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

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Homework Help Overview

The problem involves solving a system of equations that includes absolute values, specifically: |(x-1)| + |(y-3)| = 11 and |(x-3)| + |(y-17)| = 3. The context is algebra, focusing on the manipulation of absolute value expressions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to consider different cases based on the values of x and y due to the absolute value expressions. There is mention of breaking down the problem into multiple scenarios for both variables, leading to a total of nine cases to analyze.

Discussion Status

The discussion is active, with participants exploring various approaches to tackle the problem. Hints have been provided regarding the treatment of absolute values and suggestions to visualize the equations graphically. There is no explicit consensus yet on a single approach.

Contextual Notes

Participants are reminded that this is a textbook-style problem, and they are encouraged to share their thoughts and reasoning up to the point of confusion, adhering to the forum's guidelines on homework help.

Jim_
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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
 
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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.
 
Last edited:
|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!
 
@Jim_ I suggest you plot those two graphs in the xy plane and have a look.
 

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