How do you simplify #(9x + 10xy) / x#?

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Answer 1 · 2018-07-29T10:31:22

#9+10y#

#"factor numerator and cancel common factor"#

#=(cancel(x)(9+10y))/cancel(x)=9+10y#

Answer 2 · 2018-07-29T10:31:34

#9+10y#

#9x+10xy# can be factorised to become #x(9+10y)#

so #(9x+10xy)/x=(x(9+10y))/x=(cancelx(9+10y))/cancelx=9+10y#

Answer 3 · 2018-07-29T10:35:22

#=9+10y#

#(9x+10xy)/x" "larr# factorise the numerator

#=(cancelx(9+10y))/cancelx" "larr# cancel

#= 9+10y#

You can also write two fractions and simplify each.

#(9x)/x + (10xy)/x#

#=9+10y#