To answer this question we will use the following property:
[tex]\ln a=b\text{ if and only if }e^b=a.[/tex]Substituting the second equation in the first one we get:
[tex]23=15-\ln(x-4).[/tex]Adding ln(x-4) to the above equation we get:
[tex]\begin{gathered} 23+\ln(x-4)=15-\ln(x-4)+\ln(x-4), \\ 23+\ln(x-4)=15. \end{gathered}[/tex]Subtracting 23 from the above equation we get:
[tex]\begin{gathered} 23+\ln(x-4)-23=15-23, \\ \ln(x-4)=-8. \end{gathered}[/tex]Therefore:
[tex]x-4=e^{-8}.[/tex]Adding 4 to the above equation we get:
[tex]\begin{gathered} x-4+4=e^{-8}+4, \\ x=e^{-8}+4. \end{gathered}[/tex]Since
[tex]e^{-8}>0[/tex]we get that x>4, therefore
[tex]\ln(x-4)[/tex]is well defined for
[tex]x=e^{-8}+4.[/tex]Answer:
[tex]x=e^{-8}+4,\text{ y=23.}[/tex]