Find Value of $x-y+z$ in $x,y,z \in N$

[Albert1]

$x,y,z \in N\\
\left\{\begin{matrix}
3x-4y=0---(1)\\
x+y+z=\sqrt {x+y+z-3}+15---(2)
\end{matrix}\right.$
find the value of $x-y+z=?$

 

[kaliprasad]

$x,y,z \in N\\
\left\{\begin{matrix}
3x-4y=0---(1)\\
x+y+z=\sqrt {x+y+z-3}+15---(2)
\end{matrix}\right.$
find the value of $x-y+z=?$

Spoiler

from 2nd relation we get $x+y+z=19$

from (1) and above

$ x = 4, y = 3, z = 12 $ giving $x-y+z = 13$

or
$x = 8, y = 6, z = 5$ giving $x-y+z = 7$

 

[HallsofIvy]

Spoiler

from 2nd relation we get $x+y+z=19$

Spoiler

That's clever. The 2nd relation is $x+ y+ z= \sqrt{x+ y+ z- 3}+ 15$. Subtracting 3 from both sides, $x+ y+ z- 3= \sqrt{x+ y+ z- 3}+ 13$. Letting u= x+ y+ z- 3, We can write that as $u= \sqrt{u}+ 12$. And that is the same as $\sqrt{u}= u- 12$. Squaring both sides, $u= (u- 12)^2= u^2- 24u+ 144$ s that $u^2- 25u+ 144= (u- 16)(u- 9)= 0$. The roots of that are u= 16 and u= 9. If u= 16 then x+ y+ z= u+ 3= 19. If u= 9 the x+ y+ z= u+ 3= 12.

Having squared both sides we need to check for spurious solutions that might have been introduced. If x+ y+ z= 19, then the equation becomes $19= \sqrt{19- 3}+15= 4+ 15$ which is correct. If x+ y+ z= 12, then the equation becomes $12= \sqrt{12- 3}+ 15= 3+ 15$ which is not correct (we got this "spurious solution" because -3+ 15 is equal to 12.)

I'm impressed!

from (1) and above

$ x = 4, y = 3, z = 12 $ giving $x-y+z = 13$

or
$x = 8, y = 6, z = 5$ giving $x-y+z = 7$