How do you graph and solve #|x+4| + |2x - 1| <4#?

1 Answer
Oct 11, 2017

Answer:

See below.

Explanation:

This formulation is equivalent to

#sqrt((x+4)^2)+sqrt((2x-1)^2) + e^2 =4# or

#(x+4)^2+(2x-1)^2+2sqrt((x+4)^2(2x-1)^2) = (4-e^2)^2# or

#4(x+4)^2(2x-1)^2-((4-e^2)^2-(x+4)^2-(2x-1)^2)^2=0# or

#(e^2 - 3 x-7) (e^2 - x+1) (e^2 + x-9) (e^2 + 3 x-1) = 0# or

#{(3x+7 ge 0),(x-1 ge 0),(9-x ge 0),(1-3x ge 0):}# or

#{(x ge -7/3),(x ge 1),(x le 9),(x le 1/3):}#

then we conclude that there is no solution for

#|x+4| + |2x - 1| <4#