simplify (y–2/y8)(1/y–3) please. I cannot figure it out.

**plumbob** · August 24th 2008, 10:00 AM

**skeeter** · August 24th 2008, 10:16 AM

how come I get the feeling that the -2, 8, and -3 are all exponents ?

is this what you mean ?

$\frac{y^{-2}}{y^8} \cdot \frac{1}{y^{-3}}$

**Simplicity** · August 24th 2008, 10:19 AM

Originally Posted by plumbob

simplify (y–2/y8)(1/y–3) please. I cannot figure it out.

For fractions, when multiplying, you mupltiply the numerator with the numerator and the denominator with the denominator.

Is your question:

$\left(\frac{y^{-2}}{y^8}\right)\left(\frac{1}{y^{-3}}\right)$

Or

$\left(\frac{y-2}{8y}\right)\left(\frac{1}{y-3}\right)$

Or

Something else?

**plumbob** · August 24th 2008, 10:38 AM

The Prof. wrote the problem out exactly like that. They may be exponents, I am not sure, thanks for the replys.

Chris

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simplify (y–2/y8)(1/y–3) please. I cannot figure it out.

Those may be exponents,....

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