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Lineaire gelyktydige vergelykings

Module by: Free High School Science Texts Project. E-mail the author

Gelyktydige Lineêre Vergelykings

Tot dusver het alle vergelykings slegs een onbekende veranderlike gehad wat ons moes vind. Wanneer 2 onbekendes bepaal moet word, het ons 2 vergelykings nodig. Hierdie vergelykings word gelyktydige vergelykings genoem. Die oplossing vir die stelsel van gelyktydige vergelykings is die waardes van die veranderlikes wat die stelsel van vergelykings gelyktydig sal bevredig. In die algemeen beteken dit indien daar nn onbekende veranderlikes is, benodig ons nn vergelykings om 'n oplossing vir elk van die nn veranderlikes te vind.

’n Voorbeeld van stel gelyktydige vergelykings is:

2 x + 2 y = 1 2 - x 3 y + 1 = 2 2 x + 2 y = 1 2 - x 3 y + 1 = 2
(1)

Oplos van Gelyktydige Vergelykings

Om 'n numeriese waarde vir onbekende veranderlikes te vind, moet ons ten minste soveel onafhanklike vergelykings as veranderlikes hê. Ons kan gelyktydige vergelykings algebraïes of grafies oplos.

Figure 1
Khan Akademie video oor gelyktydige vergelykings - 1

Grafiese Oplossing

Gelyktydige vergelykings kan grafies opgelos word. Die oplossing van die gelyktydige vergelykings word gegee deur die koördinate van die punt waar die twee grafieke mekaar sny.

x = 2 y y = 2 x - 3 x = 2 y y = 2 x - 3
(2)

Trek die grafieke van die 2 vergelykings in Equation 2.

Figure 2
Figure 2 (MG10C10_006.png)

Die snypunt van die 2 grafieke is (2,1)(2,1). Dus, die oplossing van die stel gelyktydige vergelykings in Equation 2 is y=1y=1 and x=2x=2.

Dit kan algebraïes getoon word as:

x = 2 y y = 2 ( 2 y ) - 3 y - 4 y = - 3 - 3 y = - 3 y = 1 Stel in die eerste vergelyking in: x = 2 ( 1 ) = 2 x = 2 y y = 2 ( 2 y ) - 3 y - 4 y = - 3 - 3 y = - 3 y = 1 Stel in die eerste vergelyking in: x = 2 ( 1 ) = 2
(3)

Exercise 1: Gelyktydige Vergelykings

Los die volgende stel gelyktydige vergelykings grafies op.

4 y + 3 x = 100 4 y - 19 x = 12 4 y + 3 x = 100 4 y - 19 x = 12
(4)
Solution
  1. Step 1. Teken die grafiek wat ooreenstem met elke vergelyking. :

    Vir die eerste vergelyking:

    4 y + 3 x = 100 4 y = 100 - 3 x y = 25 - 3 4 x 4 y + 3 x = 100 4 y = 100 - 3 x y = 25 - 3 4 x
    (5)

    en vir die tweede vergelyking:

    4 y - 19 x = 12 4 y = 19 x + 12 y = 19 4 x + 3 4 y - 19 x = 12 4 y = 19 x + 12 y = 19 4 x + 3
    (6)

    Figure 3
    Figure 3 (MG10C10_007.png)

  2. Step 2. Vind die snypunt van die grafieke. :

    Die grafieke sny by (4,22)(4,22).

  3. Step 3. Skryf die oplossing neer van die stel gelyktydige vergelykings soos gegee deur die koördinate van die snypunt van die grafieke. :
    x = 4 y = 22 x = 4 y = 22
    (7)

Oplossing deur Vervanging

’n Algemene algebraïese metode is vervanging: probeer een van die vergelykings op te los vir een van die veranderlikes en vervang die resultaat in die ander vergelykings. Deur dit te doen verminder die hoeveelheid vergelykings en ook die hoeveelheid onbekende veranderlikes met 1. Hierdie proses word herhaal tot ‘n enkele vergelyking met 1 veranderlike oorbly, wat (hopelik) opgelos kan word. Terugvervanging kan dan gebruik word om die waardes van die ander veranderlikes te bevestig.

In die voobeeld Equation 1, los ons die eerste vergelyking op vir xx:

x = 1 2 - y x = 1 2 - y
(8)

en stel hierdie resultaat in die tweede vergelyking in:

2 - x 3 y + 1 = 2 2 - ( 1 2 - y ) 3 y + 1 = 2 2 - ( 1 2 - y ) = 2 ( 3 y + 1 ) 2 - 1 2 + y = 6 y + 2 y - 6 y = - 2 + 1 2 + 2 - 5 y = 1 2 y = - 1 10 2 - x 3 y + 1 = 2 2 - ( 1 2 - y ) 3 y + 1 = 2 2 - ( 1 2 - y ) = 2 ( 3 y + 1 ) 2 - 1 2 + y = 6 y + 2 y - 6 y = - 2 + 1 2 + 2 - 5 y = 1 2 y = - 1 10
(9)
x = 1 2 - y = 1 2 - ( - 1 10 ) = 6 10 = 3 5 x = 1 2 - y = 1 2 - ( - 1 10 ) = 6 10 = 3 5
(10)

Die oplossing vir die stelsel gelyktydige vergelykings Equation 1 is:

x = 3 5 y = - 1 10 x = 3 5 y = - 1 10
(11)

Exercise 2: Gelyktydige Vergelykings

Los die volgende stelsel gelyktydige vergelykings op:

4 y + 3 x = 100 4 y - 19 x = 12 4 y + 3 x = 100 4 y - 19 x = 12
(12)
Solution
  1. Step 1. Indien die vraag nie eksplisiet vra vir 'n grafiese oplossing nie, behoort die gelyktydige vergelykings algebraïes opgelos te word. :
  2. Step 2. Maak xx die onderwerp van die eerste vergelyking. :
    4 y + 3 x = 100 3 x = 100 - 4 y x = 100 - 4 y 3 4 y + 3 x = 100 3 x = 100 - 4 y x = 100 - 4 y 3
    (13)
  3. Step 3. Stel die waarde wat bereken is vir xx in die tweede vergelyking in. :
    4 y - 19 ( 100 - 4 y 3 ) = 12 12 y - 19 ( 100 - 4 y ) = 36 12 y - 1900 + 76 y = 36 88 y = 1936 y = 22 4 y - 19 ( 100 - 4 y 3 ) = 12 12 y - 19 ( 100 - 4 y ) = 36 12 y - 1900 + 76 y = 36 88 y = 1936 y = 22
    (14)
  4. Step 4. Vervang xx in die vergelyking. :
    x = 100 - 4 ( 22 ) 3 = 100 - 88 3 = 12 3 = 4 x = 100 - 4 ( 22 ) 3 = 100 - 88 3 = 12 3 = 4
    (15)
  5. Step 5. Stel die waardes van xx and yy in beide vergelykings in om die oplossing te toets. :
    4 ( 22 ) + 3 ( 4 ) = 88 + 12 = 100 4 ( 22 ) - 19 ( 4 ) = 88 - 76 = 12 4 ( 22 ) + 3 ( 4 ) = 88 + 12 = 100 4 ( 22 ) - 19 ( 4 ) = 88 - 76 = 12
    (16)

Exercise 3: Tweewielfietse en Driewiele

’n Winkel verkoop tweewielfietse en driewiele. In totaal is daar 7 fietse (fietse sluit tweewielfietse en driewiele in) en 19 wiele. Bepaal hoeveel van elke soort fiets is daar.

Solution
  1. Step 1. Identifiseer wat is gegee en wat word gevra :

    Die aantal fietse en die aantal driewiele word verlang.

  2. Step 2. Stel die nodige vergelykings op :

    As bb die aantal tweewielfietse en tt die aantal driewiele is, dan:

    b + t = 7 2 b + 3 t = 19 b + t = 7 2 b + 3 t = 19
    (17)
  3. Step 3. Los die stelsel gelyktydige vergelykings op deur vervanging :
    b = 7 - t In die tweede vergelyking: 2 ( 7 - t ) + 3 t = 19 14 - 2 t + 3 t = 19 t = 5 In die eerste vergelyking: : b = 7 - 5 = 2 b = 7 - t In die tweede vergelyking: 2 ( 7 - t ) + 3 t = 19 14 - 2 t + 3 t = 19 t = 5 In die eerste vergelyking: : b = 7 - 5 = 2
    (18)
  4. Step 4. Toets die oplossing deur vervanging in die oorspronklike stel vergelykings. :
    2 + 5 = 7 2 ( 2 ) + 3 ( 5 ) = 4 + 15 = 19 2 + 5 = 7 2 ( 2 ) + 3 ( 5 ) = 4 + 15 = 19
    (19)

Gelyktydige vergelykings

  1. Los grafies op en bevestig jou antwoord algebraïes: 3a-2b7=03a-2b7=0 , a-4b+1=0a-4b+1=0
     
    Kliek hier vir die oplossing
  2. Los algebraïes op: 15c+11d-132=015c+11d-132=0, 2c+3d-59=02c+3d-59=0
     
    Kliek hier vir die oplossing
  3. Los algebraïes op: -18e-18+3f=0-18e-18+3f=0, e-4f+47=0e-4f+47=0
     
    Kliek hier vir die oplossing
  4. Los grafies op: x+2y=7x+2y=7, x+y=0x+y=0
     
    Kliek hier vir die oplossing

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