Answer:
No solution
Step-by-step explanation:
For the equation given, we have an alternate form that is
[tex]\dfrac{y^2 + 16}{y^2-16} =\dfrac{32}{y^2-16}[/tex]
It is clear that
[tex]y^2+16 = 32 \implies y=4, y=-4[/tex]
But the denominator goes to zero and it is undefined.
If you take
[tex]$\lim _{y\to 4}\left(\dfrac{y^2+16}{y^2-16}\right)$[/tex]
you'll see that
[tex]$\lim _{y\to 4^-}\left(\dfrac{y^2+16}{y^2-16}\right) = -\infty$[/tex]
[tex]$\lim _{y\to 4^+}\left(\dfrac{y^2+16}{y^2-16}\right) =\infty$[/tex]
It diverges and the limit does not exist.
