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

1 Answer
Mar 3, 2017

#(4x)/(5)#

Explanation:

#(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.