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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?

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    For the second: you know what $3/12$ is in lowest terms, and that $a/a=1$ and that $b^2/b=b$...2011-08-30

2 Answers 2

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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:

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

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