How do you solve #\frac { t + 4} { t + 6} = \frac { t + 3} { t + 7}#?

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Answer 1 · 2017-06-27T14:11:22

You cross-multiply: #A/B=C/D->AxxD=BxxC#

#(t+4)(t+7)=(t+6)(t+3)->#

FOIL:

#t^2+7t+4t+28=t^2+6t+3t+18->#

Bring all #t#'s to one side, the numbers to the other
(note that the #t^2#'s cancel out):

#7t+4t-6t-3t=18-28->2t=-10->t=-5#

Check into the original equation:

#(-5+4)/(-5+6)=(-5+3)/(-5+7)->(-1)/1=(-2)/2# Check!