Question

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

Answer 1

Answer retracted by author due to errors.

Explanation:

Answer retracted by author due to errors.

Answer 2

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)`

Answer 3

`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.]

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.