Question

How do you graph and solve `|4x+8|>=20`?

Answer 1

The solution is a union of two intervals:
`x<=-7` or `x in (-oo,-7]`
UNION with
`x >= 3` or `x in [3,+oo)`

Explanation:

The value of `|4x+8|` is defined as follows:
if `4x+8 >= 0` then `|4x+8| = 4x+8`
otherwise (that is, id `4x+8 < 0`), `|4x+8| = -(4x+8)`.

Let's consider a point `x=-2`, where term `4x+8` changes from negative (to the left of `x=-2`) to positive (to the right of `x=-2`).

From definition of absolute value mentioned above,
if `x < -2`, 4x+8 < 0 and, therefore, `|4x+8| = -(4x+8)` and our initial inequality looks like this:
`-(4x+8) >= 20` or
`-4x-8 >= 20` or
`-8-20 >= 4x` or
`4x <= -28` or
`x <= -7`
Since `x<=-7` lies inside the interval `x < -2` that we consider, all `x<=-7` are solutions.

From definition of absolute value mentioned above,
if `x >= -2`, 4x+8 >= 0 and, therefore, `|4x+8| = 4x+8` and our initial inequality looks like this:
`4x+8 >= 20` or
`4x >= 12` or
`x >= 3`
Since `x>=3` lies inside the interval `x >= -2` that we consider, all `x>=3` are solutions.

So, we have two separate intervals that represent the solutions to this inequality:
`x <= -7` and `x >= 3`.

Graphically, it looks like this:
graph{|4x+8| [-24, 24, -23.12, 23.12]}
It can be observed that this graph is above the line `y=20` when `x<=-7` or `x >= 3`.