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    This module is approved and included inLens: Siyavula: Wiskunde (Gr 10 - 12)
    By: Siyavula

    Review Status: Approved

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Inleiding

In Graad 10 het jy geleer hoe om stelle van gelyktydige vergelykings op te los. Hierdie vergelykings was tot dusver beide lineêre (dws die hoogste krag is gelyk aan 1). In hierdie hoofstuk sal jy leer hoe om stelle van gelyktydige vergelykings op te los waar een lineêr is en een kwadratiese is. Soos in graad 10, sal die oplossing algebraïes en grafies opgelos word.

Die enigste verskil tussen 'n stelsel van lineêre gelyktydige vergelykings en​​ 'n ​​stelsel van gelyktydige vergelykings met 'n lineêre en 'n kwadratiese vergelyking, is dat die tweede stelsel sal op die meeste twee oplossings hê.

'n Voorbeeld van 'n stelsel van gelyktydige vergelykings met een lineêre vergelyking en een kwadratiese vergelyking is:

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

Grafiese Oplossing

Die metode om die oplossing vir 'n lineêre en 'n kwadratiese vergelyking grafies te vind is identies aan 'n stelsels van lineêre gelyktydige vergelykings.

Metode: Grafiese oplossing vir 'n stelsel van gelyktydige vergelykings met 'n lineêre en 'n kwadratiese vergelyking

  1. Maak yy die onderwerp van elke vergelyking.
  2. Teken die grafieke van elkeen van die vergelykings hierbo.
  3. Die oplossing van die stel van gelyktydige vergelykings word gegee deur die kruising van die twee grafieke.

Deur , yy die onderwerp van elke vergelyking te maak, kry jy:

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

Deur die grafiek van elke vergelyking te plot, kry jy 'n reguit lyn vir die eerste vergelyking en 'n parabool vir die tweede vergelyking.

Figuur 1
Figuur 1 (MG11C9_001.png)

Die parabool en die reguit lyn sny op twee punte: (2,0) en (-4,-12). Dus, die oplossing van die stel van vergelykings in vergelyking 1 is x=2,y=0x=2,y=0 en x=-4,y=12x=-4,y=12

Exercise 1: Gelyktydige Oplossings

Los grafies op:

y - x 2 + 9 = 0 y + 3 x - 9 = 0 y - x 2 + 9 = 0 y + 3 x - 9 = 0
(3)
Solution
  1. Stap 1. Maak yy die onderwerp van die vergelyking :

    Vir die eerste vergelyking:

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

    en vir die tweede vergelyking:

    y + 3 x - 9 = 0 y = - 3 x + 9 y + 3 x - 9 = 0 y = - 3 x + 9
    (5)
  2. Stap 2. Teken die grafieke vir elke vergelyking:

    Figuur 2
    Figuur 2 (MG11C9_002.png)

  3. Stap 3. Vind die snypunte van die vergelykings op die grafiek:

    Die grafieke sny by (-6,27)(-6,27) en by (3,0)(3,0).

  4. Stap 4. Skryf die oplossing van die stelsel van gelyktydige vergelykings soos gegee deur grafieke:

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

Grafiese Oplossing

Los die volgende stelsels van vergelykings grafies op. Indien van pas, laat jou antwoord in wortel vorm.

  1. b 2 - 1 - a = 0 , a + b - 5 = 0 b 2 - 1 - a = 0 , a + b - 5 = 0
  2. x + y - 10 = 0 , x 2 - 2 - y = 0 x + y - 10 = 0 , x 2 - 2 - y = 0
  3. 6 - 4 x - y = 0 , 12 - 2 x 2 - y = 0 6 - 4 x - y = 0 , 12 - 2 x 2 - y = 0
  4. x + 2 y - 14 = 0 , x 2 + 2 - y = 0 x + 2 y - 14 = 0 , x 2 + 2 - y = 0
  5. 2 x + 1 - y = 0 , 25 - 3 x - x 2 - y = 0 2 x + 1 - y = 0 , 25 - 3 x - x 2 - y = 0

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