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Solving Absolute Value Equations

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anemone

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Feb 14, 2012
3,689
Solve the system below:

$\displaystyle |x+y|+|1-x|=6$

$\displaystyle |x+y+1|+|1-y|=4$

I've solved this problem and my intention is purely to gain another insights on how others would approach it and I surely hope you find this problem as an interesting one!:)
 

ZaidAlyafey

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MHB Math Helper
Jan 17, 2013
1,667
x=-2 , y=-1 .
 

topsquark

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MHB Math Helper
Aug 30, 2012
1,123
Solve the system below:

$\displaystyle |x+y|+|1-x|=6$

$\displaystyle |x+y+1|+|1-y|=4$

I've solved this problem and my intention is purely to gain another insights on how others would approach it and I surely hope you find this problem as an interesting one!:)
That system is a very pretty problem. Could you show us your solution so we can see how you did it and see what other ways we can come up with?

Without having actually solved the problem it looks like you could solve the top equation for y and stick it in the second equation and solve for x. But it's likely to be a bit of a mess.

-Dan
 
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anemone

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Feb 14, 2012
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x=-2 , y=-1 .
Yes, that's correct:)cool:) but Zaid, you missed out another pair of answer where $\displaystyle x=-\frac{14}{3}$, $\displaystyle y=\frac{13}{3}$.

By the way, do you mind to share with me of your solution (but not the final answers)?

I will post my solution in full today too!:)
 

ZaidAlyafey

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MHB Math Helper
Jan 17, 2013
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Yes, that's correct:)cool:) but Zaid, you missed out another pair of answer where $\displaystyle x=-\frac{14}{3}$, $\displaystyle y=\frac{13}{3}$.

By the way, do you mind to share with me of your solution (but not the final answers)?

I will post my solution in full today too!:)
I expected to miss other solutions because I solved it by observation , no real method ....
 
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anemone

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Feb 14, 2012
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My solution:

$\displaystyle |x+y|+|1-x|=6$

$\displaystyle |x+y+1|+|1-y|=4$

My method is to consider the cases
A. $x+y\geq0$ or
B. $x+y<0$
for the following sub cases (I know it sounds extremely tedious but it is far from it if you read on...):
i. $1-x\geq0$ for the sub cases $y<1$ or $y\geq1$
ii. $1-x<0$ for the sub cases $y<1$ or $y\geq1$

Case A (i) +$y<1$ :
$x+y\geq0$, $1-x\geq0$ and $y<1$

From $ |x+y|+|1-x|=6$:

$\displaystyle x+y+1-x=6$ which immediately implies $y=5$.

But bear in mind that in this case, $y<1$, thus, this is not a valid case.

Case A (i)+$y\geq1$:
$x+y\geq0$, $1-x\geq0$ and $y\geq1$

From $ |x+y|+|1-x|=6$:

$\displaystyle x+y+1-x=6$, which immediately implies $y=5$.

But since $y<1$, this is not a valid case either and notice that the setting of this case does not give us another solution that is different from case 1.

Case A (ii)
+$y<1$ :
$x+y\geq0$, $1-x<0$ and $y<1$


From $ |x+y|+|1-x|=6$:

$ x+y-(1-x)=6$

$ 2x+y=7$


From $|x+y+1|+|1-y|=4$:

$x+y+1+1-y=4$

$x=2$ and $ y=7-2x=7-2(2)=3$

But $y<1$, thus, this isn't a valid case either.

Case A (ii)+$y\geq1$:
$x+y\geq0$, $1-x<0$ and $y\geq1$

From $ |x+y|+|1-x|=6$, we get $ 2x+y=7$, and


From $|x+y+1|+|1-y|=4$:

$x+y+1-(1-y)=4$

$x+2y=4$

$ 2x+y=7$

Solving these for y, we obtain $y=\frac{1}{3}$.

Since $y\geq1$, we need to eliminate this case too.

So, there are no solutions if $x+y\geq0$.

Now, let's start with the case B...(I'll do it in a separate post for easy reading.)
 
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anemone

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Feb 14, 2012
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Case B (i) +$y<1$ :
$x+y<0$, $1-x\geq0$ and $y<1$

From $ |x+y|+|1-x|=6$:

$-(x+y)+1-x=6$

$2x+y=-5$


From $|x+y+1|+|1-y|=4$:

$|x-5-2x+1|+1-y=4$

$|-x-4|=y+3$ or $|x+4|=y+3$

$x+4=-(y+3)$ or $x+4=y+3$

$x+y=-7$ or $x-y=-1$

Solving $x+y=-7$ and $2x+y=-5$ simultaneously, we get $x=2$ which contradicts our assumption that $1-x\geq0$.

Solving $x-y=-1$ and $2x+y=-5$ simultaneously, we get $x=-2$, $y=-1$ and this satisfies all of the constraints and thus, this is a pair of valid solutions.

Case B (i)+$y\geq1$:
$x+y<0$, $1-x\geq0$ and $y\geq1$

From $ |x+y|+|1-x|=6$, we get $2x+y=-5$.


From $|x+y+1|+|1-y|=4$:

$|x+y+1|-(1-y)=4$

$|x-5-2x+1|=5-y$

$|-x-4|=5-y$

$|x+4|=5-y$

$x+4=-(5-y)$ or $x+4=5-y$

$x-y=-9$ or $x+y=1$

Solving $x-y=-9$ and $2x+y=-5$ simultaneously, we get $\displaystyle x=-\frac{14}{3}$, $\displaystyle y=\frac{13}{3}$ and this satisfies all of the constraints and thus, this is another pair of valid solutions.

Since $x+y=1$ contradicts the assumption that $x+y<0$, we'll just ignore this case.

Case B (ii)
+$y<1$ :$x+y<0$, $1-x<0$ and $y<1$

From $ |x+y|+|1-x|=6$:

$ -(x+y)-(1-x)=6$, this implies $y=-7$.


From $|x+y+1|+|1-y|=4$:

$|x-7+1|+|1-(-7)|=4$

$|x-6|=-4$

and this isn't right because the absolute value can't be negative.

Case B (ii)+$y\geq1$:
$x+y<0$, $1-x<0$ and $y\geq1$

We get $y=-7$ from $ |x+y|+|1-x|=6$. But we have our assumption where $y\geq1$, thus, this isn't a valid case either.

Thus, we can conclude that the only solutions to the original system are $x=-\frac{14}{3}$, $y=\frac{13}{3}$ and $x=-2$, $y=-1$.
 

ZaidAlyafey

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MHB Math Helper
Jan 17, 2013
1,667
As expected you went into all cases I wonder if there is a shortcut .
 

ZaidAlyafey

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Jan 17, 2013
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I would actually prefer to use that |x|=a then $x=\pm a$ so I don't miss one of the cases .
 

MarkFL

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Feb 24, 2012
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anemone,

Thank you for taking the time and effort to post your solution...I know that must've taken quite some time to so clearly elucidate all of the resulting cases. (Yes)

ZaidAlyafey,

If you come up with a shorter method, we're all eyes...(Nod) (Poolparty)