Find all values of $t$ such that $t/(t+4) = -9/(t+3)$ . Help?

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Answer 1 · 2018-04-12T05:39:50

$(t+6)^2$

$t/(t+4)=-9/(t+3)$

= $t(t+3)=-9(t+4)$

= $t^2+3t=-9t-36$

= $t^2+12t+36$

$t^2 + 6t + 6t + 36 = 0$

$(t^2 + 6t) (+6t + 36) = 0$

$t(t + 6) +6(t + 6) = 0$

$(t + 6) ( t + 6) = 0$

$t + 6 = 0$

$t = -6 "twice"$

Find all values of tt such that t/(t+4) = -9/(t+3)t/(t+4) = -9/(t+3). Help?