|x-3|+|y-1|=5

\(If \;\;x\ge 3\;\;and\;\;y\ge1\;\;then\\ x-3+y-1=5\\ x+y=9\\ y=-x+9\\~\\ When\;x=3\quad y=-3+9=6 \quad (3,6)\\ When\;y=1\quad x+1=9\quad x=8 \quad (8,1)\\ \text{The segment of line joining (3,6) to (8,1) satisfies this} \)

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\(If \;\;x\ge 3\;\;and\;\;y\le1\;\;then\\ x-3-y+1=5\\ x-y=7 \\~\\ When\;x=3\quad 3-y=7 \quad y=-4\quad ( 3,-4) \\ When\;y=1\quad x-1=7 \quad x=8 \quad (8,1)\\ \text{The segment of line joining (3,-4) to (8,1) satisfies this}\)

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\(If \;\;x\le 3\;\;and\;\;y\ge1\;\;then\\ -x+3+y-1=5\\ -x+y=3\\ When\;x=3\quad y=6\quad \quad (3,6)\\ When\;y=1\quad x=-2 \quad (-2,1)\;\;\\ \text{So the points on the line segment joining (3,6) to (-2,1) are solution.} \)

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\(If \;\;x\le 3\;\;and\;\;y\le1\;\;then\\ -x+3-y+1=5\\ -x-y=1\\ x+y=-1 \\~\\ When\;x=3\quad y=-4\quad (3,-4)\\ When\;y=1\quad x=-2 \quad (-2,1)\\ \text{The segment of line joining (3,-4) to (-2,1) satisfies this}\)

**So the solution set is all the points on the perimeter of this square**