How do you solve #(4x)/(x+7)<x#?

1 Answer
Jul 1, 2016

Answer:

#x > 0# or #x in (-3,-7)#

Explanation:

Note: for the given expression to be meaningful #x!=-7#
and #x!=0# (since this would make both sides equal).

Case 1: #color(black)(x > 0)#

#color(white)("XXX")(4x)/(x+7) < x#

#color(white)("XXX")rarr 4/(x+7) < 1#

#color(white)("XXX")rarr 4 < x+7#

#color(white)("XXX")rarr x > -3#

#color(white)("XXX")#but by the Case 1 limitation #x > 0# (which is more limiting).

Case 2: #color(black)(x < 0)#

#color(white)("XXX")(4x)/(x+7) < x#

#color(white)("XXX")rarr 4/(x+7) > 1# (dividing by a negative reverses the inequality)

#color(white)("XXX"){: (color(black)("Case 2a: "),color(white)("XX"),color(black)("Case 2b: ")), (color(white)("X")color(black)(x < -7),,color(white)("X")color(black)(x > -7)), (color(white)("XX")4/(x+7) > 1,,color(white)("XX")4/(x+7) > 1), (color(white)("XX")rarr 4 < x+7,,color(white)("XX")4 > x+7), (color(white)("XX")rarr -3 < x,,color(white)("XX")-3 > x), (color(white)("XX")"impossible; since ",,color(white)("XX")rarr x in (-7,-3)), (color(white)("XX")color(white)("XXXX")x < 7,,) :}#

The following graph image might help understand this relationship.
(Note the asymptote for #color(red)((4x)/(x+7))#)

The region for which #color(red)((4x)/(x+7)) < color(blue)(x)# is shaded in #color(green)("green")#
enter image source here