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Revision §1.1 + §1.2 + §1.3
C1 Arithmetic with Letters
Mr. Fintelman (FNL)
Monday October 7th
2024
1 / 30
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Slide 1:
Slide
Wiskunde
Middelbare school
vwo
Leerjaar 2
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Lesson duration is:
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C1 Arithmetic with Letters
Mr. Fintelman (FNL)
Monday October 7th
2024
Slide 1 - Slide
Date
Monday October 7th 2024
Paragraph
§1.1 Expanding Brackets
§1.2 Simplifying Fractions
Pages from the handbook
Pag. 12-22
Subject
Revision
Today is the day...
Slide 2 - Slide
I can already…
… simplify an expression with the following rule: a(b + c) = ab + ac
… simplify an expression with the following rule: (a + b)(c + d) = ac + ad + bc + bd
… simplify expressions that contain brackets.
… simplify fractions with letters.
… add and subtract fractions with letters.
… multiply and divide fractions with letters.
Prior Knowledge
Slide 3 - Slide
Examples
(
a
+
2
)
(
b
+
5
)
=
=
(
a
⋅
b
)
+
(
a
⋅
5
)
+
(
2
⋅
b
)
+
(
2
⋅
5
)
=
(
a
b
)
+
(
5
a
)
+
(
2
b
)
+
(
1
0
)
=
a
b
+
5
a
+
2
b
+
1
0
Slide 4 - Slide
Examples
(
c
+
2
)
(
d
−
4
)
=
=
(
c
⋅
d
)
+
(
c
⋅
−
4
)
+
(
2
⋅
d
)
+
(
2
⋅
−
4
)
=
(
c
d
)
+
(
−
4
c
)
+
(
2
d
)
+
(
−
8
)
=
c
d
−
4
c
+
2
d
−
8
Slide 5 - Slide
Examples
(
x
+
2
)
(
x
−
3
)
=
=
(
x
⋅
x
)
+
(
x
⋅
−
3
)
+
(
2
⋅
x
)
+
(
2
⋅
−
3
)
=
(
x
2
)
+
(
−
3
x
)
+
(
2
x
)
+
(
−
6
)
=
x
2
−
3
x
+
2
x
−
6
=
x
2
−
x
−
6
Slide 6 - Slide
Examples
(
6
a
+
1
)
(
4
a
+
2
)
−
(
5
a
−
4
)
(
a
+
3
)
=
=
2
4
a
2
+
1
2
a
+
4
a
+
2
−
(
5
a
2
+
1
5
a
−
4
a
−
1
2
)
=
2
4
a
2
+
1
2
a
+
4
a
+
2
−
5
a
2
−
1
5
a
+
4
a
+
1
2
=
1
9
a
2
+
5
a
+
1
4
Slide 7 - Slide
Examples
9
a
−
6
a
=
3
⋅
3
a
−
2
⋅
3
a
=
−
3
2
b
2
a
b
=
b
⋅
b
a
⋅
b
=
b
a
5
a
2
5
a
b
c
=
1
⋅
5
a
5
b
c
⋅
5
a
=
5
b
c
Slide 8 - Slide
Examples
x
2
+
x
4
=
x
6
3
a
+
b
5
=
3
b
a
b
+
3
b
1
5
=
3
b
a
b
+
1
5
Slide 9 - Slide
Examples
x
3
−
y
7
=
x
y
3
y
−
x
y
7
x
=
x
y
3
y
−
7
x
4
a
b
+
c
5
=
4
c
a
b
c
+
4
c
2
0
=
4
c
a
b
c
+
2
0
5
+
b
2
a
=
b
5
b
+
b
2
a
=
b
5
b
+
2
a
Slide 10 - Slide
Examples
9
2
⋅
9
4
=
9
⋅
9
2
⋅
4
=
8
1
8
1
1
6
⋅
7
3
=
1
1
⋅
7
6
⋅
3
=
7
7
1
8
3
2
÷
5
4
=
3
⋅
4
2
⋅
5
=
1
2
1
0
=
6
5
Slide 11 - Slide
Examples
3
a
⋅
b
a
=
3
⋅
b
a
⋅
a
=
3
b
a
2
3
b
a
⋅
2
c
6
b
=
3
b
⋅
2
c
a
⋅
6
b
=
6
b
c
6
a
b
=
c
a
b
a
÷
5
4
a
=
b
⋅
4
a
a
⋅
5
=
4
a
b
5
a
=
4
b
5
Slide 12 - Slide
Worktime
You work neatly by…
… reading the
theory
(again) before asking a question to your classmate.
…
raising a hand
before asking a question to the teacher.
… if the teacher is busy, remember your question and
move on
.
Help:
Exercises: -
Assignments:
Choose from: Diagnostic Test, Revision or Practice Test.
Assignments from the planning of WEEK -:
Extra:
Exercises: -
Slide 13 - Slide
Now I can...
… simplify an expression with the following rule: a(b + c) = ab + ac
… simplify an expression with the following rule: (a + b)(c + d) = ac + ad + bc + bd
… simplify expressions that contain brackets.
… simplify fractions with letters.
… add and subtract fractions with letters.
… multiply and divide fractions with letters.
Reflection
Slide 14 - Slide
Date
Wednesday October 9th 2024
Paragraph
§1.3 Multiplying, Adding and Subtracting Powers
Pages from the handbook
Pag. 23-25
Subject
Revision
Today is the day...
Slide 15 - Slide
I can already…
… simplify a product of powers.
… simplify a sum and a difference between powers.
Prior Knowledge
Slide 16 - Slide
From the Online Revision
a
(
2
a
+
3
)
a
⋅
2
a
+
a
⋅
3
2
a
2
+
3
a
Slide 17 - Slide
From the Online Revision
(
a
−
3
)
(
2
a
+
3
)
=
a
⋅
2
a
+
a
⋅
3
−
3
⋅
2
a
−
3
⋅
3
=
2
a
2
+
3
a
−
6
a
−
9
=
2
a
2
−
3
a
−
9
Slide 18 - Slide
From the Online Revision
(
2
a
−
2
)
(
4
a
2
−
a
+
3
)
=
2
a
⋅
4
a
2
+
2
a
⋅
−
a
+
2
a
⋅
3
−
2
⋅
4
a
2
−
2
⋅
−
a
−
2
⋅
3
=
8
a
3
−
1
0
a
2
+
8
a
−
6
8
a
3
−
2
a
2
+
6
a
−
8
a
2
+
2
a
−
6
=
8
a
3
−
2
a
2
−
8
a
2
+
6
a
+
2
a
−
6
=
Slide 19 - Slide
From the Online Revision
b
a
+
d
c
=
b
⋅
d
a
⋅
d
+
d
⋅
b
c
⋅
b
=
b
d
a
d
+
b
d
b
c
=
b
d
a
d
+
c
b
Slide 20 - Slide
From the Online Revision
b
8
a
2
+
2
a
+
b
=
b
2
⋅
2
8
a
⋅
2
+
2
⋅
b
2
b
2
(
a
+
b
)
=
2
b
2
1
6
a
+
2
b
2
a
b
2
+
b
3
=
2
b
2
1
6
a
+
a
b
2
+
b
3
Slide 21 - Slide
From the Online Revision
3
1
6
a
⋅
9
a
4
=
3
⋅
9
a
1
6
a
⋅
4
=
3
1
6
a
÷
4
9
a
=
2
7
a
6
4
a
=
2
7
6
4
=
2
2
7
1
0
Slide 22 - Slide
From the Online Revision
4
a
3
⋅
−
3
a
2
=
−
1
2
a
5
−
1
2
a
3
+
2
Slide 23 - Slide
From the Online Revision
5
a
2
(
3
a
+
2
)
=
5
a
2
⋅
3
a
+
5
a
2
⋅
2
=
1
5
a
2
+
1
+
1
0
a
2
=
1
5
a
3
+
1
0
a
2
Slide 24 - Slide
From the Online Revision
(
3
a
2
−
5
a
)
(
2
a
+
4
)
=
3
a
2
⋅
2
a
+
3
a
2
⋅
4
−
5
a
⋅
2
a
−
5
a
⋅
4
=
6
a
2
+
1
+
1
2
a
2
−
1
0
a
1
+
1
−
2
0
a
=
6
a
3
+
1
2
a
2
−
1
0
a
2
−
2
0
a
=
6
a
3
+
2
a
2
−
2
0
a
Slide 25 - Slide
Examples
x
2
⋅
x
2
=
x
⋅
x
⋅
x
⋅
x
=
x
4
y
3
⋅
y
4
=
y
⋅
y
⋅
y
⋅
y
⋅
y
⋅
y
⋅
y
=
y
7
3
x
2
⋅
2
x
2
=
3
⋅
x
⋅
x
⋅
2
⋅
x
⋅
x
=
6
x
4
Slide 26 - Slide
Examples
4
a
4
+
6
a
4
=
1
0
a
4
8
a
7
−
3
a
7
=
5
a
7
7
a
5
−
5
a
2
=
k
.
n
.
Slide 27 - Slide
Examples
2
a
2
(
3
a
−
4
b
)
=
(
a
−
1
)
(
a
2
+
a
)
=
6
a
3
−
8
a
2
b
a
3
+
a
2
−
a
2
−
a
=
a
3
−
a
Slide 28 - Slide
Worktime
You work neatly by…
… reading the
theory
(again) before asking a question to your classmate.
…
raising a hand
before asking a question to the teacher.
… if the teacher is busy, remember your question and
move on
.
Help:
Exercises: -
Assignments:
Choose from: Diagnostic Test, Revision or Practice Test.
Assignments from the planning of WEEK -:
Extra:
Exercises: -
Slide 29 - Slide
Now I can...
… simplify a product of powers.
… simplify a sum and a difference between powers.
Reflection
Slide 30 - Slide
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