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!

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