Question

Question #114fc

Answer 1

`(y>=abs(x+1)-3)nn(2y < x+5)`

Explanation:

For reference lets call the "upper" line that crosses the Y-axis at `2.5` Line A
and the "lower" line that crosses the Y-axis at `(-2)` Line B

For Line A
We note the following points are on the line:
`color(white)("XXX")(x,y) in {(0,5/2), (-5,0)}`
Since this is a straight line, the slope is constant for all pairs of points:
`color(white)("XXX")(y-0)/(x-5)=(5/2-0)/(0-5)=1/2`

`color(white)("XXX")rarr 2y=x-5`

The identified area is below this line and the hollow "bubbles" indicate the points on the line are not to be included.

Therefore Line A provides the constraint
`color(white)("XXX")color(red)(2y < x-5)`

For Line B
The V shape of Line B indicates that the equation for this line involves the absolute value of a linear expression of `x`.
Since the "turning point" occurs at `x-1 rarr x+1=0`,
the absolute expression is `abs(x+1)`

Since the "turning point" is at `(x,y)=(-1,-3)`
we note that relative to `y=abs(x+1)`, all the `y` values have been reduced by `3`;
that is the equation for Line B `y=abs(x+1)-3`

The identified area is above (and includes, since there are no "bubbles") Line B.

Therefore Line B provides the constraint
`color(white)("XXX")color(blue)(y >= abs(x+1)-3`

Combining the Areas
The identified area meets both Line A and Line B constraints.

Therefore the identified area is constrained by
`color(white)("XXX")color(red)(2y < x-5) and color(blue)(y >= abs(x+1)-3`
or (in set notation)
`color(white)("XXX")color(red)((2y < x-5))nncolor(blue)((y >= abs(x+1)-3)`