nb: kyk elke dag die video vir verduideliking van die dag ... · oefening/ exercise 1 p 47 no a) 2,...

Graad 10 Hoofstuk 4 - Vergelykings en Ongelykhede Grade 10 Chapter 4 - Equations ______________________________________________________________________________________

Dag 22 / Day 22 NB: Kyk elke dag die video vir verduideliking van die dag se werk

en die memo van elke oefening.

NB: Watch the video each day for explanation of the work of the

day and the memo of the exercise.

Lineêre Vergelykings Linear Equations

Voorbeelde Examples

Los op vir x: Solve for x:

a) 2 3 2 3( 3)x x+ = − +

2 3 2 3 9

2 3 3 7

5 10

2

x x

x x

x

x

+ = − −

+ = − −

= −

= −

b) 5(1 3 ) 2(10 ) 10x x x− − − = −

5 15 20 2 10

15 13 10

3 15

5

x x x

x x

x

x

− − + = −

− − = −

− =

= −

Voorbeelde

Reël: Om ‘n breuk in ‘n vergelyking weg te kry moet met die KGV aan beide kante gemaal word.

Examples

Rule: To solve a fraction in an equation, multiply with the LCD to get rid of the fraction

1. Los op vir y:

10 1 4(7 1)

2 5

y y+ +=

KGV = 10:

5(10 1) 4(2)(7 1)

50 5 56 8

6 3

1

2

y y

y y

y

y

+ = +

+ = +

− =

= −

2. Los op vir x:

3 1 6 LCD = 4

2 4

2(3) (1) 6(4)

6 24

18

18

xx x

x

x

x

x

− =

− =

− =

− =

= −

Oefening/ Exercise 1 p 47 no a) 2, 3, 6, b) 2, 4, 6

Graad 10 Hoofstuk 4 Vergelykings

Grey Kollege 2


Grey Kollege 3

Dag 23 / Day 23 Kwadratiese vergelykings Quadratic equations

Om die vergelyking op te los: Maak die RK = 0

Faktoriseer die LK

Stel die faktore = 0

To solve the equation: Put RHS = 0

Factorise LHS

Let the factors = 0

Voorbeelde Examples


a) 2 2 8 0x x− − =

( 4)( 2) 0

4 or 2

x x

x x

− + =

= = −

b) (2 1)( 2) 2x x− + = −

2

2

2 3 2 2

2 3 0

(2 3) 0

30 or

2

x x

x x

x x

x x

+ − = −

+ =

+ =

= = −

Oefening / Exercise 2 p 49 no a) 2, 3, 6, b) 4, 13, 16, 22,24


Grey Kollege 4

Dag 24 / Day 24 Breuke vergelykings: Fraction equations:

Voorbeelde: Examples:

1. 3𝑥−4

𝑥2−3𝑥−4+

4

4−𝑥=

5

2𝑥+2

3𝑥−4

(𝑥−4)(𝑥+1)−

4

(𝑥−4)=

5

2(𝑥+1) 𝑥 ≠ 4; 𝑥 ≠ −1

KGV = 2(𝑥 − 4)(𝑥 + 1)

3𝑥−4

(𝑥−4)(𝑥+1)×

2(𝑥−4)(𝑥+1)

1−

4

(𝑥−4)×

2(𝑥−4)(𝑥+1)

1=

5

2(𝑥+1)×

2(𝑥−4)(𝑥+1)

1

2(3𝑥 − 4) − 8(𝑥 + 1) = 5(𝑥 − 4)

6𝑥 − 8 − 8𝑥 − 8 = 5𝑥 − 20

6𝑥 − 8𝑥 − 5𝑥 = −20 + 8 + 8

−7𝑥 = −4

𝑥 =4

7

Onthou om altyd jou antwoord te toets in breuk vergelykings

Remember to always test your answer in fraction equations


Grey Kollege 5

2) 2 2

3 5 4

3 2 4 1

x

x x x x

−− =

+ + − − −

3 5 4

;( 2)( 1) ( 2)( 2) 1

x

x x x x x

− −− =

+ + + − + 2; 2; 1x x x − − KGV = (x + 2)(x – 2)(x + 1)

( 3)( 2) 5( 1) 4( 2)( 2)x x x x x− − − + = − + −

2 2

2 2

2

2

5 6 5 5 4( 4)

10 1 4 16

5 10 15 0

5( 2 3) 0

5( 3)( 1) 0

3 or 1 but 1

x x x x

x x x

x x

x x

x x

x x x

− + − − = − −

− + = − +

− − =

− − =

− + =

= = − −

Slegs x = 3.

Oefening / Exercise 3 p 51 no a) 2, 6, 7, 8, b) 2, 5, 7


Grey Kollege 6


Grey Kollege 7

Dag 25 / Day 25 Gelyktydige Vergelykings Simultaneous Equations

Oplossing van gelyktydige vergelyking deur

substitusie:

Solution of simultaneous equations by

substitution:

1. Los op vir x en y:

Solve for x and y:

x + 2y = 5 en x – y = - 1

2 5 ........(1)

1 ........(2)

5 2 ........(3)

x y

x y

x y

+ =

− = −

= −

Stel (3) in (2)

5 2 1

5 3 1

3 6

2

5 2(2)

5 4

1

y y

y

y

y

x

x

x

− − = −

− = −

− = −

=

= −

= −

=

(1; 2) is die oplossing. Is the solution

Los op vir x en y deur eliminasie:

Solve for x and y by elimination:

1. Los op vir x en y:

Solve for x and y:

a) x + 2y = 5 en x – y = - 1

2 5 ........(1)

1 ........(2)

(1) - (2): 3 6

2

x y

x y

y

y

+ =

− = −

=

=

x + 2(2) = 5

x = 1

(1; 2) is die oplossing. Is the solution.

b) 3x + 2y = 2 en 5x – 2y = - 18

3 2 2 ........(1)

5 2 18 ........(2)

(1) + (2): 8 = 16

2

x y

x y

x

x

+ =

− = −

−

= −

3( 2) 2 2

6 2 2

2 8

4

y

y

y

y

− + =

− + =

=

=

(- 2 ; 4) is die oplossing. Is the solution


Grey Kollege 8

Oefening / Exercise 4 p 53 no b), d), e), f) .

Dag 26 / Day 26

Oefening / Exercise 5 p 55 no a) 2, 3, 4, b) 2, 4, 5


Grey Kollege 9

Dag 27 / Day 27 Letter Vergelykings Letter Equations

1. Los op vir x: Solve for x: a)

mx p q

mx q p

q px

m

+ =

= −

−=

b) A = 2 2( )R r −

Maak r die onderwerp van die formule.

Make r the subject of the formula:

Oplossing:

2 2

2 2

2 2

2 2

2 2

2

( )R r A

AR r

Ar R

Ar R

Ar R

Ar R

− =

− =

− = −

= − −

= −

= −

Radius kan slegs positief wees.

Radius can only be positive.

c) Los op vir x:

Solve for x:

2( ) ( )

2 2

2 2

(2 ) 2

2

2

(2 )

2

x a b x a

x a bx ab

x bx a ba

x b a ba

a bax

b

a bx

b

+ = −

+ = −

− = − −

− = − −

− −=

−

− +=

−


Grey Kollege 10

Oefening / Exercise 6 p 57 no a) 2, b) 2, 3, c) 3, d), e) 2


Grey Kollege 11

Dag 28 / Day 28

Lineêre Ongelykhede Linear Inequalities

Voorbeelde: Examples:


1. 2 3

3( 1)2

xx

− + LCD = 2

2 3 6( 1)

2 3 6 6

4 9

9

4

x x

x x

x

x

− +

− +

−

−

Getallelyn:

Number line:

2. 4 3 1 5x− −

3 3 6

1 2

x

x

−

−

Getallelyn:

Number line:

1. 2 3

3( 1)2

xx

− + KGV = 2

2 3 6( 1)

2 3 6 6

4 9

9

4

x x

x x

x

x

− +

− +

−

−

Getallelyn:

Number line:

2. 4 3 1 5x− −

3 3 6

1 2

x

x

−

−

Getallelyn:

Number line:

Oefening / Exercise 7 p 60 no a) 2, 4, b) 5, c) 2, 4, 6

-4 -3 -2 -1 9

4−

-1 2

-4 -3 -2 -1 9

4−

-1 2


Grey Kollege 12

Hersiening Oefening p 70 – Als op jou eie (opsioneel)

Revision Exercise p 70 - Everything on your own (optional)


Grey Kollege 13

Dag 29 / Day 29

GRAAD 10 GREY KOLLEGE

WISKUNDE Werkskaart 1 TOTAAL: 34

GRADE 10 GREY COLLEGE

MATHS WORKSHEET 1 TOTAL: 34

VRAAG 1 QUESTION 1


1.1. ( )( ) 357 =−− xx (4)

1.2. 33

14

−−

x (4)

1.3. −4 < 2 − 2𝑥 < 6 (3)

1.4. (x – 2)(2x + 1) = x + 4 (4)

1.5. 3( 3) 2( 1)x x− = + (4)

1.6. 5( 8) (8 3 ) 5 3( 8)x x x x+ − − = − + (6)

1.7.

5 3 413

4 9

x x− −− =

(5)

1.8. 2 36x = (4)

[34]


Grey Kollege 14

Dag 30 / Day 30





Vraag 1 QUESTION 1


Los op vir x en y gelyktydig:

Solve for x and y simultaneously:


Grey Kollege 15

Dag 31 / Day 31





VRAAG 2 / QUESTION 2

Los op vir x : / Solve for x:

2.1. 53𝑥 = 25𝑥+1

(3)

2.2. 3.2𝑥+1 − 192 = 0 (4)

2.3. (3𝑥 − 2)(𝑥 + 1) = 0 (4)

2.4. 2𝑥2 = 6𝑥 (4)

2.5. 2(𝑥 − 2)(𝑥 + 4) = 32 (5)

2.6. 2𝑥

3−

5

3=

3𝑥

2−

5

2 (6)

2.7. 3(𝑥 − 2) − (4𝑥 + 1) < −3 (4) [30]

VRAAG 3 / QUESTION 3

3.1. Los op vir x en y gelyktydig: / Solve x and y simultaneously:

6𝑥 − 𝑦 − 4 = 0 9𝑥 + 1 = −2𝑦 (5)

3.2. Maak a die onderwerp van die volgende: / Make a the subject of the following:

𝑥2 = 𝑎2 − 𝑦2 (3) [8]

nb: kyk elke dag die video vir verduideliking van die dag ... · oefening/ exercise 1 p 47 no a) 2,...

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