MATHEMATICS LESSON PLAN
GRADE 9
TERM 3: July – September 2015
PROVINCE:
DISTRICT:
SCHOOL:
TEACHER’S NAME:
DATE:
DURATION:
1
Hour
Grade 9 Lesson Plan: 1+4 Intervention – Term 3
(Draft)
Name: ………………………………………..Date: ………………………
Class: …………………………………………Time: 60 minutes
Question 1
Solve the given equations and circle only the correct answer:
Solve for 𝑥
1.
3(𝑥 − 1) =
A
15
-6
B
16
D
9
-4
C 2
D
-5
5
C 12,5
D
±5
B
5
C 4
D
30
C 0 or 1
D
4
C 5
D
-1
4𝑎 − 4𝑎² = 0 , then the solution for 𝑎 is:
-4
6.
A
C 6
Solve for 𝑥 𝑖𝑛 2𝑥 = 32
5.
A
B
-5
4.
A
1
Solve for 𝑥 𝑖𝑛 𝑥² = 25
3.
A
B
Solve for 𝑥: 2𝑥+1 = 2−5
2.
A
1
𝑥 + 12
2
B
6. If
3
𝑥+3
2
-1 or 0
= 4, then 𝑥 equals:
B
11
Grade 9 Lesson Plan: 1+4 Intervention – Term 3
(Draft)
Page 2 of 7
7. If 𝑥² + 12𝑥 + 35 = 0 , then the solution for 𝑥 is:
7.
A
5 or 7
B
5 or -7
C -5 or 7
D
-5 or -7
D
2 or 5
C 16
D
4
C 3
8
D
-2
Let 𝑥 2 - 3𝑥 - 15 = 0 then the solution for 𝑥 is:
8.
A
5 or -2
1
𝐿𝑒𝑡
9.
A
B
B
If
2𝑥+5
6
𝑥−5
+
2
1
2
A
C -5 or -2
𝑥 + 3 = 5, then the solution for 𝑥 is:
2
1
10.
2 or -5
30
= 0, then 𝑥 is:
B
2
Question 2 (14)
Solve the following equations:
1
3(𝑥 – 1) – 4𝑥 = 5 – 2(𝑥+ 1)
(3)
2
𝑥−2
4
(4)
3
3𝑎² = 2𝑎
(3)
4
1
2
(4)
−
𝑥+1
3
=
𝑥−2
12
𝑥² − 8 = 0
Question 3 (15)
Solve the following equations:
1
2
𝑏² − 6𝑏 + 9 = 0
𝑘² = 7𝑘 – 6
(2)
(4)
3
𝑎² + 3𝑎 – 10 = 0
(3)
4
5
9𝑥−3 = 81
52𝑥+2 = 252
(3)
(3)
Grade 9 Lesson Plan: 1+4 Intervention – Term 3
(Draft)
Page 3 of 7
Question 4 (11)
(a) Determine the value for y for the given values of 𝑥 if y = -𝑥²
-3
-𝑥
y
-1
0
(2)
2
(b) Write the outcome as ordered pairs:
(2)
(c) Study the table and determine which equation is representing the values as
indicated in the table:
(1)
-3
7
-𝑥
y
A
B
C
D
-2
5
0
1
1
-1
5
-9
y = -2𝑥 - 1
y=-2𝑥 +1
y= 2𝑥 +1
y= 2𝑥 –1
(d) Sipho is 6 years older than Betty. In 3 years’ time Sipho will be twice as old as
Betty. How old is Sipho now.
(6)
MEMORANDUM
Question 1 (10)
1
2
3
4
5
6
7
8
9
10
C
A
D
B
C
C
D
A
D
B
Grade 9 Lesson Plan: 1+4 Intervention – Term 3
(Draft)
Page 4 of 7
Question 2 (14)
Solve the following equations:
1. 3(𝑥 – 1) – 4𝑥 = 5 – 2(𝑥+ 1)
3𝑥 – 3 – 4𝑥 = 5 – 2𝑥 – 2 √
−𝑥 – 3 = 3 – 2𝑥 √
𝑥 = 6√
2.
𝑥−2
4
𝑥+1
𝑥−2
−
=
3
12
3𝑥 – 6 – (4𝑥 + 4) = 𝑥 – 2 √
3𝑥 – 6 – 4𝑥 – 4 – 𝑥 + 2 = 0
−2𝑥 – 8 = 0 √√
−2𝑥 = 8
𝑥 = −4 √
3. 3𝑎2 = 2𝑎
3𝑎2 - 2𝑎 = 0
𝑎(3𝑎 – 2) = 0 √
𝑎 = 0 or 3𝑎 - 2 = 0 √
𝑎 = 0 or 3𝑎 = 2
2
𝑎=3 √
4.
1
2
(3)
(4)
(3)
𝑥2 – 8 = 0
𝑥 2 - 16 = 0√
(𝑥 − 4)(𝑥 + 4) = 0 √
𝑥 = 4 𝑜𝑟 𝑥 = −4 √√
(4)
Question 3 (15)
Solve the following equations:
1. 𝑏² − 6𝑏 + 9 = 0
(𝑏 – 3)(𝑏 – 3) = 0 √
𝑏 = 3√
(2)
Grade 9 Lesson Plan: 1+4 Intervention – Term 3
(Draft)
Page 5 of 7
2. 𝑘² = 7𝑘 – 6
𝑘² − 7𝑘 + 6 = 0√
(𝑘 − 1)( 𝑘 – 6) = 0 √
𝑘 − 1 = 0 𝑜𝑟 𝑘 – 6 = 0
𝑘 = 1 𝑜𝑟 𝑘 = 6√√
3.
(4)
𝑎² + 3𝑎 – 10 = 0
(𝑎 + 5)(𝑎 – 2) = 0 √
𝑎 + 5 = 0 𝑜𝑟 𝑎 – 2 = 0
𝑎 = −5
𝑎 = 2 √√
(3)
4. 9𝑥−3 = 81
32𝑥−6 = 34 √
2𝑥 – 6 = 4 √
2𝑥 = 10
𝑥 = 5 √
(3)
5. 52𝑥+2 = 252
52𝑥+2 = 54 √
2𝑥 + 2 = 4 √
2𝑥 = 2
𝑥 = 1 √
(3)
Question 4 (11)
a) Determine the value for y for the given values of 𝑥 if y = -𝑥²
-3
-𝑥
y
-9√
(half marks)
-1
-1√
0
0√
2
-4√
(2)
b) Write the outcome as ordered pairs:
(-3; -9) (-1; -1) (0; 0) (2; -4) √√√√
(half marks)
(2)
c) Study the table and determine which equation is representing the values as
Grade 9 Lesson Plan: 1+4 Intervention – Term 3
(Draft)
Page 6 of 7
indicated in the table:
-𝑥
𝑦
A
B
C
D
-3
7
-2
5
0
1
1
-1
5
-9
y = -2𝑥 - 1
y=-2𝒙+1
y= 2𝑥 +1
y= 2𝑥 –1
(1)
Ans = B
d) Sipho is 6 years older than Betty. In 3 years’ time Sipho will be twice as old as
Betty. How old is Sipho now.
(6)
Let Betty’s age be 𝑥, then Sipho’s age will be 𝑥 + 6 √
In 3 years time Betty’s age = 𝑥 + 3 √
Sipho’s age = (𝑥 + 6) + 3 √
(𝑥 + 6) + 3 = 2(𝑥 + 3) √
𝑥 + 9 = 2𝑥 + 6
𝑥 – 2𝑥 = 6 − 9
−𝑥 = − 3
𝑥 = 3√
Sipho’s age is 3 + 6 = 9 years √
Grade 9 Lesson Plan: 1+4 Intervention – Term 3
(Draft)
Page 7 of 7