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

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Answer 1 · 2018-03-11T14:34:47

#b=5#

We got: #(b+1)/3=2#

Multiplying both sides by #3#, we get

#color(red)cancelcolor(black)3*(b+1)/(color(red)cancelcolor(black)3)=2*3#

#b+1=6#

Subtracting #1# from both sides, we have

#b+color(red)cancelcolor(black)1-color(red)cancelcolor(black)1=6-1#

#b=5#

Answer 2 · 2018-03-11T14:35:12

#b = 5#

#(b+1)/3 =2#

Cross multiply

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

#b+1 = 6#

Subtract #1# on both sides

#b + 1 - 1 = 6 - 1#

#b = 5#