Question

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

Answer 1

`y=\pm i √(7)`

Explanation:

Given equation is
`(y)/(\left(y-4\right))-(4)/(\left(y+4\right))=(3^2)/((y^2-16))`

Factor denominators then solve by making denominators equal

`(y)/(\left(y-4\right))-(4)/(\left(y+4\right))=(3^2)/((y^2-16))`

`(y)/(\left(y-4\right))-(4)/(\left(y+4\right))=(9)/((y+4)\left(y-4\right))`

`(y\left(y+4\right)-4\left(y-4\right))/(\left(y-4\right)\left(y+4\right))=(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 √(-7)`

`y=\pm i √(7)`

Hence final answer is
`y=\pm i √(7)`.