How do you simplify #1+1/2+1/3+1/6+1/6#?

2 Answers
May 12, 2018

1 + #1/2 + 1/3 + 1/6 + 1/6#

=#3/2 + 1/3 + 1/6 + 1/6#

=#11/6 + 1/6 + 1/6#

= 2 + #1/6#

= #13/6#

May 12, 2018

Answer:

#13/6 -> 2 1/6#

Explanation:

#color(blue)("The structure of a fraction is:")#

#("numerator")/("denominator")->("count")/("size indicator of what is being counted")#

#color(brown)("You can not"ul(" DIRECTLY ")"add the counts unless the size indicators are the same")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering the question")#

Multiply by 1 and you do not change the value. However, 1 comes in many forms.

Given: #1+1/2+1/3+1/6+1/6#

#color(white)(".")color(green)([1 color(red)(xx1) ]color(white)("d")+color(white)(".")[1/2color(red)(xx1)]color(white)(".")+[1/3color(red)(xx1)]color(white)("d")+1/6+1/6)#

#color(green)([1 color(red)(xx6/6) ]+[1/2color(red)(xx3/3)]+[1/3color(red)(xx2/2)]+1/6+1/6)#

#color(green)(color(white)("dd")[6/6]color(white)("d")+color(white)("dd")[3/6]color(white)("dd.")+color(white)("dd")[2/6]color(white)("dd")+1/6+1/6)#

Now we can directly add the numerators (counts). This does not change the denominators (size indicators). Note that 'size indicators' behave like units of measurement. In a way thay are just that.

You would not directly add the count of centimetres to the count of inches. You would convert them fist. This is what we have done in this solution.

#(6+3+2+1+1)/6 = 13/6#