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Strategie vir die oplos van vergelykings

Hierdie hoofstuk handel oor die oplos van verskillende soorte vergelykings met twee veranderlikes. Die algemene strategie is om die onbekende veranderlike alleen aan die linkerkant van die gelykaanteken te kry, met al die konstantes aan die regterkant van die gelykaanteken. Byvoorbeeld, in die vergelyking x - 1 = 0 , wil ons die vergelyking skryf as x = 1 .

Soos ons gesien het in hersiening van vorige werk, in die afdeling wat handel oor herrangskikking van vergelykings, is 'n vergelyking soos 'n weegskaal wat altyd gebalanseerd moet bly. Wanneer ons vergelykings oplos, moet ons in gedagte hou dat wat aan die een kant gedoen word, ook aan die ander kant gedoen moet word.

Metode: herrangskikking van vergelykings

Jy kan 'n getal optel, 'n getal aftrek, met 'n getal vermenigvuldig of met 'n getal deel, solank jy dieselfde bewerking aan beide kante doen.

Byvoorbeeld, in die vergelyking x + 5 - 1 = - 6 , wil ons x alleen aan die linkerkant van die vergelyking kry. Dit beteken ons moet 5 aftrek en 1 optel aan die linkerkant. Omdat ons die skaal gebalanseerd moet hou, moet ons ook 5 aftrek en 1 optel aan die regterkant.

x + 5 - 1 = - 6 x + 5 - 5 - 1 + 1 = - 6 - 5 + 1 x + 0 + 0 = - 11 + 1 x = - 10

In 'n ander voorbeeld, 2 3 x = 8 , moet ons deel met 2 en vermenigvuldig met 3 aan die linkerkant om x alleen te kry. Om die vergelyking gebalanseerd te hou moet ons ook aan die regterkant deel met 2 en vermenigvuldig met 3.

2 3 x = 8 2 3 x ÷ 2 × 3 = 8 ÷ 2 × 3 2 2 × 3 3 × x = 8 × 3 2 1 × 1 × x = 12 x = 12

Hierdie is die basiese reëls wat gevolg moet word wanneer 'n vergelyking vereenvoudig word. In die meeste gevalle moet die reëls herhaaldelik toegepas word voor ons die gewenste veranderlike as onderwerp het.

Die volgende moet ook in gedagte gehou word:
  1. Deling deur 0 is ontoelaatbaar.
  2. As x y = 0 , dan x = 0 en y 0 , want deling deur 0 gee 'n ongedefinieerde uitdrukking.

Ons is nou gereed om 'n paar vergelykings op te los!

Ondersoek: strategie vir die oplos van vergelykings

Identifiseer wat is verkeerd in die volgende bewerking:

4 x - 8 = 3 ( x - 2 ) 4 ( x - 2 ) = 3 ( x - 2 ) 4 ( x - 2 ) ( x - 2 ) = 3 ( x - 2 ) ( x - 2 ) 4 = 3

Oplos van lineêre vergelykings

Die eenvoudigste vergelyking om op te los is 'n lineêre vergelyking. ‘n Vergelyking word lineêr genoem indien die hoogste mag van die veranderlike (letter, byvoorbeeld x ) 1 (een) is. Die volgende is voorbeelde van lineêre vergelykings:

2 x + 2 = 1 2 - x 3 x + 1 = 2 4 3 x - 6 = 7 x + 2

In hierdie afdeling sal ons leer om te bepaal wat die waarde van 'n veranderlike moet wees om 'n vergelyking waar te maak. Byvoorbeeld, watter waarde van x maak beide kante van die baie eenvoudige vergelyking, x + 1 = 1 waar.

Aangesien die definisie van 'n lineêre vergelyking is dat die hoogste mag van die veranderlike een (1) moet wees, is daar hoogstens een oplossing of wortel vir die vergelyking.

Hierdie afdeling berus op al die metodes wat ons reeds bespreek het: uitvermenigvuldiging van uitdrukkings met hakies, groepering van terme en faktorisering. Maak seker dat jy goed vertroud is met hierdie metodes voordat jy die werk in die res van hierdie hoofstuk aanpak.

2 x + 2 = 1 2 x = 1 - 2 ( groepeer soortgelyke terme saam ) 2 x = - 1 ( so ver as moontlik vereenvoudig )

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Source:  OpenStax, Siyavula textbooks: wiskunde (graad 10) [caps]. OpenStax CNX. Aug 04, 2011 Download for free at http://cnx.org/content/col11328/1.4
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