Algebra with absolute value. Please help me to to solve.

[Jim_]

I came across this algebra problem, can someone please help me solve this problem? Please show the steps as well. Much appreciated.

|(x-1)| + |(y-3)| = 11
|(x- 3)| + |(y-17)| = 3

Find the nearest/possible x and y

 

[Fredrik]

I moved the thread to a homework forum, because it's a textbook-style problem, and we treat all textbook-style problems as homework. We only give hints here, not complete solutions. You are required to post your own thoughts on how to solve the problem, up to the point where you're stuck.

I can give you one hint right now. You can get rid of the absolute value signs by considering several possibilities separately.

 

[HallsofIvy]

|x- 1| is 1- x for x less than 1, x- 1 for x greater than or equal to 1. Similarly for |x-3|. So there are 3 possiblities to consider: x< 1, 1< x< 3, and x> 3.

For y, we have the same situation: y< 3, 3< y< 17, and y> 17. Since x and y are independent, you have to consider each of the three x situations with all three of the y situations, a total of 3x3= 9 cases. You had better get busy!

 

[LCKurtz]

@Jim_ I suggest you plot those two graphs in the xy plane and have a look.