MHB Unique Solution for 3 Simultaneous Equations with Integers: x, y, z

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The discussion focuses on solving three simultaneous equations involving integers x, y, and z. Participants confirm that there is indeed a unique solution, which is x = -3, y = 2, and z = -4. The method involves analyzing cases based on the signs of x, y, and z, leading to the conclusion that the case where x is positive yields no valid solutions. The conversation highlights the elegance of the solution method and the importance of adhering to the conditions set by the equations. The unique solution is established through careful consideration of the equations' constraints.
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Consider the 3 simultaneous equations

$$|x|-y-z=5$$

$$x-|y|+z=-9$$

$$x-y+|z|=-1$$

Find the unique solution of the simultaneous equation in which $x, y, z$ are all integers.
 
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My solution
Maybe not the most elegant but effective. Due to the presense of the absolutes we can consider 8 cases depending on whether $x, y$ and $z \ge 0$ or $< 0$. However we can rule out 6 of these immediately due to inconsistencies in the system. For example, if $x,y,z \ge 0$ the the system is

\begin{eqnarray}
x - y - 5 &=& 5\\
x-y+z &=& -9\\
x - y + z &=& -1
\end{eqnarray}

and clearly the last 2 equations are inconsistent. The only cases I found that were consistent where

\begin{eqnarray}
(i) & \;\;&x \ge 0,\;\;& y < 0,\;\; &z \ge 0\\
(ii)& \;\;&x <0.\;\; &y \ge 0,\;\;& z < 0
\end{eqnarray}

In each case, the systems were easily solved giving solutions $(-2.-4,-3)$ and $(-3,2,-4)$.
 
Jester said:
My solution

But you have 2 solutions, when surely there should only be one?! (Nerd)
Can you scratch the first one please?
(I selected the same method.)
 
Last edited:
Jester said:
My solution
Maybe not the most elegant but effective. Due to the presense of the absolutes we can consider 8 cases depending on whether $x, y$ and $z \ge 0$ or $< 0$. However we can rule out 6 of these immediately due to inconsistencies in the system. For example, if $x,y,z \ge 0$ the the system is

\begin{eqnarray}
x - y - z &=& 5\\
x-y+z &=& -9\\
x - y + z &=& -1
\end{eqnarray}

and clearly the last 2 equations are inconsistent. The only cases I found that were consistent where

\begin{eqnarray}
(i) & \;\;&x \ge 0,\;\;& y < 0,\;\; &z \ge 0\\
(ii)& \;\;&x <0.\;\; &y \ge 0,\;\;& z < 0
\end{eqnarray}

In each case, the systems were easily solved giving solutions $(-2.-4,-3)$ and $(-3,2,-4)$.
As I like Serena pointed out, there is a unique solution. Clearly my first solution violates the condition $x \ge 0$ and $z \ge 0$.
 
Hi Jester,

Thanks for participating and I think your method is quite elegant and I used the same concept to solve it too.

Hi I like Serena,

Thanks for everything!

But I have thought of another method to tackle the problem and the idea only hit me moment ago.

We're given to solve for the unique solution of the simultaneous equation below in which $x, y, z$ are all integers.

$$|x|-y-z=5$$---(1)

$$x-|y|+z=-9$$---(2)

$$x-y+|z|=-1$$---(3)

[TABLE="class: grid, width: 500"]
[TR]
[TD]Adding the equations (1) and (2) gives[/TD]
[TD]Adding the equations (1) and (3) gives
[/TD]
[TD]Adding the equations (2) and (3) gives
[/TD]
[TD][/TD]
[/TR]
[TR]
[TD]$$|x|+x=|y|+y-4$$[/TD]
[TD]$$|x|-x=|z|+z+6$$[/TD]
[TD]$$|y|-y=z-|z|+8$$[/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Let $x<0$.

We then have

$$0=|y|+y-4$$

This implies $y=2$.[/TD]
[TD][/TD]
[TD]When $y=2$, we have

$$0=z-|z|+8$$

This gives $z=-4$.[/TD]
[TD]When $x<0$, $y=2$, $z=-4$,

$$|x|-x=|z|+z+6$$ becomes

$$-2x=6$$

$x=-3$[/TD]
[/TR]
[TR]
[TD][/TD]
[TD]Now, we have to consider the case when $x>0$.

When $x>0$, we have

$$0=|z|+z+6$$

And obviously the equation above has no solution and we're done! Yeah!(Sun)
[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[/TABLE]
 
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