Find all values of x such that \(\left| |x-17| - 3 \right| = \left| |x-15| - 5 \right|\)(Enter your answer in interval notation.)

Guest Feb 27, 2019

#2**+1 **

**Find all values of x such that **

\(\large{ \mathbf{\Big| ~ |x-17| - 3~ \Big| = \Big|~ |x-15| - 5 ~\Big|} }\)

\(\begin{array}{|rcll|} \hline \Big| ~ |x-17| - 3~ \Big| &=& \Big|~ |x-15| - 5 ~\Big| \quad \boxed{\text{Formula: } |a|^2 = (a)^2} \\ \left( ~ |x-17| - 3~ \right)^2 &=& \left(~ |x-15| - 5 ~\right)^2 \\ (x-17)^2-6|x-17|+9 &=& (x-15)^2-10|x-15|+25 \\ 10|x-15| &=& 6|x-17|+ (x-15)^2-(x-17)^2 + 16 \\ 10|x-15| &=& 6|x-17| +4x-48 \quad | \quad : 2 \\ \mathbf{5|x-15|} &\mathbf{=}& \mathbf{3|x-17| +2x-24} \\ \hline \end{array} \)

Here we have two different amounts.

To dissolve them, we must make a double case distinction.

Usually we did this one after the other.

For reasons of space, we start with the first case in which the content of the left amount

is greater than or equal to zero.

\(\begin{array}{|rcll|} \hline & \underline{x\ge 15:}& \\\\ & 5(x-15)\\ & =3|x-17|+2x-24 \\ \\ \underline{\text{for } x\ge 17:} && \underline{\text{for } x \lt 17: } \\\\ 5(x-15)=3(x-17)+2x-24 && 5(x-15)=-3(x-17)+2x-24 \\ 5x-75=3x-51+2x-24 && 5x-75=-3x+51+2x-24 \\ -75=-75~\text{true} && 6x=102 \\ x\ge 15 \wedge x\ge 17 && x=17 \quad (x\lt 17) ~\text{no solution} \\ \boxed{ x\ge 17 } && \\ \hline \hline & \underline{x\lt 15:}& \\\\ & -5(x-15)\\ & =3|x-17|+2x-24 \\ \\ \underline{\text{for } x\ge 17:} && \underline{\text{for } x \lt 17: } \\\\ -5(x-15)=3(x-17)+2x-24 && -5(x-15)=-3(x-17)+2x-24 \\ -5x+75=3x-51+2x-24 && -5x+75=-3x+51+2x-24 \\ -10x=-150 && -4x=-48 \\ x=15\quad (x\ge 17) ~\text{no solution} && x=12 \quad (x\lt 17) ~\text{solution}\\ && \boxed{ x=12 } \\ \hline \end{array}\)

\(\mathbf{\boxed{x=12 \lor x\ge 17}}\)

heureka Feb 28, 2019