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

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Answer 1 · 2018-04-12T05:39:50

${(t + 6)}^{2}$ $(t+6)^2$

$\frac{t}{t + 4} = - \frac{9}{t + 3}$ $t/(t+4)=-9/(t+3)$

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

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

= $t^{2} + 12 t + 36$ $t^2+12t+36$

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

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

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

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

$t + 6 = 0$ $t + 6 = 0$

$t = - 6 twice$ $t = -6 "twice"$

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