Reducing fractions?

Question

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?

Answer 1

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:

$$\begin{align} \frac{3ab^2}{12ab} &= \frac{3ab \times b}{3ab \times 4}\\ &= \frac{b}{4} \quad\text{ assuming } 3ab \neq 0 \text{ and dividing both numerator and denominator by (3ab)} \end{align}$$

Answer 2

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:$${3ab^2\over 12ab}= {3ab \times b\over3 ab \times 4}$$

Answer 3

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! :)