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"#