Solve 0=9/I-/xzfrac{z{1}}

$$0 = \frac { 9 } { I } - \frac { } { x z \frac { z } { 1 } }$$

Solve for I

$$I=9xz^{2}$$

$$x\neq 0\text{ and }z\neq 0$$

Steps for Solving Linear Equation

Variable I cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by I.

$$0=9-Ix^{-1}z^{-2}$$

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

$$9-Ix^{-1}z^{-2}=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-Ix^{-1}z^{-2}=-9$$

Reorder the terms.

$$-z^{-2}\times \frac{1}{x}I=-9$$

Divide both sides by -1.

$$z^{-2}\times \frac{1}{x}I=\frac{-9}{-1}$$

Fraction \frac{-9}{-1} can be simplified to 9 by removing the negative sign from both the numerator and the denominator.

$$z^{-2}\times \frac{1}{x}I=9$$

Multiply both sides of the equation by x.

$$z^{-2}\times 1I=9x$$

Reorder the terms.

$$z^{-2}I=9x$$

The equation is in standard form.

$$\frac{1}{z^{2}}I=9x$$

Divide both sides by z^{-2}.

$$\frac{\frac{1}{z^{2}}Iz^{2}}{1}=\frac{9xz^{2}}{1}$$

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

$$I=\frac{9xz^{2}}{1}$$

Divide 9x by z^{-2}.

$$I=9xz^{2}$$

Variable I cannot be equal to 0.

$$I=9xz^{2}\text{, }I\neq 0$$

Solve for x

$$x=\frac{I}{9z^{2}}$$

$$z\neq 0\text{ and }I\neq 0$$