Solve for x (2+6x)/2=3x+3

$\frac{2 + 6 x}{2} = 3 x + 3$

Cancel the common factor of $2 + 6 x$ and $2$ .

Factor $2$ out of $2$ .

$\frac{2 \cdot 1 + 6 x}{2} = 3 x + 3$

Factor $2$ out of $6 x$ .

$\frac{2 \cdot 1 + 2 (3 x)}{2} = 3 x + 3$

Factor $2$ out of $2 (1) + 2 (3 x)$ .

$\frac{2 (1 + 3 x)}{2} = 3 x + 3$

Cancel the common factors.

Factor $2$ out of $2$ .

$\frac{2 (1 + 3 x)}{2 (1)} = 3 x + 3$

Cancel the common factor.

$\frac{2 (1 + 3 x)}{2 \cdot 1} = 3 x + 3$

Rewrite the expression.

$\frac{1 + 3 x}{1} = 3 x + 3$

Divide $1 + 3 x$ by $1$ .

$1 + 3 x = 3 x + 3$

Solve for $x$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $3 x$ from both sides of the equation.

$1 + 3 x - 3 x = 3$

Combine the opposite terms in $1 + 3 x - 3 x$ .

Subtract $3 x$ from $3 x$ .

$1 + 0 = 3$

Add $1$ and $0$ .

$1 = 3$

Since $1 \neq 3$ , there are no solutions.

No solution