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.
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. $$2 \leq y \leq 3 \\ \text{ and }\\ x = 4-y \text{ or } y-2$$
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
$$2\le y\le 3,\text{ and }x=4-y\text{ or }x=y-2\;.$$
Clearly, for |x- 1| we need to consider x< 1 and $x\le 1$ separately. Similarly for |y- 2| we need to consider y< 2 and $y\le 2$ separately.
That means that, here, we need to consider 4 cases: 1) x< 1 and y< 2. The equations become 1- x+ 2- y= 3- x- y= 1 and y= 3- (1- x)= 2+ x. So 3- x- y= 3- x- (2+ x)= 5 for all x. It is never 1. This case is impossible.
2) x> 1 and y< 2. The equations become x- 1+ 2- y= 1+ x- y= 1 and y= 3- (x- 1)= 4- x. So 1+ x- y= 1+ x- (4- x)= -3+ 2x= 1. 2x= 4, x= 2 and then y= 4- 2= 2. This case is impossible because y= 2 is not less than 2.
3) x< 1 and y> 2. The equations become 1- x+ y- 2= y- x- 1= 1 or y= x+ 2 and y= 3- (1- x)= 2+ x. So x+ 2= 2+ x. That is true for [b]all[/b] x. But since y= x+ 2, in order to have y> 2, we must have x positive. This case is satisfied for all 0< x< 1, y= x+ 2.
4) x> 1 and y> 1. The equations become x-1+ y- 2= x+ y- 3= 1 or x+ y= 4 and y= 3- (x- 1)= 2- x. x+ y= x+ (2- x)= 2 for all x which is never equal to 1. This case is impossible.
These equations are satisfied if and only if x is between 0 and 1 and y= x+ 2.
$$\begin{align}|x-1|+|y-2|&=1\\|x-1| + y &= 3\end{align}$$
Rewrite this as
$$\begin{align}|x-1|+|y-2|&=1\\|x-1| + (y-2) &= 1\end{align}$$
Then $y-2 = |y-2|$; which implies $y \ge 2$ and we now have
$$\begin{align}y &\ge 2\\|x-1| &= 3-y\end{align}$$
So $3-y \ge 0$; which implies $2 \le y \le 3$.
If $x \ge 1$ then $x-1=3-y \iff x=4-y$.
But $x=4-y$ and $2 \le y \le 3 \implies 1 \le x \le 2$.
So, when $1 \le x \le 2$, then $y = 4-x$.
If $x < 1$, then $1-x = 3-y \iff x=y-2$
But $x=y-2$ and $x < 1$ and $2 \le y \le 3 \implies 0 \le x < 1$.
So, when $0 \le x < 1$, then $y = x+2$.
We conclude that $0 \le x \le 2$ and $y = \begin{cases} x+2 &\text{when $0 \le x < 1$} \\ 4-x &\text{when $1 \le x < 2$} \end{cases}$
