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I appreciate this is very simple, but I'm experiencing a very basic problem with fractions containing variables and I'd just like to check I'm along the right lines. In the following instance:

$ \frac{2(x+7)(3x+1)}{2} = \frac{2}{2} \cdot \frac{2(x+7)(3x+1)}{1} = (x+7)(3x+1)$

Does the $\frac{2(x+7)(3x+1)}{1}$ have a denominator of 1 because we have factored out the 2 in the prior step?

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You made a mistake, you have:

$$\frac{2(x+7)(3x+1)}{2}=\frac 22\cdot\frac {(x+7)(3x+1)}{1}=1\cdot(x+7)(3x+1)=(x+7)(3x+1)$$

instead of what you wrote.

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  • $\begingroup$ Ah that's interesting - I've copied it from this book: amazon.co.uk/… and they don't have the 1 * part, but that definitely makes sense. In terms of the question, does the denominator become 1 in the second part because I have factored out the two? $\endgroup$ – Hemmed Dec 23 '16 at 13:55
  • $\begingroup$ @Hemmed Yes, absolutely. $\endgroup$ – E. Joseph Dec 23 '16 at 13:59
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    $\begingroup$ Thank you! Just gone through the questions with this in mind and got them right. Thank you :) $\endgroup$ – Hemmed Dec 23 '16 at 14:01

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