Answer:
The solution of the given equation is:
y=6
Step-by-step explanation:
The expression is given by:
[tex]\dfrac{-8}{2y-8}=\dfrac{5}{y+4}-\dfrac{7y+8}{y^2-16}[/tex]
Now on solving for the given equation
[tex]\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}-\dfrac{7y+8}{(y-4)(y+4)}[/tex]
since,
[tex]a^2-b^2=(a-b)(a+b)\\\\so,\\\\y^2-16=y^2-4^2\\\\i.e.\\\\y^2-16=(y-4)(y+4)[/tex]
Hence, we get:
[tex]\dfrac{-4}{y-4}=\dfrac{5\times (y-4)-(7y+8)}{(y+4)(y-4)}\\\\i.e.\\\\\dfrac{-4}{y-4}=\dfrac{5y-20-7y-8}{(y-4)(y+4)}[/tex]
[tex]-4\times (y-4)(y+4)=(-2y-28)(y-4)\\\\i.e.\\\\(y-4)(-4\times (y+4))=(-2y-28)(y-4)\\\\i.e.\\\\(y-4)(-4y+16)=(-2y-28)(y-4)\\\\i.e.[/tex]
[tex](y-4)(-4y+16)-(-2y-28)(y-4)=0\\\\i.e.\\\\(y-4)(-4y-16+2y+28)=0\\\\i.e.\\\\(y-4)(-2y+12)=0\\\\i.e.[/tex]
[tex]y-4=0\ or\ -2y+12=0\\\\i.e.\\\\y=4\ or\ 2y=12\\\\i.e.\\\\y=4\ or\ y=6[/tex]
but y≠ 4
since, the denominator of the term in the left side of the given equality and the second term in the right side of the given equality will be zero and hence, the expression will be not defined.
Hence, the value of y is: 6
