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

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Answer 1 · 2017-06-27T14:11:22

You cross-multiply: $\frac{A}{B} = \frac{C}{D} \to A \times D = B \times C$ $A/B=C/D->AxxD=BxxC$

$(t + 4) (t + 7) = (t + 6) (t + 3) \to$ $(t+4)(t+7)=(t+6)(t+3)->$

FOIL:

$t^{2} + 7 t + 4 t + 28 = t^{2} + 6 t + 3 t + 18 \to$ $t^2+7t+4t+28=t^2+6t+3t+18->$

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

$7 t + 4 t - 6 t - 3 t = 18 - 28 \to 2 t = - 10 \to t = - 5$ $7t+4t-6t-3t=18-28->2t=-10->t=-5$

Check into the original equation:

$\frac{- 5 + 4}{- 5 + 6} = \frac{- 5 + 3}{- 5 + 7} \to \frac{- 1}{1} = \frac{- 2}{2}$ $(-5+4)/(-5+6)=(-5+3)/(-5+7)->(-1)/1=(-2)/2$ Check!

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