How do solve #3/(x2)<=3/(x+3)# algebraically?
1 Answer
Oct 5, 2016
Answer:
Explanation:
Slower way:

Bring everything to the right:
#3/(x2)  3/(x+3) \leq 0# 
Lowest common denominator:
#(3(x+3)3(x2))/((x2)(x+3))\leq 0# 
Expand the numerator:
#(3x+93x+6)/((x2)(x+3))\leq 0# 
#(9+6)/((x2)(x+3))\leq 0# 
#(15)/((x2)(x+3))\leq 0# 
Since
#15# is always positive, the sign of the fraction is decided by the sign of the denominator.#(x2)(x+3)# represents a parabola with zeros in#x=3# and#x=2# . Such a parabola is negative between its solutions, as you can see in the graph here:
graph{(x3)(x+2) [4.96, 6.146, 2.933, 2.614]}