How do you simplify #6 / (2/3)#?

2 Answers
Apr 7, 2016

#9#

Explanation:

To divide a number by a fraction, you can multiply it by the inverse of the fraction.

Think of dividing something by #1/2#. There are two halves in every whole, so the number of halves in, say, #3# will be #3xx2=6#. So we can see that #3/(1/2) = 3xx2/1 = 3xx2 = 6#

So, we can put this into action with our calculation:

#6/(2/3) = 6xx3/2 =6xx1.5 = 9 #

Hope this helps; let me know if I can do anything else:)

Apr 7, 2016

#6/(2/3)= 9#

Explanation:

Using the shortcut rule of: to divide turn the divisor upside down (invert) and multiply.

We have #6/(2/3)# which is the same as#" " 6 -: 2/3#

Applying the rule

#6xx3/2#

It is perfectly correct to write 6 as #6/1#. It just not very common to see it.

#6xx3/2" " ->" " 6/1xx3/2" " ->" " (6xx3)/(1xx2) = 18/2#

But #18/2 =9#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("EDIT By Tarik:")#
Any number written as a whole number is actually a fraction: so 6 is #6/1#. Accordingly, let me rewrite the task:

#(6/1)/(2/3)=#
Now, multiply inner part with inner part and external part with external part:
Tarik

When multiplied, the product of multiplication of external parts will stay as nominator, and the product of multiplication of inner parts will stay as denominator.

#(6/1)/(2/3)=18/2=9#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Additional comment by Tony B")#

#color(red)("I will leave Tarik's edit in place as it sometimes useful to see a different approach!")#

Using Tarik's terminology; multiplying the inner numbers together and then the outer numbers together is the same process as inverting and then multiplying.