Question 278135
x/x-4-4/x+4=32/x^2-16
IF you mean x/(x-4)-4/(x+4)=32/(x^2-16) then:
[x(x+4)-4(x-4)]/(x-4)(x+4)=32/(x+4)(x-4) cancel out the denominators
[x^2+4x-4x+16]=32
x^2=32-16
x^2=16
x=sqrt16
x=+-4
Proof: is not possible because of dividing by zero.