How do you solve #5/(x+3) ≥ 3/x#?

1 Answer
Apr 18, 2017

Answer:

Solution: # -3 < x < 0 or x >= 4.5 #. In interval notation: #(-3,0) uu[4.5,oo)#

Explanation:

# 5/(x+3) >= 3/x or 5/(x+3) - 3/x >= 0 or (5x - 3(x+3))/(x(x+3)) >=0# or
# (2x-9)/( x(x+3)) >=0# . Critical points are #x=-3 , x =0 ,x =4.5#
#x != -3 , x != 0# (since denominator should not be zero.)

Check for #>=0#
1) #x < -3 ; (2x-9)/( x(x+3)) (-)/(-*-) i.e< 0#

2) # -3 < x < 0# ; #(2x-9) / ( x(x+3)) (-)/(-*+) i.e> 0 #

3) # 0 < x < 4.5 ; (2x-9)/( x(x+3)) (-)/(+*+) i.e<0 #

4) #x > 4.5 ; (2x-9)/ ( x(x+3)) >=0 (+) / (+*+) i.e >0#

Solution: # -3 < x < 0 or x >= 4.5 #. In interval notation: #(-3,0) uu[4.5,oo)# [Ans]