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# Gelyktydige vergelykings: algebraïese oplossing

## Algebraïese oplossings

Die oplos van gelyktydige vergelykings in algebra is deur middel van substitusie

Byvoorbeeld die oplossing van

y - 2 x = - 4 x 2 + y = 4 y - 2 x = - 4 x 2 + y = 4
(1)

is:

y = 2 x - 4 in tweede vergelyking x 2 + ( 2 x - 4 ) = 4 x 2 + 2 x - 8 = 0 Faktoriseer : ( x + 4 ) ( x - 2 ) = 0 Die 2 oplossings vir x is : x = - 4 en x = 2 y = 2 x - 4 in tweede vergelyking x 2 + ( 2 x - 4 ) = 4 x 2 + 2 x - 8 = 0 Faktoriseer : ( x + 4 ) ( x - 2 ) = 0 Die 2 oplossings vir x is : x = - 4 en x = 2
(2)

Die ooreenstemmende oplossings vir yy word verkry deur substitusie van die xx-waardes in die eerste vergelyking

y = 2 ( - 4 ) - 4 = - 12 vir x = - 4 en : y = 2 ( 2 ) - 4 = 0 vir x = 2 y = 2 ( - 4 ) - 4 = - 12 vir x = - 4 en : y = 2 ( 2 ) - 4 = 0 vir x = 2
(3)

Soos verwag, is hierdie oplossings identies aan die waardes verkry deur die grafiese oplossing

### Oefening 1: Gelyktydige vergelykings

Los op algebraïes:

y - x 2 + 9 = 0 y + 3 x - 9 = 0 y - x 2 + 9 = 0 y + 3 x - 9 = 0
(4)

#### Oplossing

1. Stap 1. Maak yy die onderwerp van die lineêre vergelyking :
y + 3 x - 9 = 0 y = - 3 x + 9 y + 3 x - 9 = 0 y = - 3 x + 9
(5)
2. Stap 2. Vervang in die kwadratiese vergelyking :
( - 3 x + 9 ) - x 2 + 9 = 0 x 2 + 3 x - 18 = 0 Faktoriseer : ( x + 6 ) ( x - 3 ) = 0 die 2 oplossings vir x is : x = - 6 and x = 3 ( - 3 x + 9 ) - x 2 + 9 = 0 x 2 + 3 x - 18 = 0 Faktoriseer : ( x + 6 ) ( x - 3 ) = 0 die 2 oplossings vir x is : x = - 6 and x = 3
(6)
3. Stap 3. Vervang die waardes vir xx in die eerste vergelyking om ooreenstemmende vergelyking te bereken yy-waardes. :
y = - 3 ( - 6 ) + 9 = 27 vir x = - 6 en : y = - 3 ( 3 ) + 9 = 0 vir x = 3 y = - 3 ( - 6 ) + 9 = 27 vir x = - 6 en : y = - 3 ( 3 ) + 9 = 0 vir x = 3
(7)
4. Stap 4. Skryf die oplossing vir die probleem :

Die eerste waarde is x=-6x=-6 en y=27y=27. Die tweede waarde is x=3x=3 en y=0y=0.

### Algebraïese oplossing

Los op die volgende probleme van algebraïese vergelykings. Waar toepaslik, los jou antwoord in wortelvorm.

 1. a+b=5a+b=5 a - b 2 + 3 b - 5 = 0 a - b 2 + 3 b - 5 = 0 2. a-b+1=0a-b+1=0 a - b 2 + 5 b - 6 = 0 a - b 2 + 5 b - 6 = 0 3. a-(2b+2)4=0a-(2b+2)4=0 a - 2 b 2 + 3 b + 5 = 0 a - 2 b 2 + 3 b + 5 = 0 4. a+2b-4=0a+2b-4=0 a - 2 b 2 - 5 b + 3 = 0 a - 2 b 2 - 5 b + 3 = 0 5. a-2+3b=0a-2+3b=0 a - 9 + b 2 = 0 a - 9 + b 2 = 0 6. a-b-5=0a-b-5=0 a - b 2 = 0 a - b 2 = 0 7. a-b-4=0a-b-4=0 a + 2 b 2 - 12 = 0 a + 2 b 2 - 12 = 0 8. a+b-9=0a+b-9=0 a + b 2 - 18 = 0 a + b 2 - 18 = 0 9. a-3b+5=0a-3b+5=0 a + b 2 - 4 b = 0 a + b 2 - 4 b = 0 10. a+b-5=0a+b-5=0 a - b 2 + 1 = 0 a - b 2 + 1 = 0 11. a-2b-3=0a-2b-3=0 a - 3 b 2 + 4 = 0 a - 3 b 2 + 4 = 0 12. a-2b=0a-2b=0 a - b 2 - 2 b + 3 = 0 a - b 2 - 2 b + 3 = 0 13.a-3b=0a-3b=0 a - b 2 + 4 = 0 a - b 2 + 4 = 0 14. a-2b-10=0a-2b-10=0 a - b 2 - 5 b = 0 a - b 2 - 5 b = 0 15. a-3b-1=0a-3b-1=0 a - 2 b 2 - b + 3 = 0 a - 2 b 2 - b + 3 = 0 16. a-3b+1=0a-3b+1=0 a - b 2 = 0 a - b 2 = 0 17. a+6b-5=0a+6b-5=0 a - b 2 - 8 = 0 a - b 2 - 8 = 0 18.a-2b+1=0a-2b+1=0 a - 2 b 2 - 12 b + 4 = 0 a - 2 b 2 - 12 b + 4 = 0 19. 2a+b-2=02a+b-2=0 8 a + b 2 - 8 = 0 8 a + b 2 - 8 = 0 20. a+4b-19=0a+4b-19=0 8 a + 5 b 2 - 101 = 0 8 a + 5 b 2 - 101 = 0 21. a+4b-18=0a+4b-18=0 2 a + 5 b 2 - 57 = 0 2 a + 5 b 2 - 57 = 0

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