Need help with two rational expressions

**Skynt** · October 16th 2007, 07:18 PM

I'm having problems with two rational expressions that need to be worked out. I really need a step by step example so that I can do it on the test I'll be having soon.

is it $\frac {8(y - 4)}{20-y}$ with y not equal to -3 for the domain?

$\frac {4y - 16}{5y+15} * \frac {2y + 6}{4-y}$

and

$\frac {x - 13}{x^3(3 - x)} * \frac {x(x - 3)}{5}$

**ThePerfectHacker** · October 16th 2007, 07:23 PM

Originally Posted by Skynt

$\frac {x - 13}{x^3(3 - x)} * \frac {x(x - 3)}{5}$

$-\frac{(x-13)}{x^3(x-3)} \cdot \frac{x(x-3)}{5}$

$- \frac{(x-13)}{5x^2(x-3)}$

**Jhevon** · October 16th 2007, 07:24 PM

Originally Posted by Skynt

I'm having problems with two rational expressions that need to be worked out. I really need a step by step example so that I can do it on the test I'll be having soon.

$\frac {4y - 16}{5y+15} * \frac {2y + 6}{4-y}$

note that we can factor these:

$\frac {4y - 16}{5y+15} \cdot \frac {2y + 6}{4-y} = \frac {4(y - 4)}{5(y+3)} \cdot \frac {2(y + 3)}{4-y}$

can you continue?

and

$\frac {x - 13}{x^3(3 - x)} * \frac {x(x - 3)}{5}$

note that $\frac {x - 13}{x^3(3 - x)} \cdot \frac {x(x - 3)}{5} = \frac {x - 13}{x^3(3 - x)} \cdot \frac {-x(3 - x)}{5}$

can you continue?

**Skynt** · October 16th 2007, 07:29 PM

I'm quite confused lol, can't you cancel out x-3 (sorry) and end up with
$\frac {-(x-13)}{5x^2}$ ?

**Jhevon** · October 16th 2007, 07:34 PM

Originally Posted by Skynt

I'm quite confused lol, can't you cancel out x-3 (sorry) and end up with
$\frac {-(x-13)}{5x^2}$ ?

yes, that is correct

Originally Posted by Skynt

I'm having problems with two rational expressions that need to be worked out. I really need a step by step example so that I can do it on the test I'll be having soon.

is it $\frac {8(y - 4)}{20-y}$ with y not equal to -3 for the domain?

no, it is for $y \ne 20$ , since y = 20 is the problem spot here

**Skynt** · October 16th 2007, 07:47 PM

Sorry for more questions, but I really need to grasp this completely. Thanks for the help so far, but what if I have:
$\frac {2x^2y}{xy - y}$ lost on this one.

**ThePerfectHacker** · October 16th 2007, 07:58 PM

Originally Posted by Skynt

Sorry for more questions, but I really need to grasp this completely. Thanks for the help so far, but what if I have:
$\frac {2x^2y}{xy - y}$ lost on this one.

$\frac{2x^2y}{y(x-1)} = \frac{2x^2}{x-1}$

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