Question

Solve the following system of equation:
$|x-1|+|y-2|=1,y=3-|x-1|$

Answer 1

Given equation,

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

$y=3-|x-1|$

If $x\ge 1$,

If $y\ge 2$

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

$x+y=4$

$y=3-x+1$

$x+y=4$

$4-x\ge 2$

$x\le 2$

$x=1,y=3$

$x=2,y=2$ are solutions.

If $y<2$,

$x-1-y+2=1$

$x-y=0$

$y=3-x+1$

$x+y=4$

$x=2,y=2$ (already a solution)

If $x<1$,

If $y\ge 2$,

$1-x+y-2=1$

$y-x=2$

$y=3+x-1$

$y-x=2$

$y=x+2$

$x+2\ge 2$

$x\ge 0$

$x=0,y=2$

If $y<2$,

$1-x+2-y=1$

$x+y=2$ and $y-x=2$

$\therefore$ $x=0,y=2$ (already a solution)

$\therefore$ $(x=0,y=2)$ and $(x=1,y=3$ and $(x=2,y=2)$ are solutions of given equations.