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

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

Add $3 y$ and $y$ .

$4 y - 2 = 2 (2 y - 1)$

Simplify $2 (2 y - 1)$ .

Apply the distributive property.

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

Multiply.

Multiply $2$ by $2$ .

$4 y - 2 = 4 y + 2 \cdot - 1$

Multiply $2$ by $- 1$ .

$4 y - 2 = 4 y - 2$

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

Subtract $4 y$ from both sides of the equation.

$4 y - 2 - 4 y = - 2$

Combine the opposite terms in $4 y - 2 - 4 y$ .

Subtract $4 y$ from $4 y$ .

$0 - 2 = - 2$

Subtract $2$ from $0$ .

$- 2 = - 2$

Since $- 2 = - 2$ , the equation will always be true for any value of $y$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$