Qualification > Math
Pure 3 help.
Deadly_king:
I am back now!
So you need to use the two equations. By elimination method, you eliminate one of the variables.
Eqn 1 : 2x - y - 3z = 7
Eqn 2 : x + 2y + 2z = 0
Step 1
Multiply Eqn by 2 and you'll be getting 4x - 2y -6z =14 while Eqn 2 is still x +2y +2z = 0.
Add both equations and you'll note that -2y cancels 2y. You'll thus be getting 5x - 4z = 14
Make z subject of formula to get z = (5x - 14)/4
Now you need to find z in terms of y. So you should eliminate the values of x.
Step 2
Multiply Eqn 2 by -2 and you'll be getting -2x - 4y - 4z = 0 while Eqn 1 remains 2x - y -3z = 7
Add both equations and this time 2x will cancel -2x. Hence you'll obtain -5y - 7z = 7
Again make z subject of formula to get z = (-5y - 7)/7
Now you equate all the equations of z you obtained just like Requiem previously did.
Then you should make the coefficients of both x and y = 1
I guess by now, you'll be able to understand the workings of Requiem :)
Deadly_king:
Nov 09 No 2, 6, 10
2.
3(x+2) = 3x + 32
Divide everything by 3x and you'll get : ( dividing implies powers are subtracted)
32 -1 = 3(2-x)
Therefore 3(2-x) = 9 - 1
Apply ln on both sides
ln 3(2-x) = ln 8
(2-x)ln 3 = ln 8
Hence x = 2 - (ln 8/ln3) = 0.107
Sorry but I need to go to tuition now. Will complete the other numbers once am back :)
S.M.A.T:
Nov 09 No 2, 6
S.M.A.T:
DK!! ...that was really helpful ;D
+rep
mousa:
God bless you all. 8) thanks for all reponses!!
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