Solve by Factoring 2(x+1)+4x=3(2x-1)+8

$2 (x + 1) + 4 x = 3 (2 x - 1) + 8$

Step 1

Move all the expressions to the left side of the equation.

Subtract $3 (2 x - 1)$ from both sides of the equation.

$2 (x + 1) + 4 x - 3 (2 x - 1) = 8$

Subtract $8$ from both sides of the equation.

$2 (x + 1) + 4 x - 3 (2 x - 1) - 8 = 0$

Step 2

Simplify $2 (x + 1) + 4 x - 3 (2 x - 1) - 8$ .

Simplify each term.

Apply the distributive property.

$2 x + 2 \cdot 1 + 4 x - 3 (2 x - 1) - 8 = 0$

Multiply $2$ by $1$ .

$2 x + 2 + 4 x - 3 (2 x - 1) - 8 = 0$

Apply the distributive property.

$2 x + 2 + 4 x - 3 (2 x) - 3 \cdot - 1 - 8 = 0$

Multiply $2$ by $- 3$ .

$2 x + 2 + 4 x - 6 x - 3 \cdot - 1 - 8 = 0$

Multiply $- 3$ by $- 1$ .

$2 x + 2 + 4 x - 6 x + 3 - 8 = 0$

Add $2 x$ and $4 x$ .

$6 x + 2 - 6 x + 3 - 8 = 0$

Combine the opposite terms in $6 x + 2 - 6 x + 3 - 8$ .

Subtract $6 x$ from $6 x$ .

$0 + 2 + 3 - 8 = 0$

Add $0$ and $2$ .

$2 + 3 - 8 = 0$

Add $2$ and $3$ .

$5 - 8 = 0$

Subtract $8$ from $5$ .

$- 3 = 0$

Step 3

Since $- 3 \neq 0$ , there are no solutions.

No solution