Need help solving a college algebra equation

2x+1 + 16 = 3x
____
3

( the 2x+1 is over the 3)

Can someone show me the steps to this? I'm really rusty at these and my book doesn't give examples for this one.
Thank you!!

[(2x + 1) / 3] + 16 = 3x
(2x + 1) / 3 = 3x - 16 subtract 16 from each side
2x + 1 = 9x - 48 multiply each side by 3
2x + 1 + 48 = 9x add 48 to each side
1 + 48 = 9x - 2x subtract 2x from each side
49 = 7x simplify each side by doing arithmetic
49 / 7 = x divide each side by 7
7 = x perform the division on the left side
x = 7 by symmetry

(2x+1)/3 + 16 = 3x multiply everything by 3
2x+1 + 48 = 9x now collec like terms
2x-2x + 1 + 48 = 9x - 2x
49 = 7x
x= 7

check (2(7)+1)/3 + 16 = 3(7)
(14+1)/3 + 16 = 21
15/3 + 16 = 21
5+16 = 21
21= 21

simply--you have to multiply everything by 3

3 X (2x+1) + 3 X 16 = 3 X 3x
________
3

the first 3s cancel out and the next 3s have to be multiplied

Multiply through by 3 to clear the fractions

2x + 1 + 48 = 9x

subtract 2x from both sides

7x = 49

Divide both sides by 7

x = 7

take Lcm of 2x +1 and 16
u would get (2x+1 + 48) / 3 = 3x
take the 3 in denominator on right hand side by mulptiplying it with 3x
we get 2x+1+48=9x
s0 7x=49
x=7..

Hope this helped..:)

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