Simplifying the rational expressions-Help!

**ep78** · July 31st 2008, 04:40 PM

(t+7)(t-9)/(t+3)(t+7)*(t+3)(t+13)/(9-t)(t-13)

and 3m= -9/(m+4)

Thank you!

**Jonboy** · July 31st 2008, 05:18 PM

$\frac{(t+7)(t-9)}{(t+3)(t+7)}\cdot \frac{(t+3)(t+13)}{(9-t)(t-13)}$

look to cancel things. start off just from the top and bottom of the two fractions.

you'll see: $\frac{\rlap{--------}(t+7)(t-9)}{(t+3)\rlap{--------}(t+7)}\cdot \frac{(t+3)(t+13)}{(9-t)(t-13)}$

so now you're left with: $\frac{(t-9)}{(t+3)}\cdot \frac{(t+3)(t+13)}{(9-t)(t-13)}$

now you can do some cross canceling: $\frac{(t-9)}{\rlap{--------}(t+3)}\cdot \frac{(\rlap{--------}t+3)(t+13)}{(9-t)(t-13)}$

so you're left with: $\frac{(t-9)(t+13)}{(9-t)(t-13)}$

no more canceling you can do. so just multiply out the top and bottom of the fraction.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~

$3m = - \frac{9}{(m + 4)}$

Get rid of that fraction on the right.

Just simply multiply by $m + 4$ : $(3m)(m + 4) = - 9$

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.

**Soroban** · July 31st 2008, 06:27 PM

Hello, ep78!

It's already factored for you . . . Can't you cancel ??

$\frac{(t+7)(t-9)}{(t+3)(t+7)}\cdot\frac{(t+3)(t+13)}{(9-t)(t-13)}$

In the bottom of the second fravtion, we factor: . $9 - t \:=\:-(t-9)$

We have: . $\frac{(t+7)((t-9)}{(t+3)(t+7)} \cdot\frac{(t+3)(t+13)}{-(t-9)(t-13)}$

And reduce: . $\frac{{\color{red}\rlap{/////}}(t+7){\color{blue}\rlap{/////}}(t-9)}{{\color{green}\rlap{/////}}(t+3){\color{red}\rlap{/////}}(t+7)} \cdot\frac{{\color{green}\rlap{/////}}(t+3)(t+13)}{-{\color{blue}\rlap{/////}}(t-9)(t-13)} \;\;=\;\;-\frac{t+13}{t-13}$

**Jonboy** · August 1st 2008, 12:57 PM

I didn't see that Soroban, thanks for giving a more accurate answer.

**Soroban** · August 1st 2008, 03:33 PM

Hello, Jonboy!

It took me an embarrassingly long time to catch the relationship
. . between $a - b$ and $b - a$ .

Now (many years later), it's like a flashing red light.
. . Just one more thing to watch for . . .

**Jonboy** · August 2nd 2008, 06:46 PM

Glad I'm not the only one.

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