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4. Los die vergelyking op.

5. Beantwoord dan die vraag.

Kom ons kyk na ’n paar voorbeelde:

Voorbeeld 1:

Probleem : Die lengte van ’n reghoek is 5 cm meer as sy breedte.


1) Teken ’n reghoek

2) Vra jouself: “Van watter een, lengte of breedte, weet jy niks?”in die geval: breedte = x cm

3) Druk nou die lengte uit in terme van x : in die geval: ( x + 5)(onthou 5 meer : d.w.s. x + 5)

4) Nou kan die vraag ook sê as die omtrek 80 cm is, bereken die lengte en breedte van die reghoek.(Nou kan jy die gegewens as ’n vergelyking voorstel)

Doen dit nou:

5) Los nou die vergelyking op en beantwoord die vraag in (4).

Voorbeeld 2:

Probleem : ’n Ma is vier keer so oud soos haar dogter. Hulle gesamentlike ouderdomme is 60. Hoe oud is hulle elk?


1) Vra jouself: “Van watter een weet jy niks? ” en laat dit x wees.

2) Maak ’n voorstelling van jou denke tot dusver:

Ma Dogter

4 x x

3) Stel die gegewens nou as ’n vergelyking voor, los die vergelyking op en beantwoord die vraag.

Voorbeeld 3:

Die probleem met ouderdomme is altyd ’n toffie, maar doen dit stap vir stap, skryf die plan neer van hoe jy dink en dit is baie maklik.....

Probleem: Milandre is 30 jaar ouer as Filandre. Oor 15 jaar sal Milandre twee keer so oud soos Filandre wees. Hoe oud is elk nou?


1) Begin met ’n plan en skryf dit neer:

2) Vra jouself: “Van watter een weet ek niks?” en maak dit x .

ouderdom: nou

ouderdom: oor 15 jaar

(d.w.s + 15)

Milandre x + 30; ( x + 30) + 15

Filandre x; x + 15

3) Nou kom die moeilikste deel: oor 15 jaar sal Milandre twee keer so oud soos Filandre wees. Milandre is reeds dan twee keer so oud soos Filandre en jy sal Filandre se ouderdom (in die kolom oor 15 jaar) met 2 moet vermenigvuldig sodat jy ’n vergelyking kan kry (d.w.s linkerkant = regterkant)

4) Skryf jy nou die vergelyking hier neer, los dit op en beantwoord die vraag.

Kyk op jy nou die volgende op jou eie kan doen. Onthou jy moet ’n “plan” soos jy dink telkens neerskryf.

Hier kom hulle .....

1. Die som van twee getalle is 15. Skryf die twee getalle in terme van x , neer.

2. 140 mense woon die Steve Hofmeyer konsert by.

Die volgende kaartjies is beskikbaar:

Kinders: R 20

Volwassenes: R 45

As die toegangsgeld R 5 580 beloop het, bereken hoeveel volwassenes en hoeveel kinders teenwoordig was.


1. Gegee: a ∈ {-4; -3; -2; -1; 1; 2; 3}Kies elke keer jou antwoord uit bogenoemde versameling.

1.1 -3 a + 20 = 23

1.2 8 a = -32

2. Die waarde van x word in elke geval gegee. Toets die korrektheid van elke vergelyking.

2.1 8 x - 2( x - 5) = 28 x = 3

2.2 5 x - 10 = 10 x - 10; x R

3. Los elk van die volgende vergelykings op. Wys al jou berekeninge.

3.1 1 / z = 1 / 18 ; z = ?

3.2 1 - 5 z = 11

3.2.1 waar z N

3.2.2 waar z ∈ Q

3.3 z + 3[ z + 2( z - 6)] = 45

3.4 4(6 z - 8) - 2( z + 7) = 37

3.5 z - 5( z - 8) = -48

4. Skryf elk van die volgende as algebraïese vergelykings en los dit dan op.

4.1 Ses keer ’n getal, verminder met 8 is gelyk aan 55. Bereken die getal.

4.2 ’n Negatiewe getal is nege keer die ander getal. Die som van die twee getalle is -64. Bepaal die twee getalle.

4.3 Die som van drie opeenvolgende negatiewe heelgetalle is -90.Bepaal die drie negatiewe getalle.

Questions & Answers

how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
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I'm not sure why it wrote it the other way
I got X =-6
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Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Differences Between Laspeyres and Paasche Indices
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Wiskunde graad 8. OpenStax CNX. Sep 11, 2009 Download for free at http://cnx.org/content/col11033/1.1
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