Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

up vote 6 down vote accepted

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. $$y \leq 2 \\ \text{ and }\\ x = 4-y \text{ or } y-2$$

share|improve this answer
    
@ 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

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

share|improve this answer

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 $x <1$:

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

$$y=3+x-1$$

share|improve this answer
    
So whats the final solution? –  mgh Jun 5 '12 at 18:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.