How do you simplify #(2n)/5 + (-n/6)#?

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Answer 1 · 2016-09-11T21:02:36

#(7n)/30#

Consider #+(-n/6)#

This is like #(+1)xx(-1)xxn/6#

Multiply plus and minus and the answer is minus

So #(+1)xx(-1)xxn/6 = -n/6#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Giving:#" "(2n)/5-n/6#

To be able to directly add or subtract the top numbers of fraction (count) the bottom numbers (size indicators) must be the same.

#("top number")/("bottom number")->("count")/("size indicator")->("numerator")/("denominator")#

So we need to make the bottom numbers the same. I chose 30

#color(brown)("Write as: ")((2n)/5xx1)-(n/6xx1)#

#color(brown)("But 1 comes in many forms")#

#((2n)/5xx6/6)-(n/6xx5/5)#

#(2nxx6)/(5xx6) -(nxx5)/(6xx5)#

#(12n)/30-(5n)/30 color(brown)(larr" Now we can directly subtract the counts")#

#color(brown)("but ")12n-5n= 7n#

#=>(12n)/30-(5n)/30 = (7n)/30#