How do you simplify and find the restrictions for #(x^2+x)/(x^2+2x)#?

2 Answers
Jul 14, 2017

Answer:

#(x+1)/(x+2); x!=-2, 0#

Explanation:

#(x^2+x)/(x^2+2x)#

#=(cancelx(x+1))/(cancelx(x+2))# #-># factor and cancel

#=(x+1)/(x+2)#

The restrictions consist of #x#-values that make the denominator zero.

Set the original expression's denominator equal to #0#:

#x^2+2x != 0#
#x(x+2)!= 0#
#x!= 0# and #x+2!= 0 => x!=-2#

Set the simplified expression's denominator equal to #0#:
#x+2!=0#
#x!=-2#
This is the same answer as above, but sometimes there are additional restricted values that result from the simplified expression.

#(x+1)/(x+2); x!=-2, 0#

Jul 14, 2017

Answer:

#(x+1)/(x+2)# [#x!=0# and #x!=-2#]

Explanation:

Expression #=(x^2+x)/(x^2+2x)#

#= (x(x+1))/(x(x+2))#

If #x!=0#

Expression #=(x+1)/(x+2)#

Hence, Expression is defined #forall x in RR != -2#

#:.#Expression #=(x+1)/(x+2)# [#x!=0# and #x!=-2#]