What is the solution of (-8/2y-8)=(5/y++8/y^2-16) a) y = –4 b) y = –2 c) y = 4 d) y = 6 not b or c

D) y = 6

Step-by-step explanation:

Add the opposite of the left side:

... 0 = 8/(2y -8) +5/(y +4) -(7y +8)/(y^2 -16)

... 0 = (4(y +4) +5(y -4) -(7y +8)) / (y^2 -16) . . . . use a common denominator

... 0 = 4y +16 +5y -20 -7y -8 . . . . . multiply by (y^2 -16), eliminate parentheses

... 0 = 2y -12 . . . . . collect terms

... 0 = y -6 . . . . . . . divide by 2

... 6 = y . . . . . . . . . add 6

Comment on the solution

Keeping the expression rational and using a common denominator can avoid introducing the extraneous roots that you get when you multiply by y^2-16 at the beginning of the process. That multiplication creates a polynomial that has the roots of y^2-16 along with the solutions to the original equation.

In the above, we have a step "multiply by (y^2 -16)", but that multiplication does not have the effect of adding extraneous roots. We recognize that that step can only be valid for values of y that are not ±4. The equation itself is not defined for y = ±4. Effectively, we're ignoring the denominator and looking only for values of y that make the numerator zero.