How do you solve #1/(b+2)+1/(b+2)=3/(b+1)#?

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Answer 1 · 2017-07-13T07:44:07

By arranging equation, you will get #b=-4#

When you arrange your equation, you will get

#(1+1)/(b+2) = 3/(b+1)#

#2/(b+2) = 3/(b+1)#

#2(b+1) = 3(b+2)#

#2b + 2 = 3b + 6#

#2b-3b = 6-2#

#-b = 4#

or

#b=-4#

This is your answer #b=-4#