Question

How do you solve `|3f + 2| <= 7`?

Answer 1

Consider cases `f < -2/3` and `f >= -2/3` separately to simplify the inequality, then combine the results to find:

`-3 <= f <= 5/3`

Explanation:

Split into cases `f < -2/3` and `f >= -2/3`.

Case: `bb(f < -2/3)`

If `f < -2/3` then `3f < -2` and `3f+2 < 0`

In this case, `abs(3f+2) = -(3f+2) = -3f-2`, so our inequality becomes:

`-3f-2 <= 7`

Add `3f` to both sides to get:

`-2 <= 3f + 7`

Subtract `7` from both sides to get:

`-9 <= 3f`

Divide both sides by `3` to get:

`-3 <= f`

So this case gives us solutions `-3 <= f < -2/3`

Case: `bb(f >= -2/3)`

If `f >= -2/3` then `3f >= -2` and `3f + 2 >= 0`

In this case, `abs(3f+2) = 3f+2`, so our inequality becomes:

`3f+2 <= 7`

Subtract `2` from both sides to get:

`3f <= 5`

Divide both sides by `3` to get:

`f <= 5/3`

So this case gives us solutions `-2/3 <= f <= 5/3`

Summary

Putting both solutions together, the original inequality is satisfied for all `f` with `-3 <= f <= 5/3`

Here are graphs of `y = abs(3f+2)` and `y=7`. The range of `f` values for which the inequality is satisfied are between the two points of intersection of these curves:

graph{(abs(3x+2)-y)(y+0.001x-7) = 0 [-14.28, 13.81, -1.85, 12.2]}