Question

What is the solution to he equation (y/y-4)-(4/y+4)=3^2/y^2-16

Answer 1

Answer:

$y=\pm i \sqrt{7}$

Step-by-step explanation:

Given equation is $\frac{y}{\left(y-4\right)}-\frac{4}{\left(y+4\right)}=\frac{3^2}{(y^2-16)}$

Factor denominators then solve by making denominators equal

$\frac{y}{\left(y-4\right)}-\frac{4}{\left(y+4\right)}=\frac{3^2}{(y^2-16)}$

$\frac{y}{\left(y-4\right)}-\frac{4}{\left(y+4\right)}=\frac{9}{(y+4)\left(y-4\right)}$

$\frac{y\left(y+4\right)-4\left(y-4\right)}{\left(y-4\right)\left(y+4\right)}=\frac{9}{(y+4)\left(y-4\right)}$

$y\left(y+4\right)-4\left(y-4\right)=9$

$y^2+4y-4y+16=9$

$y^2=9-16$

$y^2=-7$

take squar root of both sides

$y=\pm \sqrt{-7}$

$y=\pm i \sqrt{7}$

Hence final answer is $y=\pm i \sqrt{7}$.