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- Feb 14, 2012

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\(\displaystyle |x|-y-z=5\)

\(\displaystyle x-|y|+z=-9\)

\(\displaystyle x-y+|z|=-1\)

Find the unique solution of the simultaneous equation in which $x, y, z$ are all integers.

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- Thread starter
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- #1

- Feb 14, 2012

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\(\displaystyle |x|-y-z=5\)

\(\displaystyle x-|y|+z=-9\)

\(\displaystyle x-y+|z|=-1\)

Find the unique solution of the simultaneous equation in which $x, y, z$ are all integers.

- Jan 26, 2012

- 183

\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 $(-3,2,-4)$.

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- Mar 5, 2012

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But you have 2 solutions, when surely there should only be one?!My solution

Can you scratch the first one please?

(I selected the same method.)

Last edited:

- Jan 26, 2012

- 183

As I like Serena pointed out, there is a unique solution. Clearly my first solution violates the condition $x \ge 0$ and $z \ge 0$.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 $(-3,2,-4)$.

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- #5

- Feb 14, 2012

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

Hi

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.

\(\displaystyle |x|-y-z=5\)---(1)

\(\displaystyle x-|y|+z=-9\)---(2)

\(\displaystyle x-y+|z|=-1\)---(3)

Adding the equations (1) and (2) gives | Adding the equations (1) and (3) gives | Adding the equations (2) and (3) gives | |

\(\displaystyle |x|+x=|y|+y-4\) | \(\displaystyle |x|-x=|z|+z+6\) | \(\displaystyle |y|-y=z-|z|+8\) | |

Let $x<0$. We then have \(\displaystyle 0=|y|+y-4\) This implies $y=2$. | When $y=2$, we have \(\displaystyle 0=z-|z|+8\) This gives $z=-4$. | When $x<0$, $y=2$, $z=-4$, \(\displaystyle |x|-x=|z|+z+6\) becomes \(\displaystyle -2x=6\) $x=-3$ | |

Now, we have to consider the case when $x>0$. When $x>0$, we have \(\displaystyle 0=|z|+z+6\) And obviously the equation above has no solution and we're done! Yeah! |