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My question is : Solve simultaneously-

$$\left\{\begin{align*}&|x-1|+|y-2|=1\\&y = 3-|x-1|\end{align*}\right.$$

I tried to solve this question by the method told by Marvis as I had understood that method (its here: Solve an absolute value equation simultaneously

But the solution set i got for the above question is not correct. My solution was: $y \geq 2$, and x=3 or x=y-2. I would like to know the final correct solution.

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3-y-y+2=1 => -2y = -4 => y=2 –  Santosh Linkha Jun 5 '12 at 10:12
    
|x-1|=1-|y-2| I substituted this in y=3-|x-1| .And then after solving it i got y=y or y=2 –  mgh Jun 5 '12 at 10:29
    
abstract-algebra? –  Frank Science Jun 5 '12 at 12:09

2 Answers 2

up vote 1 down vote accepted

We shall proceed on similar lines as the answer here.

You have that $\lvert x - 1 \rvert + \lvert y - 2 \rvert = 1$. This gives us that $$\lvert x - 1 \rvert = 1 - \lvert y-2 \rvert.$$Plugging this into the second equation gives us $$y = 3 - \left( 1 - \lvert y-2 \rvert \right) = 2 + \lvert y - 2 \rvert$$ This gives us that $$y - \lvert y - 2 \rvert = 2.$$ If $y > 2$, then we get that $$y - y+ 2 =2,$$ which is true for all $y >2$.

If $y \leq 2$, then we get that $$y + y-2 =2 \implies y=2$$ Hence we get that $$y \geq 2$$ From the second equation, we get that $$\lvert x - 1 \rvert = 3 - y$$ Since $\lvert x - 1 \rvert \geq 0$, we need $$3-y \geq 0$$ This means that $y \leq 3$. Hence, we have that $2 \leq y \leq 3$.

If $x \geq 1$, then $x-1 = 3-y \implies x = 4-y$. Note that since $y \in [2,3]$, $x = 4-y \geq 1$.

If $x < 1$, then $x-1 = y-3 \implies x = y-2$. Note that since $y \in [2,3]$, $x = y-2 < 1$.

Hence, the solution set is given as follows. $$2 \leq y \leq 3 \\ \text{ and }\\ x = 4-y \text{ or } y-2$$

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:Hey i have a doubt while considering cases for x why did you take 'equal to' sign in 'both' the cases? –  mgh Jun 5 '12 at 20:48
    
Though it is right but how do we come to know that whether we should take 'equal to' sign in both cases for 'x'? –  mgh Jun 5 '12 at 20:52
    
i think it does not matter even if we not consider the 'equal to' sign in both cases .Am i right? –  mgh Jun 5 '12 at 20:58
    
@meg_1997 It doesn't matter. I will anyway change it though. –  user17762 Jun 5 '12 at 21:01

As in my solution to the previous question, let $u=|x-1|$. Then $u+|y-2|=1$ and $y=3-u$, so $y-2=1-u$, and $u+|1-u|=1$.

If $u>1$, $|1-u|=u-1$, and this is $u+(u-1)=1$, or $u=1$, which is impossible since we assumed that $u>1$. If $u\le 1$, it’s $u+(1-u)=1$, or $1=1$, which is always true. Thus, all solutions must have $u\le 1$. On the other hand, $u=|x-1|$, so $u\ge 0$. Thus, we’re limited to those values of $u$ satisfying $0\le u\le 1$. This implies that $y=3-u$ must satify $2\le y\le 3$.

Pick any $y$ such that $2\le y\le 3$. Recall that $|x-1|=u=3-y$, so $x-1=\pm(3-y)$, and $x=4-y$ or $x=y-2$. Thus, the solutions are the pairs such that

$$2\le y\le 3,\text{ and }x=4-y\text{ or }x=y-2\;.$$

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