Question

o

Answer 1

y ≥ -x, in the range -6 ≤ x ≤ 0
The boundary of this is a segment of the line
y = -x
and in this range it runs from (-6, 6) to (0, 0). I got the equation by just looking at the "equals" case of the inequality, and the endpoints by "plugging in" the extreme values of x (-6 and 0) into the equation.

2x + 3y ≥ 6, in the range -12 ≤ x ≤ -6
Again focusing on the "equals" case, the boundary of this is a segment of the line
2x + 3y = 6
and using -12 and -6 for x, we see the segment runs from
(-12, 10) to (-6, 6)
where it meets the previous segment.

x ≤ 0 puts a boundary at the y-axis (x = 0).
y ≤ 10 puts a boundary at the horizontal line y = 10.
These boundaries meet, of course, at (0, 10), and they meet the other boundaries at
(-12, 10) and (0, 0)
which are precisely the endpoints of the conjoined line segments.

The area in question is below the line y=10 and left of the y-axis, and above the two slanted line segments running from
(-12, 10) to (-6, 6) to (0, 0).

It's area can most easily be computed by treating it as two right triangles and a rectangle.

The triangle defined by
(-12, 10), (-6, 6), and (-6, 10)
has area 6 * 4 / 2 = 12.

The triangle defined by
(-6, 6), (0, 0), and (0, 6)
has area 6 * 6 / 2 = 18.

The rectangle defined by
(-6, 6), (-6, 10), (0, 10), and (0, 6)
has area 6 * 4 = 24

Total area: 12 + 18 + 24 = 54.

Answer 2

so is the answer 54?

Answer 3

Yes.