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

Question

asked May 4, 2020 26.9k views

1 Answer

Abdul Hamid · Answer 1 · 2020-05-09T15:26:56+0000

Answer:

$y=\pm i √(7)$

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
$y=\pm i √(7)$ .