# 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?

• tutorial.math.lamar.edu/Classes/Alg/RationalExpressions.aspx Aug 30, 2011 at 15:09
• For the first: factor out the 2 so you can see what to cancel. Aug 30, 2011 at 15:09
• For the second: you know what $3/12$ is in lowest terms, and that $a/a=1$ and that $b^2/b=b$... Aug 30, 2011 at 15:10

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}

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

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! :)