How do you solve #(t+4)/t+3/(t-4)=-16/(t^2-4t)#?

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Answer 1 · 2018-05-03T21:38:15

#t = -3# is the sole solution.

We can assume #t^2-4t = t(t-4) ne 0,# i.e. # t \ne 0 and t ne 4# so we can cancel the common denominator when we get them.

#(t+4)/t+3/(t-4)=-16/(t^2-4t)#

#{(t+4)(t-4) + 3t}/{t(t-4)} = {-16}/{t(t-4)}#

#(t+4)(t-4) + 3t = -16#

#t^2 -16 + 3t = -16 #

#t(t+3) = 0#

#t = 0# was ruled out at the start

#t = -3# is the sole solution

Check:

# {-3+4}/-3 + 3/{-3 -4} = -1/3 -3/7 = -16/21 #

# -16/{(-3)^2 - 4(-3) } = -16/{21} quad sqrt #