Solve 5/z+2+9=kquad[z]

$$\frac { 5 } { z + 2 } + 9 = k \quad [ z ]$$

Solve for k

$$k=\frac{9z+23}{z\left(z+2\right)}$$

$$z\neq -2\text{ and }z\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by z+2.

$$5+\left(z+2\right)\times 9=kz\left(z+2\right)$$

Use the distributive property to multiply z+2 by 9.

$$5+9z+18=kz\left(z+2\right)$$

Add 5 and 18 to get 23.

$$23+9z=kz\left(z+2\right)$$

Use the distributive property to multiply kz by z+2.

$$23+9z=kz^{2}+2kz$$

Swap sides so that all variable terms are on the left hand side.

$$kz^{2}+2kz=23+9z$$

Combine all terms containing k.

$$\left(z^{2}+2z\right)k=23+9z$$

The equation is in standard form.

$$\left(z^{2}+2z\right)k=9z+23$$

Divide both sides by z^{2}+2z.

$$\frac{\left(z^{2}+2z\right)k}{z^{2}+2z}=\frac{9z+23}{z^{2}+2z}$$

Dividing by z^{2}+2z undoes the multiplication by z^{2}+2z.

$$k=\frac{9z+23}{z^{2}+2z}$$

Divide 9z+23 by z^{2}+2z.

$$k=\frac{9z+23}{z\left(z+2\right)}$$

Solve for z (complex solution)

$$\left\{\begin{matrix}z=-\frac{\sqrt{4k^{2}+56k+81}+2k-9}{2k}\text{; }z=-\frac{-\sqrt{4k^{2}+56k+81}+2k-9}{2k}\text{, }&k\neq 0\\z=-\frac{23}{9}\text{, }&k=0\end{matrix}\right.$$

Solve for z

$$\left\{\begin{matrix}z=-\frac{\sqrt{4k^{2}+56k+81}+2k-9}{2k}\text{; }z=-\frac{-\sqrt{4k^{2}+56k+81}+2k-9}{2k}\text{, }&k\leq -\frac{\sqrt{115}}{2}-7\text{ or }\left(k\neq 0\text{ and }k\geq \frac{\sqrt{115}}{2}-7\right)\\z=-\frac{23}{9}\text{, }&k=0\end{matrix}\right.$$