Question

How do you multiply `\frac { 3} { 4} \times \frac { 4} { 9} \times \frac { 1} { 8}`?

Answer 1

`1/24`

Explanation:

`a/b\timesc/d\timese/f=(a\timesc\timese)/(b\timesd\timesf)`

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`\therefore 3/4\times4/9\times1/8\rArr(3\times4\times1)/(4\times9\times8)`
simplifying...
`\rarr(12\times1)/(36\times8)=12/288`
`288\div12=24`
`\rarr(12\div12)/(288\div12)=1/24`

Answer 2

`1/24" "` with a lot of explanation

Explanation:

`color(darkorchid)("Preamble")`

Demonstrating a principle by example

`2xx6` give the same answer as `6xx2=12`

So you can swap them around

Also

`1/2xx1/3 = (1xx1)/(2xx3)=(1xx1)/(3xx2)=1/3xx1/2`

So you can swap those around as well.
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`color(darkorchid)("Answering the question")`

Given:`" "color(green)(3/4)xx color(blue)(4/9)xxcolor(red)(1/8)`

What follows is the process that is behind the principle of cancelling out.

Write as:`" "(color(blue)(4))/(color(green)(4)) xx(color(green)(3))/(color(blue)(9))xxcolor(red)(1/8)`

But `4/4=1` giving:

`1xx3/9xx1/8" "larr" in cancelling form this is " 3/(cancel(4))xx(cancel(4))/9xx1/8`

`3/9xx1/8`

`(3-:3)/(9-:3)xx1/8" "larr" in canceling form this is "(cancel(3)^1)/(cancel(9)^3)xx1/8`

`1/3xx1/8 = (1xx1)/(3xx8)= 1/24`