What are the boundaries of x if #(2x-1)/(x+5) >= (x+2)/(x+3)?#

1 Answer
Mar 2, 2018

Answer:

#x = -5, x = -3, x = 1-sqrt(14), x = 1+sqrt(14)#
#>=" occurs for "x <-5" and "x>=1+sqrt(14)" and"#
#-3 < x <= 1-sqrt(14)"."#

Explanation:

#=> (2x-1)/(x+5) - (x+2)/(x+3) >= 0#

#=> ((2x-1)(x+3) - (x+2)(x+5))/((x+5)(x+3)) >= 0#

#=> (2x^2+5x-3-x^2-7x-10)/((x+5)(x+3)) >= 0#

#=> (x^2 -2x-13)/((x+5)(x+3)) >= 0#

#=> ((x - 1 - sqrt(14))(x - 1 + sqrt(14)))/((x+5)(x+3)) >= 0#

#"We have following zeros in order of magnitude : "#

#.... -5 .... -3 .... 1-sqrt(14)....1+sqrt(14).....#
#-----------0+++#
#-------0+++++++#
#-----0+++++++++#
#--0++++++++++++#
#"========================="#
#++0---0++0---0+++#

#"We see ">=0" occurs for "x <-5" and "x>=1+sqrt(14)" and"#
#-3 < x <= 1-sqrt(14)"."#