What is the answer to (4x+3)/6-2x+(3x+1)/4+1=0 ?

The answer to (4x+3)/6-2x+(3x+1)/4+1=0 is x=3

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(4x+3)/6-2x+(3x+1)/4+1=0

\frac{4 x + 3}{6} - 2 x + \frac{3 x + 1}{4} + 1 = 0

x = 3

Solution steps

\frac{4 x + 3}{6} - 2 x + \frac{3 x + 1}{4} + 1 = 0

Multiply by LCM

2 (4 x + 3) - 24 x + 3 (3 x + 1) + 12 = 0

Expand

2 (4 x + 3) - 24 x + 3 (3 x + 1) + 12 : - 7 x + 21

- 7 x + 21 = 0

Move

21

to the right side

- 7 x = - 21

Divide both sides by

- 7

x = 3

- 3 (x - 1) + 7 x = 27

\frac{x}{5} = 16

3 x - 11 + 4 x + 2 = - x + 47

\frac{2 x - 1}{2} - \frac{x + 13}{24} = 3 x + \frac{5 ( x + 1 )}{8}

2 x - 4 = - \frac{1}{3}

What is the answer to (4x+3)/6-2x+(3x+1)/4+1=0 ?
The answer to (4x+3)/6-2x+(3x+1)/4+1=0 is x=3
Solve for x: (4x+3)/6-2x+(3x+1)/4+1=0
The solution is x=3