Solve 3x/5+4/x-2x/y=2

$$\frac { 3 x } { 5 } + \frac { 4 } { x } - \frac { 2 x } { y } = 2$$

Solve for y (complex solution)

$$y=-\frac{10x^{2}}{-3x^{2}+10x-20}$$

$$x\neq 0\text{ and }x\neq \frac{5+\sqrt{35}i}{3}\text{ and }x\neq \frac{-\sqrt{35}i+5}{3}$$

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 5xy, the least common multiple of 5,x,y.

$$xy\times 3x+5y\times 4-5x\times 2x=10xy$$

Multiply x and x to get x^{2}.

$$x^{2}y\times 3+5y\times 4-5x\times 2x=10xy$$

Multiply 5 and 4 to get 20.

$$x^{2}y\times 3+20y-5x\times 2x=10xy$$

Multiply x and x to get x^{2}.

$$x^{2}y\times 3+20y-5x^{2}\times 2=10xy$$

Multiply 5 and 2 to get 10.

$$x^{2}y\times 3+20y-10x^{2}=10xy$$

Subtract 10xy from both sides.

$$x^{2}y\times 3+20y-10x^{2}-10xy=0$$

Add 10x^{2} to both sides. Anything plus zero gives itself.

$$x^{2}y\times 3+20y-10xy=10x^{2}$$

Combine all terms containing y.

$$\left(x^{2}\times 3+20-10x\right)y=10x^{2}$$

The equation is in standard form.

$$\left(3x^{2}-10x+20\right)y=10x^{2}$$

Divide both sides by 3x^{2}-10x+20.

$$\frac{\left(3x^{2}-10x+20\right)y}{3x^{2}-10x+20}=\frac{10x^{2}}{3x^{2}-10x+20}$$

Dividing by 3x^{2}-10x+20 undoes the multiplication by 3x^{2}-10x+20.

$$y=\frac{10x^{2}}{3x^{2}-10x+20}$$

Variable y cannot be equal to 0.

$$y=\frac{10x^{2}}{3x^{2}-10x+20}\text{, }y\neq 0$$

Solve for y

$$y=-\frac{10x^{2}}{-3x^{2}+10x-20}$$

$$x\neq 0$$

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 5xy, the least common multiple of 5,x,y.

$$xy\times 3x+5y\times 4-5x\times 2x=10xy$$

Multiply x and x to get x^{2}.

$$x^{2}y\times 3+5y\times 4-5x\times 2x=10xy$$

Multiply 5 and 4 to get 20.

$$x^{2}y\times 3+20y-5x\times 2x=10xy$$

Multiply x and x to get x^{2}.

$$x^{2}y\times 3+20y-5x^{2}\times 2=10xy$$

Multiply 5 and 2 to get 10.

$$x^{2}y\times 3+20y-10x^{2}=10xy$$

Subtract 10xy from both sides.

$$x^{2}y\times 3+20y-10x^{2}-10xy=0$$

Add 10x^{2} to both sides. Anything plus zero gives itself.

$$x^{2}y\times 3+20y-10xy=10x^{2}$$

Combine all terms containing y.

$$\left(x^{2}\times 3+20-10x\right)y=10x^{2}$$

The equation is in standard form.

$$\left(3x^{2}-10x+20\right)y=10x^{2}$$

Divide both sides by 3x^{2}-10x+20.

$$\frac{\left(3x^{2}-10x+20\right)y}{3x^{2}-10x+20}=\frac{10x^{2}}{3x^{2}-10x+20}$$

Dividing by 3x^{2}-10x+20 undoes the multiplication by 3x^{2}-10x+20.

$$y=\frac{10x^{2}}{3x^{2}-10x+20}$$

Variable y cannot be equal to 0.

$$y=\frac{10x^{2}}{3x^{2}-10x+20}\text{, }y\neq 0$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{5}\left(\sqrt{y\left(40-7y\right)}+\sqrt{5}y\right)}{3y-10}\text{; }x=\frac{\sqrt{5}\left(-\sqrt{y\left(40-7y\right)}+\sqrt{5}y\right)}{3y-10}\text{, }&y\neq \frac{10}{3}\text{ and }y\neq 0\\x=2\text{, }&y=\frac{10}{3}\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{5y}\left(\sqrt{40-7y}+\sqrt{5y}\right)}{3y-10}\text{; }x=\frac{\sqrt{5y}\left(-\sqrt{40-7y}+\sqrt{5y}\right)}{3y-10}\text{, }&y\neq \frac{10}{3}\text{ and }y\leq \frac{40}{7}\text{ and }y>0\\x=2\text{, }&y=\frac{10}{3}\end{matrix}\right.$$