Solve sqrt{z-y/4}=1/x

$$\sqrt { \frac { z - y } { 4 } } = \frac { 1 } { x }$$

Solve for x (complex solution)

$$x=2\left(z-y\right)^{-\frac{1}{2}}$$

$$z\neq y$$

Steps for Solving Linear Equation

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

$$x\sqrt{\frac{z-y}{4}}=1$$

Divide each term of z-y by 4 to get \frac{1}{4}z-\frac{1}{4}y.

$$x\sqrt{\frac{1}{4}z-\frac{1}{4}y}=1$$

The equation is in standard form.

$$\sqrt{\frac{z-y}{4}}x=1$$

Divide both sides by \sqrt{\frac{1}{4}z-\frac{1}{4}y}.

$$\frac{\sqrt{\frac{z-y}{4}}x}{\sqrt{\frac{z-y}{4}}}=\frac{1}{\sqrt{\frac{z-y}{4}}}$$

Dividing by \sqrt{\frac{1}{4}z-\frac{1}{4}y} undoes the multiplication by \sqrt{\frac{1}{4}z-\frac{1}{4}y}.

$$x=\frac{1}{\sqrt{\frac{z-y}{4}}}$$

Divide 1 by \sqrt{\frac{1}{4}z-\frac{1}{4}y}.

$$x=2\left(z-y\right)^{-\frac{1}{2}}$$

Variable x cannot be equal to 0.

$$x=2\left(z-y\right)^{-\frac{1}{2}}\text{, }x\neq 0$$

Solve for x

$$x=\frac{2}{\sqrt{z-y}}$$

$$z>y$$

Steps for Solving Linear Equation

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

$$x\sqrt{\frac{z-y}{4}}=1$$

Divide each term of z-y by 4 to get \frac{1}{4}z-\frac{1}{4}y.

$$x\sqrt{\frac{1}{4}z-\frac{1}{4}y}=1$$

The equation is in standard form.

$$\sqrt{\frac{z-y}{4}}x=1$$

Divide both sides by \sqrt{\frac{1}{4}z-\frac{1}{4}y}.

$$\frac{\sqrt{\frac{z-y}{4}}x}{\sqrt{\frac{z-y}{4}}}=\frac{1}{\sqrt{\frac{z-y}{4}}}$$

Dividing by \sqrt{\frac{1}{4}z-\frac{1}{4}y} undoes the multiplication by \sqrt{\frac{1}{4}z-\frac{1}{4}y}.

$$x=\frac{1}{\sqrt{\frac{z-y}{4}}}$$

Divide 1 by \sqrt{\frac{1}{4}z-\frac{1}{4}y}.

$$x=\frac{2}{\sqrt{z-y}}$$

Variable x cannot be equal to 0.

$$x=\frac{2}{\sqrt{z-y}}\text{, }x\neq 0$$

Solve for y

$$y=z-\frac{4}{x^{2}}$$

$$x>0$$

Steps for Solving Radical Equations

Multiply both sides of the equation by x.

$$x\sqrt{\frac{z-y}{4}}=1$$

Divide each term of z-y by 4 to get \frac{1}{4}z-\frac{1}{4}y.

$$x\sqrt{\frac{1}{4}z-\frac{1}{4}y}=1$$

Divide both sides by x.

$$\frac{x\sqrt{-\frac{1}{4}y+\frac{z}{4}}}{x}=\frac{1}{x}$$

Dividing by x undoes the multiplication by x.

$$\sqrt{-\frac{1}{4}y+\frac{z}{4}}=\frac{1}{x}$$

Square both sides of the equation.

$$-\frac{1}{4}y+\frac{z}{4}=\frac{1}{x^{2}}$$

Subtract \frac{1}{4}z from both sides of the equation.

$$-\frac{1}{4}y+\frac{z}{4}-\frac{z}{4}=\frac{1}{x^{2}}-\frac{z}{4}$$

Subtracting \frac{1}{4}z from itself leaves 0.

$$-\frac{1}{4}y=\frac{1}{x^{2}}-\frac{z}{4}$$

Subtract \frac{1}{4}z from \frac{1}{x^{2}}.

$$-\frac{1}{4}y=-\frac{z}{4}+\frac{1}{x^{2}}$$

Multiply both sides by -4.

$$\frac{-\frac{1}{4}y}{-\frac{1}{4}}=\frac{-\frac{z}{4}+\frac{1}{x^{2}}}{-\frac{1}{4}}$$

Dividing by -\frac{1}{4} undoes the multiplication by -\frac{1}{4}.

$$y=\frac{-\frac{z}{4}+\frac{1}{x^{2}}}{-\frac{1}{4}}$$

Divide -\frac{z}{4}+\frac{1}{x^{2}} by -\frac{1}{4} by multiplying -\frac{z}{4}+\frac{1}{x^{2}} by the reciprocal of -\frac{1}{4}.

$$y=z-\frac{4}{x^{2}}$$

Solve for y (complex solution)

$$y=z-\frac{4}{x^{2}}$$

$$arg(\sqrt{\frac{1}{x^{2}}}x)<\pi \text{ and }x\neq 0$$