How do solve #3/(x-2)<=3/(x+3)# algebraically?
1 Answer
Oct 5, 2016
Explanation:
Slower way:
-
Bring everything to the right:
#3/(x-2) - 3/(x+3) \leq 0# -
Lowest common denominator:
#(3(x+3)-3(x-2))/((x-2)(x+3))\leq 0# -
Expand the numerator:
#(3x+9-3x+6)/((x-2)(x+3))\leq 0# -
#(9+6)/((x-2)(x+3))\leq 0# -
#(15)/((x-2)(x+3))\leq 0# -
Since
#15# is always positive, the sign of the fraction is decided by the sign of the denominator.#(x-2)(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{(x-3)(x+2) [-4.96, 6.146, -2.933, 2.614]}