Solve for y 2(y+1)+4y=3(2y-1)+8

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

Simplify $2 (y + 1) + 4 y$ .

Simplify each term.

Apply the distributive property.

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

Multiply $2$ by $1$ .

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

Add $2 y$ and $4 y$ .

$6 y + 2 = 3 (2 y - 1) + 8$

Simplify $3 (2 y - 1) + 8$ .

Simplify each term.

Apply the distributive property.

$6 y + 2 = 3 (2 y) + 3 \cdot - 1 + 8$

Multiply $2$ by $3$ .

$6 y + 2 = 6 y + 3 \cdot - 1 + 8$

Multiply $3$ by $- 1$ .

$6 y + 2 = 6 y - 3 + 8$

Add $- 3$ and $8$ .

$6 y + 2 = 6 y + 5$

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

Subtract $6 y$ from both sides of the equation.

$6 y + 2 - 6 y = 5$

Combine the opposite terms in $6 y + 2 - 6 y$ .

Subtract $6 y$ from $6 y$ .

$0 + 2 = 5$

Add $0$ and $2$ .

$2 = 5$

Since $2 \neq 5$ , there are no solutions.

No solution