How do you simplify and find the excluded value of #(2b )/ (6b-2)#?

1 Answer
Jun 16, 2017

#b/(3b-1)#

Explanation:

#"factorise the denominator by taking out 2 as a "#
#color(blue)"common factor"#

#rArr(2b)/(2(3b-1))#

#=(cancel(2)^1 b)/(cancel(2)^1(3b-1))larr" cancelling"#

#"excluded values are any values of b that make the "#
#"denominator equal to zero as this would make the"#
#"fraction "color(blue)"undefined"#

#rArr"solve " 3b-1=0rArrb=1/3#

#rArrb=1/3" is the excluded value"#