chapter3 equations grade 10

Chapter 3. EQUATIONS, SYSTEMS OF EQUATION
Preparation date:
Teaching date:
Period 17. OVERVIEW OF EQUATION
I. Objectives: By the end of the lesson, Students will be able to:
- To realize some concepts as equation with one unknown, condition of an equtions,
equation with multiple unknowns and parametric equations
- To understand the condition of an equations
II. Teaching aids:
- Teacher: lesson plan, text book
- Student: review about equation that learnt at grade 9
III. Method: Raising and settling problems
IV. Procedure:
1- Orginization.
2- Review:
Student 1: given an example about equation with one unknown?
Student 2: given an example about equation with two unknowns?
3- New lesson:
Vocabulary
Equation: Phương trình
Right side: vế phải
System of equation: Hệ phương trình
Set of roots: tập nghiệm
Solve: giải ( PT)
No real root: vô nghiệm
Root/ solution : Nghiệm
Condition: điều kiện
Expression: biểu thức
One unknown: một ẩn
Left side: vế trái
Multiple unknown: nhiều ẩn

Activity 1. BASIC CONCEPTS
Teacher’s activities


Students’ activities
I – Concept of equations.

Request Students doing 1.

Introduce the concept of Equations with one Do the 1.
unknown
1) Equations with one unknown
Definition (text book )
Given ex 1 in order to aks student to
determine Left side and right side.
Ex 1: 3x – 2 = x + 2
Compte the value of each side when x = 2 ? When x = 2, we have:
compare ?
Left side : 3.2 – 2 = 4
Right side: 2 + 2 = 4.
Thus, x = 2 is a root of this equation.
Solve the equation?
Solution :


Teacher’s activities
Given ex 2 and aks student to find the roots
.
Comment about each side ?

Students’ activities
3x – 2 = x + 2 <=> 3x – x = 2 + 2 =>
2x = 4 <=> x = 2.
EX 2: solve the following equation:
5x + 1 = 5x – 3
<=> 5x – 5x = –3 – 1 <=> 0x = – 4
No value of x that satisfy this equation.
So the equation has no real equation
Ex3:Solve the following equation:

Given ex 3 and aks student to find the roots
3
≈ 0,866
Request students convert the roots to
3
2
2x =
<=> x =
demical
Given notes
Activity 2. CONDITIONS FOR AN EQUATIONS
Teacher’s activities
Request students doing



2.

Students’ activities
2) Conditions for an equation

(text book)


Remark and correct the solution?
Do the 2
Given the concept of condition for an Concept: text book
equation?
In order to find the condition of equation
x +1
= x −1
x−2

what will we do ?

Nominate student do the ex.
Remark

The equation:
x–2

≠

≥

x +1
= x −1
x−2

0 => x

≠
≥

2

x – 1 0 => x 1
the definite condition for the equation
[1;+

∞

) \ {2}

Activity 3. EQUATION WITH MULTIPLE UNKNOWNS
Teacher’s activities
Students’ activities
Introduce about Equation with multiple 3)Equation with multiple unknowns:
unknowns
Ex
a) 3x + 2y = x2 – 2xy + 8 is an equation
Given example about Equation with two with two unknowns ( x and y )
unknowns x,y.
( x ; y ) = ( 2 ; 1 ) is a root of the equation.
Compte the value of each side when x =


2, y=1 ? compare ? conclude?
Given example about Equation with b) 4x2 – xy + 2z = 3z2 + 2xz + y2
three unknowns x,y,z.
is an equation with three unknowns ( x ,y

Compte the value of each side when x = and z )
-1c, y=1, z=2 ? compare ? conclude?
( x ; y ; z ) = (–1 ; 1 ; 2 ) is a root of the
equation.
Activity 3. CONDITIONS FOR AN EQUATIONS
Teacher’s activities
Students’ activities
Introduce about parametric equations 4) parametric equations
Given example about parametric Ex :
equations
a) 3x + m = 0
Remark
b) (m – 2 )x2 + 5x – 6 = 0
4- Consolidate
Request student recall the main point of the lesson.
5- Homework:
Memorize the lesson.
Do the exercise 1,2 text book page 57


Preparation date:
Teaching date:
Period 18. OVERVIEW OF EQUATION
I. Objectives: By the end of the lesson, Students will be able to:
- To realize some concepts as Equivalent equations, Resulting equations,
- To understand the Equivalent transformation
- applying the Equivalent transformation to solve equations
II. Teaching aids:
- Teacher: lesson plan, text book
- Student: review about equation that learnt at last lesson

III. Method: Raising and settling problems
IV. Procedure:
1- Orginization.
2- Review:
Student 1: recall the concept of equations with one unknown, given example?
Student 2: definite the condition of an equation ?
3- New lesson:
Vocabulary
Equivalent equations: phương trình tương Unique solution: nghệm duy nhất
đương
Resulting equations: phương trình hệ quả
Sign: đổi dấu
Transform: biến đổi PT
Extraneous solution: nghiệm ngoại lai
Equivalent transformation: biến đổi tương Check the sollution: kiểm tra nghiệm
đương
Changing sides: chuyển vế
Polynomial: đa thức
Activity 1. EQUIVALENT AND RESULTING EQUATIONS
Teacher’s activities
Request student doing



4

Given concept in textbook
Nominate student find the set of roots of
each equations, then compare them.
Remark .

Students’ activities
II-Equivalent and resulting equations.
1) Equivalent equations.
a. Concept 1 : ( text book)
b. Ex : given two equations :
3x + 2 = 0
(1)
2x +

4
3

=0

(2)


−

2
3

S1 = S 2 = {
} so ( 1 ) and ( 2) are
equivalent
Nominate student find the set of roots of 2. resulting equations
a. Concept 2 : ( text book)
each equations, then compare them.
b. Ex : given two equation

(1)
Remark
(2)
S1 = { 3}
S2 = { 3;8}
Given concept in textbook

⊂

S1 S2
so ( 1 ) and ( 2) are resulting equations
Activity 2. EQUIVALENT TRANFORMATION
Teacher’s activities
Students’ activities
Introduce the concept of Equivalent 2) Equivalent transformation:
transformation:
a- concept : ( text book )
Introduce the symbol: equivalent.
b- theorem : ( text book)
c- notes ( text book )
Request student doing



5
* symbol : “

⇔

”

Remark
Activity 3. RESULTING EQUATIONS
Teacher’s activities
Introduce the concept of
equations

Students’ activities
resulting 3) Resulting equations:
* Concept : ( text book )
f(x) = g(x) => f1(x) = g1(x)
The concept of extraneous solution
Request student doing EX 3.

Ex 3: solve the following equations:
x2
1
1
=
+
2
x −4 x−2 x+2

Nominate student go to write down the


solution

Conditions for the equations : x

≠ ±2

x2
1
1
=
+
2
x −4 x−2 x+2

Compare the value which have found => x2 = x + 2 + x – 2
with the condition
=> x2 = 2x => x2 – 2x = 0
=> x(x – 2) = 0
remark
 x = 0 (satisfy)
 x = 2 (no satisfy)


=>
So, the equation has a unique solution: x =
0
Ex 4:
a the equation has unique solution: x=1
b set of the roots: s={2.3}
4- Consolidate
Request student recall the main point of the lesson.
5- Homework:
Memorize the lesson.
Do the exercise 3,4 text book page 57

Preparation date:
Teaching date:
Period 19: EQUATIONS CONVERTED INTO LINEAR AND QUADRATIC
EQUATIONS
I. Objectives: By the end of the lesson, Students will be able to:
- Consolidate the way to solve and argue the linear equations, quadratic equations,
biquadratic equations, redical equations
- Solve the linear equations, quadratic equations, biquadratic equations, redical equations. - Find the condition of equations, eliminate extraneous solutions that not satisfy the
condition
II. Teaching aids:
- Teacher: lesson plan, text book
- Student: review about equation that learnt at grade 9
III. Method: Raising and settling problems
IV. Procedure:
1- Orginization.
2- Review:
Student 1: recall the way to solve and argue the linear equations
Student 2: state the Vieta’s formulas ?
3- New lesson:
Vocabulary
Converted into: đưa về (dạng)
Vieta’s formulas: định lý viet
Co-efficient: hệ số
Absolute – value bar: giá trị tuyệt đối
True roots with every x: có nghiệm với mọi x Eliminated; loại ( nghiệm)
Solve and justify: giải và biện luận
Squaring both sides: bình phương 2 vế
Radical equations: phương trình chứa căn

Positive/ negative: dương/âm
Activity 1. REVIEW OF LINEAR EQUATIONS
Teacher’s activities

Students’ activities


Recall the method to solve and jusify the 1. Review of linear equations
linear equations

Summarized in the table (text book)

Applying to the ex 1:

EX1: solve and jusify the following equation:

Determine a,b in equations

(1)
Solution: (1)

Remark and consolidate

•
•

(1) becomes ( has no real roots)
(1)

Activity 2. REVIEW OF QUADRATIC EQUATIONS

Teacher’s activities
Students’ activities
Recall the method to solve and jusify the 2. Review of quadratic equations
linear equations

Summarized in the table (text book)

Applying to the ex 2:

EX1: solve and jusify the following equation:

Determine a,b and c in equations

(1)
Solution: (1)

Remark and consolidate

•
•

(1) becomes ( has no real roots)
(1)

Activity 3 .RADICAL EQUATIONS
Hoạt động của GV
Form 1.

Nội dung
3 . Radical equations

EX 2: solve the equation

 g ( x) ≥ 0

f ( x) = g ( x) ⇔  f ( x) ≥ 0
 f ( x) = g 2 ( x)


x–3=

3x + 1

Solution

Form 2
f ( x) =

  g ( x) ≥ 0

 f ( x) = g ( x)
g ( x) ⇔ 
  f ( x) ≥ 0

  f ( x) = g ( x)

Given Ex 2( textbook)

x≥−

Condition :

x – 3 =

1
3

3 x + 1 ⇒ ( x − 3) 2 = 3 x + 1

x = 1
⇒ x2 − 9x + 8 = 0 ⇒ 
x = 8


+ with x = 1, we have :
LS : 1 – 3 = – 2
Find the codition of that equation
Guiding students to square to sides
Check the result: if x=1, x = 8 was a root
of that equation

RS:

3.1 + 1 = 4 = 2

x = 1 isn’t a root of equation
+ With x = 8 , we have
LS : 8 – 3 = 5
RS:

3.8 + 1 = 25 = 5

x = 8 is a root of this equation
so, the root of equation: x = 8
4- Consolidate
Request student recall the main point of the lesson.
5- Homework:
Memorize the lesson.
Do the exercise 7,8 page 62


Preparation date:
Teaching date:
Period 20 : REVIEW
I. Objectives: By the end of the lesson, Students will be able to:
- Consolidate the way to solve and argue the linear equations, quadratic equations,
biquadratic equations, redical equations
- Solve the linear equations, quadratic equations, biquadratic equations, redical equations. - Find the condition of equations, eliminate extraneous solutions that not satisfy the
condition
II. Teaching aids:
- Teacher: lesson plan, text book
- Student: review about equation that learnt at last lesson
III. Method: Raising and settling problems
IV. Procedure:
1- Orginization.
2- Review:
Student 1: recall the way to solve and argue the linear equations
Student 2: state the Vieta’s formulas ?
3- New lesson:
Activity 1 : solve EX 1/ text book page 62
Teacher’s activities
Guiding student to recognize the form

of each equations.

Students’ activities
EX 1: solve the following equation
a)

Request student to solve equations.

x 2 + 3x + 2 2 x − 5
=
2x + 3
4
x≠−

3
2

condition
2
Nominate 4 students go to write down 4(x + 3x + 2) = (2x – 5)(2x + 3)
23
the solution
−
=> 16x + 23 = 0 <=> x =
b)

2x + 3
4
24
−

= 2
+2
x −3 x+3 x −9

16


≠±

condition : x
3
Guiding student to compare the value
(2x + 3)(x + 3) – 4(x – 3) = 24 + 2(x2 – 9)
of root with condition, then eliminate
=> 5x = –15 <=> x = –3 ( not satisfy )
extraneous solution
So this equation has no real root
Request students to comment about
3x − 5 = 3
solutions.
c)
Remark and summarize
5
x≥

condition :

3

3x – 5 = 9 <=> x =

d)

14
3

2x + 5 = 2
x≥−

condition :

5
2
−

2x + 5 = 4 <=> x =

1
2

Activity 2 : solve EX 7/ text book page 63
Teacher’s activities
Students’ activities
Guiding student to recognize form of EX 7: solve the following equations
each equations.
5x + 6 = x − 6 ;
x≥6
a)
condition:
=> 5x + 6 = (x – 6)2 => x2 – 17x + 30 = 0.
Request student to solve equations.

X = 15 (satisfy) ; x = 2 (not satisfy)
thus: x = 15
Nominate 4 students go to write down
x ∈ [−2;3]
3− x = x + 2 +1
b)
; condition:
the solution
=> 3 – x = x + 3 + 2

x+2

x+2

=> – x =
=> x2 – x – 2 = 0
=> x = – 1 (satisfy) ; x = 2 (not satisfy)
hence : x = – 1
Guiding student to compare the value
2 x2 + 5 = x + 2
x ≥ −2
; ĐK:
of root with condition, then eliminate c)
=> 2x2 + 5 = x2 + 4x + 4 => x2 – 4x + 1 = 0
extraneous solution
x1,2 = 2 ± 3
Request students to comment about
=>
( satisfy )
solutions.

Remark and summarize.

x≥−

4 x + 2 x + 10 = 3 x + 1
2

d)
; condition:
2
2
=> 4x + 2x + 10 = 9x + 6x + 1
=> 5x2 + 4x – 9 = 0 => x1 = 1 ( satisfy )
−

1
3

9
5

x2 =
(not satisfy )
thus : x = 1
Activity 3 : solve EX 8/ text book page 63
Teacher’s activities

Students’ activities

EX 8: given the equation:
3x2 – 2(m + 1)x + 3m – 5 = 0
Determine m such that the equations has a solution
three times as larger as the other solution. Find the
solutions in this case.
Guiding student to compute the value Solution: let x1, x2 be the roots of equations .
of .
applying the vieta’s formulas, we have
Appying the Vieta’s formulas

x1 + x2 =

2(m + 1)
3

x1.x2 =

3m − 5
3

và
Moreover x1 = 3x2 , so we obtain : m2 – 10m + 21 =
0
=> m = 3 ; m = 7.
+ if m = 3, then : x1 = 2 ; x2 =
+ if m = 7, then : x1 = 4 ; x2 =

4- Consolidate
Request student recall the main point of the lesson.
5- Homework:

Memorize the lesson.

2
3
4
3


Preparation date:
Teaching date:
Period 21: Equations and system of linear equations with multiple unknowns
I. Objectives: By the end of the lesson, Students will be able to:
- Review the concept of quadratic equations and systems of to linear equations with two
unknown.
- Solve and jusify systems of quadratic equations
- solve systems of three linear equations with three unknown
II. Teaching aids:
- Teacher: lesson plan, text book
- Student: review about systems of equation that learnt at last lesson.
III. Method: Raising and settling problems
IV. Procedure:
1- Orginization.
2- Review:
Student 1: solve equation:
Student 2: solve equation:
3- New lesson:

x−2 = x
x = x−2

Vocabulary
Linear equations: Phương trình bậc nhất
System of linear equations with multiple
unknown:hệ phương trình bậc nhất nhiều
ấn
Coefficient: hệ số
Not all equal to 0; Không đồng thời bằng 0
0rdered pair: cặp số

Triangular form: Dạng tam giác
Substitute + into: Thay( giá trị)
Corressponding: tương ứng
Eliminated: khử ẩn

Activity 1 : Linear equations with two unknowns:
Teacher’s activities
Students’ activities
Introduce the concept of liner equations with I- Review of equations and system of linear
two unknowns
equations with two unknowns:
1.Linear equations with two unknowns:
Given examples
and guiding student a) concept : ( text book)
determine the value of a,b,c
form : ax + by = c
b) Ex :

3x – y = 2 (a = 3 ; b = – 1 ; c = 2)
Request student doing 1.

–2x = 6 (a = –2 ; b = 0 ; c = 6)
Nominate students go to write down the
5y = –2 (a = 0 ; b = 5 ; c = –2)
solution
Remark
Activity 2: NOTES
Teacher’s activities
Students’ activities
In case a and b equals 0, then what’s the set c) Note : ( text book)
of roots of that equation.
y=−

≠

a
c
x+
b
b

When b
0, guiding student compute the
value of y.
Determine the set of roots.
State the set of roots of liner equations with Read the notes
two unknowns.
Request student doing

Remark



2.

Drawn the line 3x – 2y = 6 in the
coordinate plane Oxy


.
Activity 3 :SYSTEM OF LINER EQUATIONS WITH TWO UNKNOWNS.
Teacher’s activities

Students’ activities
2. System of linear equations with two
Introduce the concept of systems of liner unknowns.
equations with two unknowns
a) concept ( text book)
Take examples
a1 x + b1 y = c1

How many methods two solve systems of
a2 x + b2 y = c2
liner equations with two unknowns
form :
Request student applying to solve systems
of liner equations in



3.

4 x − 3 y = 9

2 x + y = 5

b) EX:
way 1:.

Nominate students to solve that system of 4 x − 3 y = 9 ⇔ 4 x − 3(5 − 2 x) = 9 ⇒


equation by using exchange method.
2 x + y = 5
 y = 5 − 2x
 12
 x = 5
4 x − 15 + 6 x = 9 10 x = 24
⇒
⇔

 y = 5 − 2x
 y = 5 − 2x  y = 1
 5

Nominate students to solve that system of Way 2:.
equation by using adding two side method
4 x − 3 y = 9

remark .
Nominate students to solve system of

4 x − 3 y = 9
⇔
⇔

2 x + y = 5
 4 x + 2 y = 10
1
12


2x + = 5
x=


2 x + y = 5


5
5
⇔
⇔

 −5 y = − 1
y = 1
y = 1
5
5



EX 2: solve the system equation


equation

3 x − 6 y = 9

−2 x + 4 y = −3

.

3x − 6 y = 9
6 x − 12 y = 18
x ∈ ∅
⇔
⇔

−2 x + 4 y = −3 −6 x + 12 y = −9  y ∈ ∅

So the system equation has no root

Remark
4- Consolidate
- Request student recall the main point of the lesson.
- Do exercise 1/ text book page 68
5- Homework:
- Memorize the lesson.
- Do the exercise 2,3,4/ text book page 68

Preparation date:
Teaching date:
Period 22: Equations and system of linear equations with multiple unknowns
I. Objectives: By the end of the lesson, Students will be able to:
- Review the concept of quadratic equations and systems of to linear equations with two
unknown.
- Solve systems of liner equations and system of linear equations with two unknown.
II. Teaching aids:
- Teacher: lesson plan, text book
- Student: review about systems of equation that learnt at last lesson.
III. Method: Raising and settling problems
IV. Procedure:
1- Orginization.


2- Review:
Student 1: is the pair of number (2;0) a root of equation: 2x-3y=4
Student 2: solve the system equation:
3- New lesson:

 x + 2 y = −1

x − y = 2

Activity 1 : Linear equations with three unknowns
Teacher’s activities

Students’ activities
II Linear equations with three unknowns
Introduce the concept of liner equations 1. Linear equations with three unknowns

with three unknowns.
a) concept: (text book)
form : ax + by + cz = d.
b) example:
x + 2y – 3z = 5
Take some examples and request student to ( a = 1; b = 2; c = – 3; d = 5)
determine co-efficient a,b,c,d in each 5y + 2z = 0.
equation.
( a = 0; b = 5; c = 2; d = 0)
3z = 15
( a = 0; b = 0; c = 3; d = 15)

Hoạt động 2: systems of linear equations with three unknowns
Teacher’s activities
Students’ activities
2 Systems of linear equations with three
Introduce the concept of Systems of unknowns
linear equations with three unknowns
a) concept: (textbook)
form :
State the form of a root?

Introduce the system equation in triangle
form.

 a1 x + b1 y + c1 z = d1

 a2 x + b2 y + c2 z = d 2
a x + b y + c z = d
3

3
3
 3

b) example


 x + 3 y − 2 z = −1
3
4 y + 3z =
2

2z = 3


Given some examples about Systems of 

linear equations with three unknowns

 x + 2 y − 3 z = 11

 2 x + 3 y + 7 z = −6
 −3 x + y − 3 z = 5


Activity 3 : gauss’ method.
Teacher’s activities

(1)

(2)

Students’ activities
3. The method to solve system equation:

Guiding student to solve the systems
equations in triangle form. Given
example.
Nominate students to solve that system
of equation.

Remark


 x + 3 y − 2 z = −1  x + 3 y − 2 z = − 1


3
3


*  4 y + 3z = ⇔  4 y + 3 z =
2
2


2z = 3
3



z=

2
17

x = 4

3

⇔ y = −
4

3

z = 2


Guiding student to solve the systems So, the root of that system equation are:
 17 3 3 
equations that not in triangle form. Given
 ;− ; ÷
4 2
 4
example.
(x; y; z) =
Convert to the systems equations
triangle form.

in

Nominate students to solve that system
of equation .
Remark.


 x + 2 y − 3 z = 11
 x + 2 y − 3 z = 11


*  2 x + 3 y + 7 z = −6 ⇔ 
y − 13 z = 28
 −3 x + y − 3 z = 5
 7 y − 12 z = 38


 x + 2 y − 3 z = 11
 x + 2 y − 3z = 11


⇔
y − 13z = 28 ⇔ 
y − 13 z = 28


− 79z = 158
z = −2


x = 1


⇔ y = 2
 z = −2


Hence, the root of that system equation are:
(x; y; z) = (1; 2; – 2 )
4- Consolidate
- Request student recall the main point of the lesson.
5- Homework:
- Memorize the lesson.
- Do the exercise 5,6,7/ text book page 68,69


Preparation date:
Teaching date:

Period 23: Equations and system of linear equations with multiple unknowns
I. Objectives: By the end of the lesson, Students will be able to:
- Review the method to solve systems of three linear equations with three unknown
- solve the solve systems of three linear equations with three unknown by using gauss ‘
method.
II. Teaching aids:
- Teacher: lesson plan, text book
- Student: review about systems of equation that learnt at last lesson.
III. Method: Raising and settling problems
IV. Procedure:
1- Orginization.
2- Review:
State the methods to solve systems of three linear equations with three unknowns
3- New lesson:

Activity 1: solve EX 2/ text book page 68.
Teacher’s activities

Students’ activities
EX 2: solve the following system equations

Request and guiding student to solve a, b
EX2.
Nominate 2 students to write down the
solutions.

a)

2 x − 3 y = 1 2 x − 3 y = 1
⇔

x + 2 y = 3
2 x + 4 y = 6

x + 2 y = 3
 x = 11 / 7
⇔
⇔
 −7 y = −5
y = 5 / 7

Remark .
b)

3 x + 4 y = 5

3 x + 4 y = 5
⇔

4 x − 2 y = 2
8 x − 4 y = 4

3 x + 4 y = 5
 x = 9 / 11
⇔

⇔ 11x = 9
 y = 7 / 11

Guiding students transform co- efficients
into integer.
Nominate 2 students to write down the
solutions.
c)
.

1
2
2
 3 x + 2 y = 3
4 x + 3 y = 4
⇔

4 x − 9 y = 6
1 x − 3 y = 1
 3

4
2


4 x + 3 y = 4
x = 9 / 8
⇔
⇔
12 y = −2
 y = −1 / 6

.
Remark

d)

.

0,3 x − 0, 2 y = 0,5
3 x − 2 y = 5
⇔

0,5 x + 0, 4 y = 1, 2
5 x + 4 y = 12

6 x − 4 y = 10
11x = 22
⇔
⇔
5 x + 4 y = 12

5 x + 4 y = 12
x = 2
⇔
y = 1/ 2

Activity 2: solve EX 3/ text book page 68.
Teacher’s activities
Students’ activities
Request student to sum up the content of EX 3:
exercise.
Solution
Guiding student to choose unknown and We have system equation:
find it’s condition.
10 x + 7 y = 17800
10 x + 7 y = 17800

 2 x + y = 3000

Set up the system equation

⇔
10 x + 5 y = 15000

 2 x + y = 3000
 x = 800 ( TM )
⇔
⇔
 2 y = 2800
 y = 1400 ( TM )

Nominate student to write down the
solutions
remark
Activity 3: solve EX 5 text book page 68.
Teacher’s activities
Students’ activities
Request student to solve the system EX 5: solve the following system equations
equations by using Gauss’ method
x + 3 y + 2z = 8 x + 3y + 2z = 8
Nominate 2 students to write down the
solutions

Remark

a)



2 x + 2 y + z = 6 ⇔ 
3x + y + z = 6




4 y + 3z = 10
8 y + 5 z = 18

 x + 3 y + 2z = 8
x = 1



 4 y + 3z = 10 ⇔  y = 1

z = 2
z=2



hence : (x ; y ; z) = (1 ; 1 ; 2)


b)

 x − 3 y + 2 z = −7
 x − 3 y + 2 z = −7


−2 x + 4 y + 3z = 8 ⇔  − 2 y + 7 z = −6
3x + y − z = 5
 − 10 y + 7 z = −26



 x − 3 y + 2 z = −7
 x = 11 / 4


⇔  − 2 y + 7 z = −6 ⇔  y = 5 / 2

28 z = −4  z = −1 / 7



hence:(x ; y ; z) =
4- Consolidate
- Request student recall the main point of the lesson.
5- Homework:
- Memorize the lesson.
- Do the exercise in review of chapter III

1
 11 5
; ; − ÷

7
4 2


Preparation date:
Teaching date:

Period 24: Practise
I. Objectives: By the end of the lesson, Students will be able to:
- Review gauss’ method to solve systems of three linear equations with three unknown
- solve the solve systems of three linear equations with three unknown by using triangle
form, Gauss’ method
II. Teaching aids:
- Teacher: lesson plan, text book
- Student: review about systems of equation that learnt at last lesson.
III. Method: Raising and settling problems
IV. Procedure:

1- Orginization.
2- Review:
- State the methods to solve systems of three linear equations with three unknowns
- state the steps to solve an exercise by set up an system equations
3- New lesson:
Activity 1: solve EX 2 ( page 68)
Teacher’s activities

Students’ activities
EX2: Giải các hệ phương trình:

Request student to solve the system
equations by using elimination method
Nominate 2 students to write down the
solutions

a)

2 x − 3 y = 1 2 x − 3 y = 1
⇔

x + 2 y = 3
2 x + 4 y = 6

x + 2 y = 3
 x = 11 / 7
⇔
⇔
 −7 y = −5
y = 5 / 7

3x + 4y = 5 3x + 4y =5
b. 
⇔
4x
−
2y
=
2
8x − 4y = 4

Remark

 9
11x =9
 x =11

⇔  5−3x ⇔ 
 y = 4
y= 7
 11


Teacher’s activities
Guiding student transform
effiction to integer form

Students’ activities
the co-

Nominate 2 students to write down the
solutions
c)

1
2
2
x
+
y
=
 3
4 x + 3 y = 4
2
3
⇔

4 x − 9 y = 6
1 x − 3 y = 1
 3
4
2

4 x + 3 y = 4
x = 9 / 8
⇔
⇔
12 y = −2
 y = −1 / 6

Remark

d)

0,3 x − 0, 2 y = 0,5
3 x − 2 y = 5
⇔

0,5 x + 0, 4 y = 1, 2
5 x + 4 y = 12

6 x − 4 y = 10
11x = 22
⇔
⇔
5 x + 4 y = 12
5 x + 4 y = 12
x = 2
⇔
y = 1/ 2

Activity 2: solve EX 4 ( page 68).
Teacher’s activities
Students’ activities
Guiding students set up a system of EX 4.
Day 1: both assembly lines made 930 shirts
equations

Day 2: due to the first assembly line increasing
productivity by 18%, and the second one by 15%

so both assembly lines made 1083 shirts. How
many shirt did each assembly line make on the
first day.
Solution
Nominate a student to write down the Let x, y be the quantity of shirts that both
solutions
assembly line make on the first day ( x,y are
possitive integer)
Request students to comment the According to the suppsosed, we have the system
solutions
equation below

Remark

x + y = 930

1,18x + 1,15y = 1083

Solve the system equation, we obtain,

 x = 450

 y = 480


Hence : on the first day, the first assembly line
made 450 shirts , and the second one mades 480
shirts

Activity 3: solve EX 5/ textbook page 68.

Teacher’s activities
Students’ activities
Request students to solve the system EX 7: solve the following systems equations
equations by using Gauss’ method
3 x + 4 y = 5
3 x + 4 y = 5
⇔

4 x − 2 y = 2
8 x − 4 y = 4

Nominate a student to write down the a)
solutions
3 x + 4 y = 5
 x = 9 / 11
⇔


Request students to comment the
⇔ 11x = 9
 y = 7 / 11
solutions
Remark
 x − 3 y + 2 z = −7
 x − 3 y + 2 z = −7

b




 −2 x + 4 y + 3 z = 8 ⇔  − 2 y + 7 z = −6
3 x + y − z = 5
 − 10 y + 7 z = −26



4- Consolidate
- Request student recall the main point of the lesson.
5- Homework:
- Memorize the lesson.
- Do the exercise 1-7 in exercise book