What is the value of "a" so the solution to: 5ax+5a≥2ax-3x-27+19x is "all real #'s"?

Answer 1

5ax+5a>/=2ax-3x-27+19x

(3a-16)x >/=-5a-27

x >/= -(5a-27)/(3a-16)

let a =16/3, then solution set for x is any real number

substitute that a value into the original equation/inequality to get

5(16/3)x +5(16/3) >/= 2(16/3)x- 3x -27 +19x

(48/3 -16)x >/= -27-80/3

0>/= -53 2/3

0 will always be greater than -53 2/3, regardless of the value of x