Question

Solve the following inequality algebraically. 3∣x−3∣+8<20

Answer 1

The inequality to solve is:

`3\vert x-3\vert+8<20`

Let's get the absolute value isolated:

`\begin{gathered} 3|x-3|<20-8 \\ 3|x-3|<12 \\ |x-3|<(12)/(3) \\ |x-3|<4 \end{gathered}`

Now we have an absolute value equation. One thing to note here is:

If

|x - a| < b

We can write:

x - a < b

and

x - a > - b

From our equation, we can thus say:

x - 3 < 4

and

x - 3 > -4

Solving 1st:

`\begin{gathered} x-3<4 \\ x<4+3 \\ x<7 \end{gathered}`

and solving 2nd one:

`\begin{gathered} x-3>-4 \\ x>-4+3 \\ x>-1 \end{gathered}`

Together, we can write the solution set as:

[tex]-1