Graph of double absolute values

Homework Statement

|x| + |y| ≤ 1
What is the region in the plane that solves this inequality?

The Attempt at a Solution

I first tried graphing it by isolating the y variable
|y| ≤ -|x| + 1
Then I looked at the hint we were given, which was to assume that x and y are both positive, so I treated it as if it were written
y ≤ -x+1
And then constructed a graph of a line that was basically y = -x+1 with a domain of [0,3]
But then what threw me off is that the y variable does have the absolute value bars, so there can't be negative y values in the graph, right? So now I'm thinking I just have to flip the part of the graph so that it more closely resembles y=|x-1| in the domain of [0,3]. Is that correct or am I completely missing this?

eumyang
Homework Helper
I first tried graphing it by isolating the y variable
|y| ≤ -|x| + 1
Then I looked at the hint we were given, which was to assume that x and y are both positive, so I treated it as if it were written
y ≤ -x+1
And then constructed a graph of a line that was basically y = -x+1 with a domain of [0,3]
I'm not sure how can the domain be [0, 3]. Can you put any value between and including 0 & 3 in for x in |x| + |y| ≤ 1?

What I would do is graph the four inequalities. (Yes, there are four; can you figure out what they would be? You have one of them: y ≤ -x+1.) The intersection of the four graphs would serve as the graph of |x| + |y| ≤ 1.

I don't think I actually meant domain, that's my bad. I meant my line went from x=0 to x=3 just because of the size of my graph.

Are the inequalities:
x + y ≤ 1
x - y ≤ 1
-x + y ≤ 1
-x - y ≤ 1

eumyang
Homework Helper
Are the inequalities:
x + y ≤ 1
x - y ≤ 1
-x + y ≤ 1
-x - y ≤ 1
Yes.