Reducing fractions? I want to reduce the two following fractions:
$$
\frac{2x + 2y}{x + y}
$$
$$
\frac{3ab^2}{12ab}
$$
I fully understand the concept of reduce fractions of this type:
$$
\frac{15}{20}
$$
but i do not know what steps to take for reducing fractions like the two above. Anyone that can explain the steps needed, or point me to a website explaining it?
 A: For the first fraction:
$\begin{align}
         \frac{2x + 2y}{x + y}
                &= \frac{2(x + y)}{x + y} \\
                &= 2 \text{ assuming } (x+y) \neq 0 \text{ and dividing both numerator and denominator by (x + y)}

\end{align}$
For the second fraction:
$$\begin{align}
     \frac{3ab^2}{12ab}
     &= \frac{3ab \times b}{3ab \times 4}\\
     &= \frac{b}{4} \quad\text{ assuming } 3ab \neq 0 \text{ and dividing both numerator and denominator by (3ab)}

\end{align}$$
A: HINT $\ \ \ $You need the knowledge of how you multiply two algebraic expressions and the distributive property (also factoring). 
So note how $2x + 2y = 2(x + y)$. For the second, note that you can express the fraction like so:$${3ab^2\over 12ab}= {3ab \times b\over3 ab \times 4}$$
A: For the first example, note that $2x+2y = 2(x+y)$.
So, $\dfrac{2(x+y)}{x+y}= 2$.
Simple enough?
For the second, $\dfrac{3ab^2}{12ab}$, you can do a similar thing. You can cancel like terms by expanding the numerator as follows:
$3ab^2= 3abb= 3(ab)b$.
So, $3(ab)b / 12ab$ means you cancel "$ab$" and you get $3b / 12$.
Also notice that $3$ and $12$ are both divisible by 3. So divide both numerator and denominator by three to get $\dfrac{3}{3}b$ which is $b$, and $12/3$ which is $4$.
So the answer is $b/4$.
Hope this helped! :)
