Question #114fc

1 Answer
Nov 11, 2017

Answer:

#(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)#