Solve 18(g^2-my^2)/y^2=3

$$\frac { 18 ( g ^ { 2 } - m y ^ { 2 } ) } { y ^ { 2 } } = 3$$

Solve for m

$$m=\left(\frac{g}{y}\right)^{2}-\frac{1}{6}$$

$$y\neq 0$$

Steps for Solving Linear Equation

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

$$18\left(g^{2}-my^{2}\right)=3y^{2}$$

Use the distributive property to multiply 18 by g^{2}-my^{2}.

$$18g^{2}-18my^{2}=3y^{2}$$

Subtract 18g^{2} from both sides.

$$-18my^{2}=3y^{2}-18g^{2}$$

The equation is in standard form.

$$\left(-18y^{2}\right)m=3y^{2}-18g^{2}$$

Divide both sides by -18y^{2}.

$$\frac{\left(-18y^{2}\right)m}{-18y^{2}}=\frac{3y^{2}-18g^{2}}{-18y^{2}}$$

Dividing by -18y^{2} undoes the multiplication by -18y^{2}.

$$m=\frac{3y^{2}-18g^{2}}{-18y^{2}}$$

Divide 3y^{2}-18g^{2} by -18y^{2}.

$$m=\frac{g^{2}}{y^{2}}-\frac{1}{6}$$

Solve for g (complex solution)

$$g=-\frac{\sqrt{6\left(6m+1\right)y^{2}}}{6}$$

$$g=\frac{\sqrt{6\left(6m+1\right)y^{2}}}{6}\text{, }y\neq 0$$

Solve for g

$$g=\frac{\sqrt{36m+6}y}{6}$$

$$g=-\frac{\sqrt{36m+6}y}{6}\text{, }m\geq -\frac{1}{6}\text{ and }y\neq 0$$