3) = 5
5x +6x – 6 = 5
11x = 5+6
11x = 11
x=1
Substitute 1 to x in the equation 2 and solve
for y.
y = -3x + 3
y = -3(1) + 3
y=0
The solution for the equation 5x -2y = 5
and y = -3x + 3 is (1, 0). Very good!
Check. Substitute 1 to x and 0 to y in both
equations.
5x -2y = 5
5(1) – 2(0) = 5
5=5
y = -3x + 3
0 = -3(1) + 3
0=0
How about number 2?
The solution is (1, 0) or SS = (1, 0).
2. y = 2x – 1
3y = 2x -3
Substitute 2x -1 to equation 2 and solve for x.
3y = 2x -3
3(2x -1) = 2x – 3
6x -3 = 2x – 3
6x -2x = -3 + 3
4x = 0
x=0
Substitute 0 to x in the equation 1 and solve
for y.
y = 2x -1
y = 2(0) -1
y = -1
Check and substitute 0 to x and 1 to y in both
equations.
Yes, the solution of the equations y = 2x – 1
and 3y = 2x -3 is (0, -1). Good job!
y = 2x –1
3y = 2x – 3
-1 = 2(0) -1
3(-1) = 2(0) -3
-1 = -1
-3 = -3
The solution is (0, -1) or SS = (0, -1)
How for number 3?
3. 2x -3y = -6
x=y+3
Substitute y + 3 to equation 1 and solve for y.
2x – 3y = -6
2(y +3) -3y = -6
2y +6 -3y = -6
2y-3y = -6 -6
-y = -12
-Multiply both sides by -1
y = 12
Substitute 12 to y in equation 2 and solve for
x.
x=y+3
x = 12+ 3
x = 15
Check and substitute 15 to x and 12 for y in
both equations.
2x – 3y = -6
x=y+3
2(15) – 3(12) = -6
15 = 12 + 3
30 – 36 = -6
15 = 15
-6 = -6
Therefore, the solution is (15, 12).
Yes, the solution of the equations 2x -3y = -6
and x = y + 3 is (15, 12).
I’m glad that you all understand our lesson this
morning. Good job!
F. GENERALIZATION
Did you learn something today?
What have you learned in our lesson?
IFVL:
I am glad that you have learned something Yes, Ma’am
I have learned on how to graph of the system
today.
and substitute the given values.
G. EVALUATION
Get 1 whole sheet of paper and answer the
following and I will give you 15 minutes to
answer.
Instruction: Use substitution to solve the
system.
1. x = 2y
x + 3y = 15
2. -6x – 7y = 2
-8x + y = -18
3. 7x -8y = -13
x – 6y = 3
4. y = 2x – 18
-2x + 2y = 6
5. –x – 4y = 7
y = 3x + 8
Are you all done? Pass your papers forward.
That’s all for today class.
Good bye and thank you class.
Good bye and thank you Ma’am.