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

1 Answer

3 votes

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).

User Abdul Hamid
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