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# I'm not a troll - Ask me anything! Watch

1. (Original post by hasan4life)
your the salaan to my roti
hehe
2. Why are you so jealous about biryani007?
3. (Original post by GnomeMage)
is this a real question??
Yes. from one of the STEP III papers
(Original post by Mubariz)
Loads of questions cannot be solved

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Solve for x,y and z, this can be in terms of a and b (a and b are constants) so 3 main variables, 3 equations
4. (Original post by plusC)
Yes. from one of the STEP III papers

Solve for x,y and z, this can be in terms of a and b (a and b are constants) so 3 main variables, 3 equations
Oh I thought it meant solve in terms of real numbers.

Being a numpty me

Anyway it's step 3 so I'm gonna back away

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5. (Original post by Mubariz)
Oh I thought it meant solve in terms of real numbers.

Being a numpty me

Anyway it's step 3 so I'm gonna back away

Posted from TSR Mobile
Don't let titles deceive you its pretty simple
Spoiler:
Show

x+ay+z=2 (1)
x+y+az=2 (2)
2x+y+z=2b (3)
(1)=(2) by default so x+ay+z=x+y+az
subtract x and rearrange for a condition connecting y and z
ay+z=y+az
y(a-1)=z(a-1)

Assuming we get the condition y=z so using this on (3) gives 2x+2y=2b x+y=b label this (4)
using this condition on (2) x+y+ay=2 (5)

doing (5)-(4) yields ay=2-b
using this on (4) yields

and thats it (no solutions when a=1 or a=0 though)

6. (Original post by plusC)
Yes. from one of the STEP III papers

Solve for x,y and z, this can be in terms of a and b (a and b are constants) so 3 main variables, 3 equations
bugger...i thought i need to provide a number for it.!
7. Ah so this is Simmy 0_0

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