# Thread: Reduce fraction to lowest terms

1. ## Reduce fraction to lowest terms

This problem confuses me:
1. $\frac{32c^3d^3}{64c^2d}$

I know how to reduce these two:
4. $\frac{2x+6}{3ax+9a}=\frac{2(x+3)}{3a(x+3)}=\frac{2 }{3a}$

9. $\frac{y^2+2y-15}{2y^2-12y+18}=\frac{(y-3)(y+5)}{(y-3)(2y-6)}=\frac{y+5}{2y-6}$

but how can I do that process when there is only one term in the numerator and denominator, like number 1??

2. take out a common factor of 32c^2.d and then cancel those terms.

3. Originally Posted by bryang
I know how to reduce these two:
4. $\frac{2x+6}{3ax+9a}=\frac{2(x+3)}{3a(x+3)}=\frac{2 }{3a}$

9. $\frac{y^2+2y-15}{2y^2-12y+18}=\frac{(y-3)(y+5)}{(y-3)(2y-6)}=\frac{y+5}{2y-6}$

but how can I do that process when there is only one term in the numerator and denominator, like number 1??
interesting. it's like you learned how to run before you can crawl...

This problem confuses me:
1. $\frac{32c^3d^3}{64c^2d}$
think of it this way: $\frac {32c^3d^3}{64c^2d} = \frac {32}{64} \cdot \frac {c^3}{c^2} \cdot \frac {d^3}{d}$

now recall the rule: $\frac {x^a}{x^b} = x^{a - b}$

can you finish?

...you may prefer hmmmm's suggestion, since it is similar to the way you did the others

$

\frac{cd^2}{2}
$

Thanks hmmmm and Jhevon, it makes sense now.

5. Originally Posted by bryang
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