How do you combine #(6b)/(b-4)-1/(b+1)#?

1 Answer
Nov 23, 2017

#(6b(b+1))/((b+1)(b-4))-(b-4)/((b+1)(b-4))=color(blue)((6b^2+5b+4)/((b+1)(b-4))#
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Explanation:

Combine:

#(6b)/(b-4)-(1)/(b+1)#

In order to add or subtract fractions, they must have the same denominator. For denominators that are binomials, we can multiply both binomials by a form of #1# that will give them the same denominator. For example: #(x-9)/(x-9)=1#.

Multiply both binomials so that they will have the same denominator, #(b+1)(b-4)#.

#(6b)/(b-4)xxcolor(teal)((b+1)/(b+1))-(1)/(b+1)xxcolor(magenta)((b-4)/(b-4))#

Simplify.

#(6b(b+1))/((b+1)(b-4))-(b-4)/((b+1)(b-4))#

Combine.

#(6b(b+1)-(b-4))/((b+1)(b-4))#

Simplify.

#(6b^2+6b-b+4)/((b+1)(b-4))#

Simplify.

#(6b^2+5b+4)/((b+1)(b-4))#