Solve for z 2/5(z+1)=y for z

$\frac{2}{5} (z + 1) = y$ for $z$

Multiply both sides of the equation by $\frac{5}{2}$ .

$\frac{5}{2} (\frac{2}{5} \cdot (z + 1)) = \frac{5}{2} y$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{5}{2} (\frac{2}{5} \cdot (z + 1))$ .

Apply the distributive property.

$\frac{5}{2} (\frac{2}{5} z + \frac{2}{5} \cdot 1) = \frac{5}{2} y$

Combine $\frac{2}{5}$ and $z$ .

$\frac{5}{2} (\frac{2 z}{5} + \frac{2}{5} \cdot 1) = \frac{5}{2} y$

Multiply $\frac{2}{5}$ by $1$ .

$\frac{5}{2} (\frac{2 z}{5} + \frac{2}{5}) = \frac{5}{2} y$

Apply the distributive property.

$\frac{5}{2} \cdot \frac{2 z}{5} + \frac{5}{2} \cdot \frac{2}{5} = \frac{5}{2} y$

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5}{2} \cdot \frac{2 z}{5} + \frac{5}{2} \cdot \frac{2}{5} = \frac{5}{2} y$

Rewrite the expression.

$\frac{1}{2} (2 z) + \frac{5}{2} \cdot \frac{2}{5} = \frac{5}{2} y$

Cancel the common factor of $2$ .

Factor $2$ out of $2 z$ .

$\frac{1}{2} (2 (z)) + \frac{5}{2} \cdot \frac{2}{5} = \frac{5}{2} y$

Cancel the common factor.

$\frac{1}{2} (2 z) + \frac{5}{2} \cdot \frac{2}{5} = \frac{5}{2} y$

Rewrite the expression.

$z + \frac{5}{2} \cdot \frac{2}{5} = \frac{5}{2} y$

Cancel the common factor of $5$ .

Cancel the common factor.

$z + \frac{5}{2} \cdot \frac{2}{5} = \frac{5}{2} y$

Rewrite the expression.

$z + \frac{1}{2} \cdot 2 = \frac{5}{2} y$

Cancel the common factor of $2$ .

Cancel the common factor.

$z + \frac{1}{2} \cdot 2 = \frac{5}{2} y$

Rewrite the expression.

$z + 1 = \frac{5}{2} y$

Simplify the right side.

Combine $\frac{5}{2}$ and $y$ .

$z + 1 = \frac{5 y}{2}$

Subtract $1$ from both sides of the equation.

$z = \frac{5 y}{2} - 1$