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:

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

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

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