How do you solve #9-3/(1/3)+1#?

3 Answers
Aug 21, 2016

1

Explanation:

Consider #3/(1/3)#

Write as #3-:1/3#

Turn the #1/3# upside down and multiply

#3xx3/1 = 9#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Putting it all back together

#9-9+1#

#=0+1#

#=1#

May 31, 2017

Explaining why you turn upside down (invert) and multiply when dividing fractions.

#color(magenta)("Explanation part 1 of 2")#

Explanation:

Very important facts:

#color(blue)("Fact 1: ")#A fraction's structure consists of:

#("count")/("size indicator of what you are counting")" "->" "("numerator")/("denominator")#

#color(blue)("Fact 2: ")# You can not DIRECTLY DIVIDE the counts in a fraction unless the 'size indicators' are the same.

#color(blue)("Fact 3: ")# When dividing the counts and the 'size indicators' are not the same the shortcut process includes an adjustment factor. This adjustment factor converts the answer to that which you would obtain if you had made the 'size indicators' the same before you applied the division.

#color(blue)("Fact 4: ")# Multiply by 1 and you do not change an intrinsic value. However, 1 comes in many forms so you can change the way a fraction looks without changing its intrinsic value.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Demonstration of principle")#

Using numbers that we should be familiar with:

Consider #1/2-:1/4 = 2#

Making the adjustment before the division

#color(green)([1/2color(blue)(xx1)]color(magenta)(-:1/4)=2#

#color(green)([1/2color(blue)(xx2/2)]color(magenta)(-:1/4)=2#

#color(green)([2/4]color(magenta)(-:1/4)=2#

Now you can directly divide the top numbers (numerators)

#color(green)(2color(magenta)(-:1)=2#
................................................................
Just for interest lets look at the shortcut method applied to #2/4-:1/4#

Turn the #1/4# upside down and multiply

#" "color(green)(2/4xxcolor(magenta)(4/1) = 2)#

Using the principle that #axxb# is the same answer as #bxxa#
#" "->2xx3=6=3xx2#

Write #color(green)(2/4xxcolor(magenta)(4/1) = 2)# as:

#" "color(green)(2/(color(magenta)(1))xx(color(magenta)(4))/4=2)#

#" "2xx1=2#
#" "color(oliveg)(uarr)#
#color(olive)("Think of dividing the 'size indicators' as an adjustment factor.")#
In this case the adjustment factor is 1 as both of the denominators are the same.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

May 31, 2017

Explaining why you turn upside down (invert) and multiply when dividing fractions.

#color(magenta)("Explanation part 2 of 2")#

Explanation:

Lets look at the shortcut method where the denominators (size indicators) are not the same.

Consider the example: #3/5 -:1/10=6#

It is known that there are #color(brown)(ul(|bar(color(white)(2/2) 2" "|)))# of #1/10# in #1/5#

Applying the rule 'turn upside down and multiply' we have:

#color(green)(3/5-:color(magenta)(1/10) " "=" " 3/5xxcolor(magenta)(10/1))#

Grouping the denominators and numerators we have:

#color(green)(3/(color(magenta)(1))xx(color(magenta)(10))/5) #

So now we have the structure:

#("count"_1)/("count"_2)xx("size indicator"_2)/("size indicator"_1)" "->" "3xxcolor(brown)(ul(|bar(color(white)(2/2) 2" "|)))#

This is: #" division of count"xx"conversion factor"#

Hope this helps.