(t+7)(t-9)/(t+3)(t+7)*(t+3)(t+13)/(9-t)(t-13)
and 3m= -9/(m+4)
Thank you!
look to cancel things. start off just from the top and bottom of the two fractions.$\displaystyle \frac{(t+7)(t-9)}{(t+3)(t+7)}\cdot \frac{(t+3)(t+13)}{(9-t)(t-13)}$
you'll see: $\displaystyle \frac{\rlap{--------}(t+7)(t-9)}{(t+3)\rlap{--------}(t+7)}\cdot \frac{(t+3)(t+13)}{(9-t)(t-13)}$
so now you're left with: $\displaystyle \frac{(t-9)}{(t+3)}\cdot \frac{(t+3)(t+13)}{(9-t)(t-13)}$
now you can do some cross canceling: $\displaystyle \frac{(t-9)}{\rlap{--------}(t+3)}\cdot \frac{(\rlap{--------}t+3)(t+13)}{(9-t)(t-13)}$
so you're left with: $\displaystyle \frac{(t-9)(t+13)}{(9-t)(t-13)}$
no more canceling you can do. so just multiply out the top and bottom of the fraction.
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Get rid of that fraction on the right.$\displaystyle 3m = - \frac{9}{(m + 4)}$
Just simply multiply by $\displaystyle m + 4$: $\displaystyle (3m)(m + 4) = - 9$
Now just multiply the left hand side and get it equal to zero.
Then you're solving for m like a normal quadratic problem.
Keep in mind that m cannon be - 4 as it would make the bottom of the fraction zero in the original problem.
Hello, ep78!
It's already factored for you . . . Can't you cancel ??
In the bottom of the second fravtion, we factor: .$\displaystyle 9 - t \:=\:-(t-9)$$\displaystyle \frac{(t+7)(t-9)}{(t+3)(t+7)}\cdot\frac{(t+3)(t+13)}{(9-t)(t-13)}$
We have: . $\displaystyle \frac{(t+7)((t-9)}{(t+3)(t+7)} \cdot\frac{(t+3)(t+13)}{-(t-9)(t-13)} $
And reduce: .$\displaystyle \frac{{\color{red}\rlap{/////}}(t+7){\color{blue}\rlap{/////}}(t-9)}{{\color{green}\rlap{/////}}(t+3){\color{red}\rlap{/////}}(t+7)} \cdot\frac{{\color{green}\rlap{/////}}(t+3)(t+13)}{-{\color{blue}\rlap{/////}}(t-9)(t-13)} \;\;=\;\;-\frac{t+13}{t-13} $