How do you simplify #(r + 9)/4 + (r - 3)/2#?
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In a fraction the top number is the count of what have got. The bottom number is the size indicator of what you are counting.
So #3/2# is stating that you have a count of three of something and that it takes 2 of what you are counting to make a whole of something.
#color(brown)("You can not directly add or subtract the counts unless the size indicator is the same".#
#("count")/("size indicator") ->("numerator")/("denominator")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:#" "(r+9)/4+(r-3)/2#
To be able to directly add the 'counts' make the size indicator of #(r-3)/2# the same as that for #(r+9)/4#
Consider: #(r-3)/2#
Multiply by 1 but in the form of #1=2/2# giving:
#(r+9)/4+[(r-3)/2xx2/2]#
#=(r+9)/4+(2r-6)/4#
Write as:
#(r+9+2r-6)/4#
#=(3r+3)/4#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Factor out the 3 giving:
#(3(r+1))/4" "=" "3/4(r+1)#
