How do you simplify #(20x^2y)/(25xy)# and find the excluded values?

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Answer 1 · 2017-03-03T10:08:22

#(4x)/(5)#

#(20x^2y)/(25xy)#

#(4x^2y)/(5xy)#-----------Cancel out by 5.

#(4xy)/(5y)#----------------Cancel out #x#

#(4x)/(5)#-------------------Cancel out #y#

We get simplified form,

#(4x)/(5)#

Here, excluded values are:
#x!in0 & y !in0#

Because, the denominator becomes zero when # x or y =0#.
And 0 in the denominator gives #oo#. While, #oo# is not-defined number.