We are given the following function:
[tex]y=\lvert{8-2x}\rvert[/tex]
We are asked to determine the values of "x" for which "y = 16". To do that we will set the function equal to 16:
[tex]\lvert{8-2x}\rvert=16[/tex]
Now, to solve the equation we will use the fact that the absolute value has two possible values, a positive and a negative value. For the positive value we have:
[tex]8-2x=16[/tex]
Now, we solve for "x". First, we subtract 8 from both sides:
[tex]\begin{gathered} -2x=16-8 \\ -2x=8 \end{gathered}[/tex]
Now, we divide both sides by -2:
[tex]x=\frac{8}{-2}=-4[/tex]
Therefore, the first possible solution is "x = -4".
For the negative form of the function we have:
[tex]-(8-2x)=16[/tex]
Now, we multiply both sides by -1:
[tex]8-2x=-16[/tex]
Now, we subtract 8 from both sides:
[tex]\begin{gathered} -2x=-16-8 \\ -2x=-24 \end{gathered}[/tex]
Now, we divide both sides by -2:
[tex]x=-\frac{24}{-2}=12[/tex]
The second value is 12. Therefore, the solution set is:
[tex]{}{}\lbrace-4,12\rbrace[/tex]
