David L. answered • 05/31/20

Expert Math and Software Tutoring, plus Software Troubleshooting

When I first looked at this, I also thought "how exactly am I supposed to graph this?"

When doing a math problem, first try to understand it, then think about how to answer it. We can use graphing as a tool to understand it.

When graphing equations in the x-y plane with equals signs, the graph will be a straight or curved line. But here we have an inequality, so the graph will be an area of the plane, with a straight or curved line as the boundary. The boundary will be |x|+|y| = 1.

That's still a lot, so let's break the problem into pieces. This technique is called "divide and conquer".

Piece number 1: x≥0 and y≥0. Then |x|+|y| = x+y ≤ 1, so y ≤ -x+1.

This is everything below the boundary line y = -x+1, but where x≥0 and y≥0 (above the x-axis and to the right of the y-axis).

The boundary line's y-intercept is 1, so the boundary line goes through (x,y) = (0,1).

On the boundary line y = -x+1, if y=0, x=1, so the boundary line also goes through (x,y) = (1,0).

Graph this boundary line, and shade in the area below it, but above the x-axis and to the right of the y-axis.

Piece number 2: x≥0 and y≤0. Then |x|+|y| = x-y ≤ 1, so x-1 ≤ y, or y ≥ x-1.

This is everything above the boundary line y = x-1, but where x≥0 and y≤0 (below the x-axis and to the right of the y-axis).

The boundary line's y-intercept is -1, so the boundary line goes through (x,y) = (0,-1).

On the boundary line y = x-1, if y=0, x=1, so the boundary line also goes through (x,y) = (1,0).

Graph this boundary line, and shade in the area below it, but below the x-axis and to the right of the y-axis.

Piece number 3: x≤0 and y≥0. Then |x|+|y| = -x+y ≤ 1, so y ≤ x+1.

This is everything below the boundary line y = x+1, but where x≤0 and y≥0 (above the x-axis and to the left of the y-axis).

The boundary line's y-intercept is 1, so the boundary line goes through (x,y) = (0,1).

On the boundary line y = x+1, if y=0, x=-1, so the boundary line also goes through (x,y) = (-1,0).

Graph this boundary line, and shade in the area below it, but above the x-axis and to the left of the y-axis.

Piece number 4: x≤0 and y≤0. Then |x|+|y| = -x-y ≤ 1, so -x-1 ≤ y, or y ≥ -x-1.

This is everything above the boundary line y = x+1, but where x≤0 and y≤0 (below the x-axis and to the left of the y-axis).

The boundary line's y-intercept is -1, so the boundary line goes through (x,y) = (0,-1).

On the boundary line y = -x-1, if y=0, x=-1, so the boundary line also goes through (x,y) = (-1,0).

Graph this boundary line, and shade in the area above it, but below the x-axis and to the left of the y-axis.

No more pieces.

The graph of the boundary lines should look like this:

So this diamond shape is the boundary of the area where |x|+|y|≤1.

The area where |x|+|y|≤1 is true includes this diamond shape and everything inside it.