Solve for z Ag((3k)/2-z)=mgz

$A g (\frac{3 k}{2} - z) = mg z$

Simplify $A g (\frac{3 k}{2} - z)$ .

Apply the distributive property.

$z = A g \frac{3 k}{2} + A g (- z)$

Multiply $A g \frac{3 k}{2}$ .

Combine $A$ and $\frac{3 k}{2}$ .

$z = g \frac{A (3 k)}{2} + A g (- z)$

Combine $g$ and $\frac{A (3 k)}{2}$ .

$z = \frac{g (A (3 k))}{2} + A g (- z)$

Rewrite using the commutative property of multiplication.

$z = \frac{g (A (3 k))}{2} - A g z$

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

$z = \frac{3 g A k}{2} - A g z$

Add $A g z$ to both sides of the equation.

$z + A g z = \frac{3 g A k}{2}$

Factor $z$ out of $z + A g z$ .

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

$z \cdot 1 + A g z = \frac{3 g A k}{2}$

Factor $z$ out of $A g z$ .

$z \cdot 1 + z (A g) = \frac{3 g A k}{2}$

Factor $z$ out of $z \cdot 1 + z (A g)$ .

$z (1 + A g) = \frac{3 g A k}{2}$

Divide each term in $z (1 + A g) = \frac{3 g A k}{2}$ by $1 + A g$ and simplify.

Divide each term in $z (1 + A g) = \frac{3 g A k}{2}$ by $1 + A g$ .

$\frac{z (1 + A g)}{1 + A g} = \frac{\frac{3 g A k}{2}}{1 + A g}$

Simplify the left side.

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

Cancel the common factor.

$\frac{z (1 + A g)}{1 + A g} = \frac{\frac{3 g A k}{2}}{1 + A g}$

Divide $z$ by $1$ .

$z = \frac{\frac{3 g A k}{2}}{1 + A g}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$z = \frac{3 g A k}{2} \cdot \frac{1}{1 + A g}$

Multiply $\frac{3 g A k}{2}$ by $\frac{1}{1 + A g}$ .

$z = \frac{3 g A k}{2 (1 + A g)}$