Solve 2^2x-1/y^2=c^2

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

Solve for x

$$x=\frac{\left(cy\right)^{2}+1}{4}$$

$$y\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by y^{2}.

$$2^{2}x-1=y^{2}c^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x-1=y^{2}c^{2}$$

Add 1 to both sides.

$$4x=y^{2}c^{2}+1$$

The equation is in standard form.

$$4x=c^{2}y^{2}+1$$

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Solve for c (complex solution)

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

$$c=\frac{\sqrt{4x-1}}{y}\text{, }y\neq 0$$

Solve for c

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

$$c=-\frac{\sqrt{4x-1}}{|y|}\text{, }x\geq \frac{1}{4}\text{ and }y\neq 0$$