Question

How do you solve `|3x - 2| = | 4- x |`?

Answer 1

`color(red)(x=-1)` or `color(blue)(x=3)`

Explanation:

Note that if `abs(3x-2) = 0` then `x=2/3`
and that if `abs(4-x)=0` then `x=4`

We therefore want to consider the 3 options for `x`
Option [1]: `x <2/3`
Option [2]: `x in [2/3,4]`
Option [3]: `x > 4`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Option [1]
If `x <2/3`
then
`color(white)("XXX")abs(3x-2)=2-3xcolor(white)("xxx")`since `(3x-2)` would be less than zero
and
`color(white)("XXX")abs(4-x)=4-xcolor(white)("xxx")since`4-x# would be greater than zero

So `abs(3x-2)=abs(4-x)` (for this option)
means we need to solve the equation
`color(white)("XXX")2-3x=4-x`
`color(white)("XXX")rarr -2x=2`
`color(white)("XXX")rarr x=-1`
(which is consistent with the requirement that `x < 2/3`)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Option [2]
If `x in [2/3,4]`
then (using reasoning similar to above)
`color(white)("XXX")abs(3x-2)=3x-2`
and
`color(white)("XXX")abs(4-x)=4-x`

...and we need to solve
`color(white)("XXX")3x-2=4-x`
`color(white)("XXX")rarr 2x=6`
`color(white)("XXX")rarr x=3`
(consistent with the requirement `x in [2/3,4]`)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Option [3]
If `x > 4`
then
`color(white)("XXX")abs(3x-2) = 3x-2`
and
`color(white)("XXX")abs(4-x)=x-4`

...and we need to solve
`color(white)("XXX")3x-2=x-4`
`color(white)("XXX")2x=6`
`color(white)("XXX")x=3`
but this is not consistent with the requirement `x > 4`
so we can eliminate this as an extraneous result.

================================================

The only valid results are
`color(white)("XXX")x=-1 or x= 3`