MODULE 1 : GELYKTYDIGE LINEÊRE VERGELYKINGS ... - AdMaths
MODULE 1 : GELYKTYDIGE LINEÊRE VERGELYKINGS ... - AdMaths
MODULE 1 : GELYKTYDIGE LINEÊRE VERGELYKINGS ... - AdMaths
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SUBSTITUSIE-METODE BEGIN WANKEL<br />
VOORBEELD 3<br />
Los op vir x en y: 4 x − 3y<br />
= 6 (1)<br />
3 x + 2y<br />
= 13 (2)<br />
(2): 2 y = − 3x<br />
+ 13<br />
y =<br />
− 3x<br />
+<br />
2<br />
13<br />
(3)<br />
Substitueer nou hierdie ongemaklike “voorlopige antwoord” in (1):<br />
Dus: x = 3 en y = 2<br />
⎛ − 3x<br />
+ 13 ⎞<br />
(1): 4 x − 3⎜<br />
⎟ = 6<br />
⎝ 2 ⎠<br />
3(<br />
−3x<br />
+ 13)<br />
4 x −<br />
=<br />
2<br />
× 2 : 8x<br />
− 3(<br />
−3x<br />
+ 13)<br />
= 12<br />
8 x + 9x<br />
− 39 = 12<br />
17 x = 51<br />
x = 3<br />
− 3(<br />
3)<br />
+<br />
2<br />
(3): y =<br />
= 2<br />
Hierdie som is onaanvaarbaar lank! Is daar ‘n korter metode wat breuke vermy?<br />
Wag en sien...<br />
A1.2<br />
13<br />
6
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