Solve 1/sqrt{x}=1/2y+1/3z

$$\frac{ 1 }{ \sqrt{ x } } = \frac{ 1 }{ 2y } + \frac{ 1 }{ 3z }$$

Solve for y

$$y=-\frac{3\sqrt{x}z}{2\left(-3z+\sqrt{x}\right)}$$

$$z\neq 0\text{ and }\left(z<0\text{ or }x\neq 9z^{2}\right)\text{ and }x>0$$

Steps for Solving Linear Equation

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 6yz, the least common multiple of 2y,3z.

$$6yzx^{-\frac{1}{2}}=3z+2y$$

Subtract 2y from both sides.

$$6yzx^{-\frac{1}{2}}-2y=3z$$

Combine all terms containing y.

$$\left(6zx^{-\frac{1}{2}}-2\right)y=3z$$

The equation is in standard form.

$$\left(\frac{6z}{\sqrt{x}}-2\right)y=3z$$

Divide both sides by 6zx^{-\frac{1}{2}}-2.

$$\frac{\left(\frac{6z}{\sqrt{x}}-2\right)y}{\frac{6z}{\sqrt{x}}-2}=\frac{3z}{\frac{6z}{\sqrt{x}}-2}$$

Dividing by 6zx^{-\frac{1}{2}}-2 undoes the multiplication by 6zx^{-\frac{1}{2}}-2.

$$y=\frac{3z}{\frac{6z}{\sqrt{x}}-2}$$

Divide 3z by 6zx^{-\frac{1}{2}}-2.

$$y=\frac{3\sqrt{x}z}{2\left(3z-\sqrt{x}\right)}$$

Variable y cannot be equal to 0.

$$y=\frac{3\sqrt{x}z}{2\left(3z-\sqrt{x}\right)}\text{, }y\neq 0$$

Solve for x

$$x=36\times \left(\frac{yz}{2y+3z}\right)^{2}$$

$$\left(y>0\text{ and }y<-\frac{3z}{2}\right)\text{ or }\left(z>0\text{ and }y<-\frac{3z}{2}\right)\text{ or }\left(z>0\text{ and }y>0\right)$$