Solve 3/x^2=3/4x+3/x3

$$\frac { 3 } { x ^ { 2 } } = \frac { 3 } { 4 x } + \frac { 3 } { x 3 }$$

Solve for x_3

$$x_{3}=-\frac{4x^{2}}{x-4}$$

$$x\neq 0\text{ and }x\neq 4$$

Steps for Solving Linear Equation

Variable x_{3} cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4x_{3}x^{2}, the least common multiple of x^{2},4x,x_{3}.

$$4x_{3}\times 3=x_{3}x\times 3+4x^{2}\times 3$$

Multiply 4 and 3 to get 12.

$$12x_{3}=x_{3}x\times 3+4x^{2}\times 3$$

Multiply 4 and 3 to get 12.

$$12x_{3}=x_{3}x\times 3+12x^{2}$$

Subtract x_{3}x\times 3 from both sides.

$$12x_{3}-x_{3}x\times 3=12x^{2}$$

Multiply -1 and 3 to get -3.

$$12x_{3}-3x_{3}x=12x^{2}$$

Combine all terms containing x_{3}.

$$\left(12-3x\right)x_{3}=12x^{2}$$

Divide both sides by 12-3x.

$$\frac{\left(12-3x\right)x_{3}}{12-3x}=\frac{12x^{2}}{12-3x}$$

Dividing by 12-3x undoes the multiplication by 12-3x.

$$x_{3}=\frac{12x^{2}}{12-3x}$$

Divide 12x^{2} by 12-3x.

$$x_{3}=\frac{4x^{2}}{4-x}$$

Variable x_{3} cannot be equal to 0.

$$x_{3}=\frac{4x^{2}}{4-x}\text{, }x_{3}\neq 0$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{x_{3}\left(x_{3}+64\right)}-x_{3}}{8}$$

$$x=\frac{\sqrt{x_{3}\left(x_{3}+64\right)}-x_{3}}{8}\text{, }x_{3}\neq 0$$

Solve for x

$$x=\frac{-\sqrt{x_{3}\left(x_{3}+64\right)}-x_{3}}{8}$$

$$x=\frac{\sqrt{x_{3}\left(x_{3}+64\right)}-x_{3}}{8}\text{, }x_{3}>0\text{ or }x_{3}\leq -64$$