+0

# Equations

0
350
7

|x+y-4|=5

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

I'd be thankful if somebody can explain the way to solve it.

Apr 23, 2018
edited by Guest  Apr 23, 2018

#1
+12223
+1

|x+y-4|=5

Apr 23, 2018
edited by Omi67  Apr 23, 2018
edited by Omi67  Apr 23, 2018
edited by Omi67  Apr 23, 2018
edited by Omi67  Apr 23, 2018
#2
+1

I used the program called Desmos to help me with this task and I saw that there is more than one correct answer.

Image of graphs -->https://www.desmos.com/calculator/rpf7nusezm

Guest Apr 23, 2018
#3
+12223
+2

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

Apr 23, 2018
#4
+2

Thanks a lot!

Guest Apr 23, 2018
#5
+12223
+2

I'd be thankful if somebody can explain the way to solve it.

Apr 23, 2018
#6
+2

Thank you. I got it now

Guest Apr 23, 2018
edited by Guest  Apr 23, 2018
edited by Guest  Apr 23, 2018
#7
+109527
+2

|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

Apr 24, 2018