How do you simplify #1/6(3x+5y)+2/3(3/5x-6y)#?

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Answer 1 · 2017-03-25T15:03:49

#= (9x)/10 - 3 1/6y#

Note that there are twp terms, each has brackets which can be removed by using the distributive law.

#1/6(3x+5y) +2/3(3/5x-6y)#

Here are the details of the multiplication and cancelling.

#=(1xxcancel3x)/cancel6_2 + (1xx5x)/6 + (2xxcancel3x)/(cancel3xx5) -(2xxcancel6^2y)/cancel3#

#=1/2x+5/6y +2/5x-4y#

#=(1/2x+2/5x)+(5/6y -4y)" "larr# collect like terms

#=(5x+4x)/10 -3 1/6y" "larr# fraction rules, LCD

#= (9x)/10 - 3 1/6y#