How do you simplify #b/(b+3)+6/(b-2)#?

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Answer 1 · 2017-04-13T13:32:31

#(b^2+4b +18)/((b+3)(b-2))#

Finding the LCD involves application of LCM;

LCM of #(b+3) and (b-2) = (b+3)(b-2)#

Multiply each fraction as necessary to create equivalent fractions.

#b/(b+3) xx(b-2)/(b-2)+6/(b-2) xx(b+3)/(b+3]#

Now simplify each fraction:

#(b(b-2))/((b+3)(b-2))+(6(b+3))/((b+3)(b-2))#

Now combine into one fraction:

#=(b^2-2b+6b+18)/((b+3)(b-2))#

#=(b^2+4b +18)/((b+3)(b-2))#