How do you solve #abs(3x-4)>abs(x+6)#?

3 Answers
Apr 23, 2017

Answer:

Answer retracted by author due to errors.

Explanation:

Answer retracted by author due to errors.

Apr 23, 2017

Answer:

Please see below.

Explanation:

Solve the equality.
Cut the number line at those solutions.
Test each piece of the number line.

#abs(3x-4) = abs(x+6)# if and only if

#3x-4=x+6# #" "# OR #" "# #3x-4=-(x+6)#

#2x=10# #" "# OR #" "# #4x=-2#

#x=5# #" "# OR #" "# #x=-1/2#

Cut the number line at #-1/2# and at #5# and choose a test number in each piece of the line.

Test interval #" " # #" "# using #" "# #" "# True or false

#(-oo,-1/2)# #" "# #x=-3# #" "# #abs(-13) > abs(3)# is true

#(-1/2,5)# #" "# #" "# #x=0# #" "# #" "# #abs(-4) > abs(6)# is false

#(5,oo)# #" "# #" "# #x=6# #" "# #" "# #abs(14) > abs(12)# is true.

The solution set is

#(-oo,-1/2) uu (5,oo)#

Graphical solution

#abs(3x-4) > abs(x+6)# when #abs(3x-4) - abs(x+6)# is positive.

Here is the graph of #y = abs(3x-4) - abs(x+6)#

graph{abs(3x-4)-abs(x+6) [-11.79, 16.69, -7.2, 7.04]}

We can see that #y > 0# for #x# in #(-oo,-1/2) uu (5,oo)#

Apr 23, 2017

Answer:

#color(red)( x in (-oo, -1/2) uu (5, oo))#

Explanation:

First, let us calculate roots/zeros of the expressions on both sides of the equation (i.e. the value(s) of #x# for which these expressions become zero separately). The reason for doing this will become clear after a few steps.

1. #3x-4=0 implies x=4/3#
2. #x+6=0 implies x=-6#

Now let us put these points on a number line. [just for a ROUGH idea. No need to draw to scale.]

Can be reproduced with permission. Personal image.
Now observe that in the interval,
1. #(-oo, -6)# value of both the expressions is negative, i.e. #3x-4 < 0 # and #x+6 < 0#.
#therefore# in this interval, #|3x-4| = -(3x-4) = 4-3x# and #|x+6| = -(x+6) = -x-6#.

2. #[-6, 4/3)# , #3x-4# is negative , i.e. #3x-4 < 0 # and #x+6# is positive i.e. #x+6 >= 0#.
#therefore# in this interval, #|3x-4| = -(3x-4) = 4-3x# and #|x+6| = (x+6)#.

3. #[4/3, oo)# value of both expressions is positive, i.e. #3x-4 >= 0# and #x+6 > 0#.
#therefore# in this interval, #|3x-4| = (3x-4)# and #|x+6| = (x+6)#.

Now let us solve this inequality using the above-obtained results:-

1. #(-oo, -6)#
#4-3x > -x-6#
#=> 4+6 > 3x-x#
#=> 5>x => x<5# #therefore x in (-oo, 5)# BUT the interval we are considering is #(-oo, -6)#.
Hence #color(blue)(x in (-oo, -6) nn (-oo, 5))#
#=> color(red)(x in (-oo, -6))#

2. #[-6, 4/3)#
#4-3x > x+6 #
#=> 4-6 > 3x+x #
#=> -1/2 > x => x<-1/2# #therefore x in (-oo, -1/2)# BUT our interval is #[-6, 4/3)#.
Hence #color(blue)(x in [-6, 4/3) nn (-oo, -1/2))#
#=> color(red)( x in [-6, -1/2) )#

3. #[4/3, oo)#
#3x-4 > x+6#
#=> 3x-x > 4+6#
#=> x>5# #therefore x in (5, oo)# BUT similar to previous two cases, #color(blue)(x in [4/3, oo) nn (5, oo))#
#=> color(red)( x in (5, oo))#

#therefore color(blue)( x in (-oo, -6) uu [-6, -1/2) uu (5, oo))#
#=> color(red)( x in (-oo, -1/2) uu (5, oo))#
In case you need help with finding intersection or union of sets, leave a comment below or message me and I will be glad to help you out.