# Solve an absolute value equation simultaneously

My question is :

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

What I did : $y=3 - |x-1|$ is given.

Thus $y = 3-(x-1)$ or $y = 3-\left(-(x-1)\right),$ and so $$y = 4-x\qquad\mbox{ or } \qquad y = 2+x.$$

If $y = 2+x$, then $$x - 1 = 4-y \quad(1)\qquad\mbox{ or } \qquad y - 2 = x \quad(2).$$

Substituting 1), we get $$|4-y| - |y-2| = 1 \qquad\mbox{ or } \qquad |y-2-1|-|y-2| = 1.$$

I got here but I am not getting how to get the final solution. Any help would be greatly appreciated.

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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 $y \geq 2$, then we get that $$y + y- 2 =2 \implies y = 2$$ If $y < 2$, then we get that $$y + 2 - y =2$$ which is true for all $y <2$. Hence we get that $$y \leq 2$$ From the second equation, we get that $$\lvert x - 1 \rvert = 3 - y$$ Note that since $y \leq 2$, $3-y > 0$ always.

If $x >1$, then $$x - 1 = 3 - y \implies x = 4-y$$Note that $4-y > 1$, since $y \leq 2$.

If $x \leq 1$, then $$x - 1 = y-3 \implies x = y-2$$Note that $y-2 \leq 1$, since $y \leq 2$.

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

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@ Marvis:I used this method of yours to solve my question which is here: math.stackexchange.com/questions/154171/… . I am not getting the correct answer and i am not getting where i am going wrong. please will you give me hints to solve the question the way you have solved it above? –  mgh Jun 5 '12 at 19:08
@meg_1997 I have answered your other question now. –  user17762 Jun 5 '12 at 19:38

It’s entirely possible to solve this system by an exhaustive analysis of cases along the lines on which you’ve begun, but there’s an easier way. Notice that both equations have a term $|x-1|$. Let’s let $u=|x-1|$ and see whether the resulting system is a little easier to solve. After the substitution the original system

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

\left\{\begin{align*}&u-|y-2|=1\\&y=3-u\end{align*}\right.\;.\tag{2}

The second equation of $(2)$ implies that $y-2=1-u$; substituting this into the first equation yields $$u-|1-u|=1\;.\tag{3}$$ If $u<1$, then $1-u>0$, and therefore $|1-u|=1-u$, and $(3)$ reduces to $u-(1-u)=1$, which is easily solved to get $u=1$ and hence $y=3-u=2$. Recall that $u=|x-1|$; thus, $|x-1|=1$, $x-1=\pm 1$, and $x=0$ or $x=2$. You can check that if $y=2$ and $x=0$ or $x=2$, then $(1)$ is satisfied.

If $u\ge1$, then $|1-u|=u-1$, and $(3)$ reduces to $u-(u-1)=1$, or $1=1$. This is always true, so every value of $u>1$ leads to a solution. Let’s check that. Let $u$ be any number greater than $1$, and let $y=3-u$. Then $y-2=1-u<0$, so $|y-2|=2-y=u-1$, and $$u-|y-2|=u-(2-y)=u-(u-1)=1\;,$$ so $(2)$ is indeed satisfied. All we need to do now is solve for $x$ in this case. We have $|x-1|=u$, so $x-1=\pm u$, and $x=1\pm u$. In other words, for each $u\ge 1$ there are two solutions,

\left\{\begin{align*}&x=1+u\\&y=3-u=3-(x-1)=4-x\end{align*}\right.\tag{4} and

\left\{\begin{align*}&x=1-u\\&y=3-u=3-(1-x)=2+x\;.\end{align*}\right.\tag{5}

Finally, $(4)$ and $(5)$ can be simplified by getting rid of $u$. Since $u\ge 1$, $(4)$ gives us every value of $x\ge 2$, and $(5)$ gives us every value of $x\le 0$. Thus, $(4)$ and $(5)$ reduce to saying that the solutions to $(1)$ are

\begin{align*} &y=4-x\text{ for any }x\ge 2\\ &y=2+x\text{ for any }x\le 0\;, \end{align*}\tag{6}

with no solutions having $0<x<2$. You can improve a little on $(6)$ by recasting it in terms of $y$: note that both branches of the solution give you all values of $y\le 2$, so you can rephrase $(6)$ by saying that the solutions are all pairs such that

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

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Whenever you see an absolute, you need to consider the possible cases:

$$|x-1|=\begin{cases} x-1 & x \geq 1 \\-(x-1) & x<1\\\end{cases}$$

As for the first equation, you might consider four cases:

• $y \geq 2$ and $x \geq 1$: $$(x-1)-(y-2)=1 \implies y=x$$ $$y=3-(x-1) \implies y=4-x$$ $$\Downarrow$$ \left\{\begin{align*}&y=x\\&y=4-x\end{align*}\right.

• $y \geq 2$ and $x <1$:

$$-(x-1)-(y-2)=1$$ $$y=3+(x-1)$$ $$\Downarrow$$

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

• $y<2$ and $x \geq 1$: $$(x-1)+(y-2)=1$$ $$y=3-(x-1)$$
• $y<2$ and $x <1$:

$$-(x-1)+(y-2)=1$$

$$y=3+x-1$$

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So whats the final solution? –  mgh Jun 5 '12 at 18:25