Question

How do you solve `x+3=-abs(3x-1)`?

Answer 1

Consider two possibilities which are significant for the absolute value term

Possibility 1: `(x<=1/3)`
In this case `(3x-1) <= 0`
and the equation can be re-written
`x+3 = - (-(3x-1))`
`x+3 = 3x-1`
`-2x = -4`
`x= 2`
Note that this is an extraneous solution since it does not exist within the range of values for Possibility 1: `(x<=1/3)`

**Possibility 2: `(x>1/3)`
In this case `(3x-1) > 0`
and the equation can be re-written
`x+3 = -(3x -1)`
`4x = -2`
`x = -1/2`
Note that once again we have an extraneous solution sin it does not exist within the range of values for Possibility 2: (`x>1/3`)

To see why this happens consider a re-arrangement of the original equation into the form:
`x+3 +abs(3x-1) = 0`
The graph of the left side of this equation is shown below. Notice that it does not intersect the X-axis (that is it is never equal to `0`).
graph{x+3+abs(3x-1) [-16.01, 16.02, -8.01, 8]}