Simplification of the above method.

We had the step $\dfrac{2\times 9}{3\times4}.$ If there are common factors between the numerator and denominator, cancel them before multiplying the terms in numerator and denominator. Thus $\dfrac{2\times9}{3\times4}=\dfrac{1*(2)\times3*(3)}{1*(3)\times2*(2)}.~\\ \dfrac{(2)\times(3)}{(3)\times(2)}=1.~~~~\\ \therefore ~\dfrac{1*(2)\times3*(3)}{1*(3)\times2*(2)}= \dfrac{1\times3}{1\times2}=\dfrac{3}{2}.$

#Algebra

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Niranjan Khanderia Why are you making it so complex. You can do it fast by directly cancelling.

Anuj Shikarkhane - 6 years, 7 months ago

You are right. However for those who can not do that. I have come across beginners that need this.

Niranjan Khanderia - 6 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$