How do you add #3/4# and #5/9#?

2 Answers
Feb 12, 2017

Answer:

See the entire explanation below:

Explanation:

To add fractions with different denominators we need to first convert each fraction to have a common denominator.

To do this we can't change the value of either fraction and therefore need to multiply it by some form of #1# or #a/a#.

In the example you provided, #3/4 + 5/9#, we need to get both fractions over a common denominator. In this case the common denominator can be #4 xx 9 = 36#. So, we need to multiply the first fraction by #1# or #9/9# and we need to multiply the second fraction by #1# also, but this time #4/4#.

#3/4 + 5/9# becomes:

#(9/9 xx 3/4) + (4/4 xx 5/9)#

#27/36 + 20/36#

Now that the two fractions are over a common denominator we can add the numerators.

#(27 + 20)/36#

#47/36#

Hope this helps.

Feb 12, 2017

Answer:

Explanation given using your example

Explanation:

A fraction's structure is #("count")/("size indicator")->("numerator")/("denominator")#

The word 'count' speaks for itself.

The word 'size indicator' is a number indicating how many of what you are counting it takes to make a whole 1 of something.

It takes 2 of #1/2# to make 1
It takes 3 of #1/3# to make 1
it takes 45 of #1/45# to make 1 and so on.

By the way; not normally done but you may write whole numbers the same way. That is, for example, 8 may be written as #8/1# as it takes 1 of what you are counting to make 1

#color(brown)("You can not "ul("directly"color(white)(.))"add or subtract the 'counts' in a fraction")#
#color(brown)("unless the size indicators are the same.")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Demonstration using your example")#

#3/4+5/9#

It does not matter what you make the bottom numbers as long as they are the same.

Multiply by 1 and you do not change the value. However, 1 comes in many forms so you can change the way a number looks without changing its value.

#color(green)([3/4color(red)(xx1)] +[5/9color(red)(xx1)] " "=" "[3/4color(red)(xx9/9)] +[5/9color(red)(xx4/4)] )#

#" "=color(green)([27/36]color(white)(.)+color(white)(..)[20/36])#

Now that the size indicators (denominators) are the same you may directly add the counts giving:
#" "color(green)(=(27+20)/36)#

#" "color(green)(=47/36)#

But 47 can be written as 36 + 11 so we can split this as:

#" "color(green)(=36/36+11/36)#

But #36/36=1# giving:

#" "color(green)(=1color(white)(.) 11/36)#