Question

How do you solve and write the following in interval notation: `4+abs(3x+2) <=5`?

Answer 1

`x in [-1, -1/3]`

Explanation:

`4+|3x+2| <= 5`

`=> |3x+2| <= 1`

The absolute value function `|a|` is defined as

`|a| = {(a if a>=0), (-a if a<0):}`

We will use this definition to consider all possible cases.

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Case 1: `3x+2 >= 0`

Note that for this to occur, we must have `3x >= -2 => x >= -2/3`

Now, by the definition of the absolute value function, we have `|3x+2| = 3x+2` in this case. Substituting that into our initial inequality, we get

`3x+2 <= 1`

`=> 3x <= -1`

`=> (3x)/3<=(-1)/3`

Recall that when multiplying or dividing by a positive number, the direction of the inequality stays the same, and when multiplying or dividing by a negative number, we change the direction of the inequality

`=> x <= -1/3`

So, together with the initial restriction `x=>-2/3`, we have

`-2/3 <= x <= -1/3`

or, in interval notation, `x in [-2/3, -1/3]`

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Case 2: `3x+2 < 0`

As above, we first solve for `x` to find that this case occurs when `3x < -2 => x < -2/3`

By the definition of the absolute value function, in this case we have `|3x+2| = -(3x+2)`. Again, we substitute this into the initial inequality.

`-(3x+2) <= 1`

`=> 3x+2 >= -1`

We multiplied by `-1` in the prior step, and so reversed the direction of the inequality.

`=> 3x >= -3`

`=> x >= -1`

Together with the initial restriction of `x < -2/3` in this case, we have

`-1 <= x < -2/3`

or, in interval notation, `x in [-1, -2/3)`

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If we look at where the intervals of the solutions of each case lay, we find that they meet at and include `x=-2/3`, and so putting them together, we get

`-1 <= x <= -1/3`

or, in interval notation, `x in [-1, -1/3]`