How do you solve #(2x)/(x+5)<=0#?

1 Answer
Dec 16, 2016

Answer:

#-5 < x <= 0 #

Explanation:

#color(red)("There is a trap in this question in that there are excluded")##color(red)("values for "x)#

#color(blue)("Step 1")#
Solve disregarding the excluded values.

Multiply both sides by #(x+5)#

#2x<=0xx(x+5)#

#2x<=0#

#color(blue)("Condition 1: "x<=0)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2")#
An equation or expression becomes 'undefined' if you have division by 0.

#=>(x+5) != 0#

#color(blue)("Condition 2: "x!=-5)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#

Can we have #x< -5color(white)(.)?#

Suppose #x=-6# then we have:

#(2(-6))/(-6+5) = (-12)/-1=+12#

This is not an excluded value but does not satisfy #(2x)/(x+5)<=0#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all together using proper notation")#

#-5 < x <= 0 #