Solve for y Mg((3L)/2-y)=mgy

$M g (\frac{3 L}{2} - y) = mg y$

Simplify $M g (\frac{3 L}{2} - y)$ .

Apply the distributive property.

$y = M g \frac{3 L}{2} + M g (- y)$

Multiply $M g \frac{3 L}{2}$ .

Combine $M$ and $\frac{3 L}{2}$ .

$y = g \frac{M (3 L)}{2} + M g (- y)$

Combine $g$ and $\frac{M (3 L)}{2}$ .

$y = \frac{g (M (3 L))}{2} + M g (- y)$

Rewrite using the commutative property of multiplication.

$y = \frac{g (M (3 L))}{2} - M g y$

Move $3$ to the left of $g M$ .

$y = \frac{3 g M L}{2} - M g y$

Add $M g y$ to both sides of the equation.

$y + M g y = \frac{3 g M L}{2}$

Factor $y$ out of $y + M g y$ .

Factor $y$ out of $y^{1}$ .

$y \cdot 1 + M g y = \frac{3 g M L}{2}$

Factor $y$ out of $M g y$ .

$y \cdot 1 + y (M g) = \frac{3 g M L}{2}$

Factor $y$ out of $y \cdot 1 + y (M g)$ .

$y (1 + M g) = \frac{3 g M L}{2}$

Divide each term in $y (1 + M g) = \frac{3 g M L}{2}$ by $1 + M g$ and simplify.

Divide each term in $y (1 + M g) = \frac{3 g M L}{2}$ by $1 + M g$ .

$\frac{y (1 + M g)}{1 + M g} = \frac{\frac{3 g M L}{2}}{1 + M g}$

Simplify the left side.

Cancel the common factor of $1 + M g$ .

Cancel the common factor.

$\frac{y (1 + M g)}{1 + M g} = \frac{\frac{3 g M L}{2}}{1 + M g}$

Divide $y$ by $1$ .

$y = \frac{\frac{3 g M L}{2}}{1 + M g}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$y = \frac{3 g M L}{2} \cdot \frac{1}{1 + M g}$

Multiply $\frac{3 g M L}{2}$ by $\frac{1}{1 + M g}$ .

$y = \frac{3 g M L}{2 (1 + M g)}$