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This is a three-unknown equation. I have to get three lines there.

Let a = a three digit number, e.g 246, 371 //just for further thinking
Let x = a's ones place
Let y = a's tenths place
Let z = a's hundreds place

I've got these two lines already, just cannot get the third.

x + y + z = 11

3x = y

Here i have to put this into maths: If we flip the number like: 246 -> 642 or 371 -> 173 the new number is bigger than the old number by 297.

EDIT Added more information (because I messed something up there):
All the three digits of the number added must equal 11;
The ones place is 3 times bigger than the tens place;
Flipped number is 297 greater than the original number;

Can you help me figure this out?
Would really appreciate you help!!

EDIT2:The final equation system:
$$x+y+z=11$$ $$100x+10y+z−(100z+10y+x)=99(x−z)=297⟹x−z=3$$ $$x=3y$$ Thanks to Math Lover!

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Observe that the number is $100z+10y+x$ and the flipped number is $100x+10y+z$. Consequently, $$100x+10y+z - (100z+10y+x) = 99(x-z)=297 \implies x-z = 3.$$

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  • $\begingroup$ I think you've reversed the roles of $x$ and $z$. In the OP, $z$ is the hundreds digit of the original number. $\endgroup$ – G Tony Jacobs Nov 6 '17 at 19:03
  • $\begingroup$ @GTonyJacobs Thanks. Corrected! $\endgroup$ – Math Lover Nov 6 '17 at 19:05
  • $\begingroup$ I've missed a mistake I made in the second statement there. $\endgroup$ – Kerdo Nov 6 '17 at 19:06
  • $\begingroup$ It should be that the ones place is 3 times bigger than the tenths place. So is it 3x = y? $\endgroup$ – Kerdo Nov 6 '17 at 19:07
  • $\begingroup$ Because right now I get the answers (x,y,z)= (14/5;42/5;-1/5). This cannot translate to a 3 digit number :/ $\endgroup$ – Kerdo Nov 6 '17 at 19:08

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