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9. (a) х = 0 Dit is ook ‘n aanvaarbare antwoord.

(b) 2 х + 6 = 2 х + 6

2 х – 2 х = 6 – 6

0 = 0

  • Hierdie oplossing gee nie ‘n enkele waarde vir х nie.
  • Maar die stelling is waar: Nul is gelyk aan nul.
  • As ons ‘n antwoord kry wat ooglopend waar is, soos 12 = 12 of –3 = –3, ens., dan weet ons dat enige waarde van die veranderlike die vergelyking waar sal maak.
  • Ons gee dus die antwoord: х kan enige waarde aanneem .

(c) 3 – 2 х = –2 – 2 х

–2 х + 2 х = –2 – 3

0 = –5

  • Hierdie antwoord gee nie ‘n waarde vir х nie.
  • Inderwaarheid is die stelling onwaar. Nul is nie gelyk aan negatief vyf nie.
  • As ons ‘n onwaar antwoord kry, soos 5 = –5 of 2 = –9, ens., dan weet ons dat geen waarde van die veranderlike die vergelyking waar sal maak nie.
  • Dus is die antwoord: Daar is geen oplossing nie .

Van nou af moet jy jou oë oophou vir hierdie spesiale gevalle (jy sal hulle nie veel sien nie) en ‘n geskikte antwoord gee

Aktiwiteit 3

Om te bevestig dat oplossings korrek is

[lu 2.4, 2.6]

  • In Wiskunde is dit dikwels moeilik om seker te wees dat jou antwoord korrek is, maar wanneer ons vergelykings oplos, is dit baie maklik: ons kontroleer net ons antwoorde! Dit moet egter baie noukeurig op ‘n spesifieke manier gedoen word.

Dis hoe: Ons kyk weer na vraag 8 hierbo.

(a) 5( х + 1) = 20 gee die oplossing: х = 3

Begin met die oorspronklike vergelyking.

Kontroleer die linkerkant (LK) en regterkant (RK) apart .

Substitueer die oplossing vir х en vereenvoudig:

LK = 5( х + 1) = 5[( 3 ) + 1] = 5(3 + 1) = 5(4) = 20

Soos gewoonlik by substitusie is hakies baie handig.

RK = 20

Omdat die RK en die LK gelyk is, weet ons die oplossing is korrek.

(b) 8 + 4( х – 1) = 0 Veronderstel ons antwoord was х = 2. Toets die antwoord:

LK = 8 + 4( х – 1) = 8 + 4[( 2 ) – 1] = 8 + 4(2 – 1) = 8 + 4(1) = 8 + 4 = 12

RK = 0

Omdat LK ≠ RK weet ons dat 2 nie ‘n oplossing vir die vergelyking is nie.

Natuurlik is die regte antwoord: х = –1. Gaan dit na:

LK = 8 + 4( х – 1) = 8 + 4[( –1 ) – 1] = 8 + 4(–1 – 1) = 8 + 4(–2) = 8 – 8 = 0

LK = RK, en ons het bevestig dat х = –1 die korrekte oplossing is.

(c) х ( х + 3) = х 2 + 6 oplossing: х = 2

LK = х ( х + 3) = ( 2 )(( 2 ) + 3) = 2(2 + 3) = 2(5) = 10

RK = х 2 + 6 = ( 2 ) 2 + 6 = 4 + 6 = 10

LK = RK, dus is х = 2 die korrekte oplossing.

(d) ½ (4 х + 6) = 1 oplossing: х = –1

LK = ½ (4 х + 6) = ½ (4( –1 ) + 6) = ½ (–4 + 6) = ½ (2) = 1

RK = 1

LK = RK, dus is х = –1 die korrekte oplossing.

Gaan nou terug na 5, 6 en 7 en kontroleer jou oplossing op dieselfde manier.

As ons terug gaan na die spesiale gevalle in 9, kan ons hulle ook kontroleer:

(a) 2( х + 1) = х + 2 gee die oplossing: х = 0

LK = 2( х + 1) = 2(( 0 ) + 1) = 2(0 + 1) = 2(1) = 2

RK = х + 2 = ( 0 ) + 2 = 2

LK = RK, dus is х = 0 die korrekte oplossing.

(b) 2( х + 3) = 2 х + 6 Enige getal is ‘n oplossing! Toets bv. 5; of enige ander getal.

LK = 2( х + 3) = 2(( 5 ) + 3) = 2(5 + 3) = 2(8) = 16

RK = 2 х + 6 = 2( 5 ) + 6 = 10 + 6 = 16

LK = RK as х = 5. Inderdaad, LK sal gelyk wees aan RK vir enige waarde.

(c) 3 – 2 х = –2(1 + х ) Daar is geen oplossing nie; probeer 12. Jy kan ander getalle probeer.

LK = 3 – 2 х = 3 – 2( 12 ) = 3 – 24 = – 21

RK = –2(1 + х ) = –2(1 + ( 12 )) = –2(1 + 12) = –2(13) = –26

LK ≠ RK en hulle sal ongelyk wees vir enige getal.

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Source:  OpenStax, Wiskunde graad 9. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col11055/1.1
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