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

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Discussion Overview

The discussion centers around solving a set of three simultaneous equations involving absolute values, specifically seeking integer solutions for the variables x, y, and z. The equations are presented in a mathematical format, and participants explore various methods to find a unique solution.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the equations and requests a unique integer solution.
  • Multiple participants share their proposed solutions, with some claiming to find two solutions, which raises questions about the uniqueness of the solution.
  • Another participant acknowledges a previous solution that violates certain conditions (specifically, x and z being non-negative).
  • A participant introduces a new method of solving the equations by adding them in pairs and analyzing cases based on the sign of x.
  • Through their method, one participant concludes that when assuming x<0, they derive specific values for y and z, while also noting that the case for x>0 leads to no solutions.

Areas of Agreement / Disagreement

Participants express differing views on the uniqueness of the solution, with some asserting that there is a unique solution while others suggest the existence of multiple solutions. The discussion remains unresolved regarding the overall validity of the proposed solutions.

Contextual Notes

Participants note conditions such as the requirement for x, y, and z to be integers, and the implications of the absolute value operations on the equations. There are also references to specific cases based on the signs of the variables, which may affect the outcomes.

anemone
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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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