Solve f(x_{1})-f(x_{2})=x_{1}+1/2x_{1}+1/2-(x_{2}+1/2x_{2}+1/2)

$$f ( x _ { 1 } ) - f ( x _ { 2 } ) = x _ { 1 } + \frac { 1 } { 2 x _ { 1 } } + \frac { 1 } { 2 } - ( x _ { 2 } + \frac { 1 } { 2 x _ { 2 } } + \frac { 1 } { 2 } )$$

Solve for f (complex solution)

$$\left\{\begin{matrix}f=\frac{2x_{1}x_{2}-1}{2x_{1}x_{2}}\text{, }&x_{2}\neq 0\text{ and }x_{1}\neq 0\\f\in \mathrm{C}\text{, }&x_{1}=x_{2}\text{ and }x_{2}\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2x_{1}x_{2}, the least common multiple of 2x_{1},2,2x_{2}.

$$fx_{1}\times 2x_{1}x_{2}-fx_{2}\times 2x_{1}x_{2}=2x_{1}x_{2}\left(x_{1}+\frac{1}{2x_{1}}+\frac{1}{2}\right)-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

Multiply x_{1} and x_{1} to get x_{1}^{2}.

$$fx_{1}^{2}\times 2x_{2}-fx_{2}\times 2x_{1}x_{2}=2x_{1}x_{2}\left(x_{1}+\frac{1}{2x_{1}}+\frac{1}{2}\right)-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

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

$$fx_{1}^{2}\times 2x_{2}-fx_{2}^{2}\times 2x_{1}=2x_{1}x_{2}\left(x_{1}+\frac{1}{2x_{1}}+\frac{1}{2}\right)-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

Multiply -1 and 2 to get -2.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{1}x_{2}\left(x_{1}+\frac{1}{2x_{1}}+\frac{1}{2}\right)-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2x_{1} and 2 is 2x_{1}. Multiply \frac{1}{2} times \frac{x_{1}}{x_{1}}.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{1}x_{2}\left(x_{1}+\frac{1}{2x_{1}}+\frac{x_{1}}{2x_{1}}\right)-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

Since \frac{1}{2x_{1}} and \frac{x_{1}}{2x_{1}} have the same denominator, add them by adding their numerators.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{1}x_{2}\left(x_{1}+\frac{1+x_{1}}{2x_{1}}\right)-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

Use the distributive property to multiply 2x_{1}x_{2} by x_{1}+\frac{1+x_{1}}{2x_{1}}.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+2x_{1}x_{2}\times \frac{1+x_{1}}{2x_{1}}-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

Express 2\times \frac{1+x_{1}}{2x_{1}} as a single fraction.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+\frac{2\left(1+x_{1}\right)}{2x_{1}}x_{1}x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

Cancel out 2 in both numerator and denominator.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+\frac{x_{1}+1}{x_{1}}x_{1}x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2x_{2} and 2 is 2x_{2}. Multiply \frac{1}{2} times \frac{x_{2}}{x_{2}}.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+\frac{x_{1}+1}{x_{1}}x_{1}x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{x_{2}}{2x_{2}}\right)$$

Since \frac{1}{2x_{2}} and \frac{x_{2}}{2x_{2}} have the same denominator, add them by adding their numerators.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+\frac{x_{1}+1}{x_{1}}x_{1}x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1+x_{2}}{2x_{2}}\right)$$

Express \frac{x_{1}+1}{x_{1}}x_{1} as a single fraction.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+\frac{\left(x_{1}+1\right)x_{1}}{x_{1}}x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1+x_{2}}{2x_{2}}\right)$$

Cancel out x_{1} in both numerator and denominator.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+\left(x_{1}+1\right)x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1+x_{2}}{2x_{2}}\right)$$

Use the distributive property to multiply x_{1}+1 by x_{2}.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1+x_{2}}{2x_{2}}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x_{2} times \frac{2x_{2}}{2x_{2}}.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}-2x_{1}x_{2}\left(\frac{x_{2}\times 2x_{2}}{2x_{2}}+\frac{1+x_{2}}{2x_{2}}\right)$$

Since \frac{x_{2}\times 2x_{2}}{2x_{2}} and \frac{1+x_{2}}{2x_{2}} have the same denominator, add them by adding their numerators.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}-2x_{1}x_{2}\times \frac{x_{2}\times 2x_{2}+1+x_{2}}{2x_{2}}$$

Do the multiplications in x_{2}\times 2x_{2}+1+x_{2}.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}-2x_{1}x_{2}\times \frac{2x_{2}^{2}+1+x_{2}}{2x_{2}}$$

Express 2\times \frac{2x_{2}^{2}+1+x_{2}}{2x_{2}} as a single fraction.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}-\frac{2\left(2x_{2}^{2}+1+x_{2}\right)}{2x_{2}}x_{1}x_{2}$$

Cancel out 2 in both numerator and denominator.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}-\frac{2x_{2}^{2}+x_{2}+1}{x_{2}}x_{1}x_{2}$$

Multiply both sides of the equation by x_{2}.

$$fx_{1}^{2}\times 2x_{2}x_{2}-2fx_{2}^{2}x_{1}x_{2}=x_{2}\left(2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}\right)-\left(2x_{2}^{2}+x_{2}+1\right)x_{1}x_{2}$$

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

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{2}x_{1}x_{2}=x_{2}\left(2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}\right)-\left(2x_{2}^{2}+x_{2}+1\right)x_{1}x_{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{3}x_{1}=x_{2}\left(2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}\right)-\left(2x_{2}^{2}+x_{2}+1\right)x_{1}x_{2}$$

Use the distributive property to multiply x_{2} by 2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}.

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{3}x_{1}=2x_{1}^{2}x_{2}^{2}+x_{1}x_{2}^{2}+x_{2}^{2}-\left(2x_{2}^{2}+x_{2}+1\right)x_{1}x_{2}$$

Use the distributive property to multiply -1 by 2x_{2}^{2}+x_{2}+1.

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{3}x_{1}=2x_{1}^{2}x_{2}^{2}+x_{1}x_{2}^{2}+x_{2}^{2}+\left(-2x_{2}^{2}-x_{2}-1\right)x_{1}x_{2}$$

Use the distributive property to multiply -2x_{2}^{2}-x_{2}-1 by x_{1}.

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{3}x_{1}=2x_{1}^{2}x_{2}^{2}+x_{1}x_{2}^{2}+x_{2}^{2}+\left(-2x_{2}^{2}x_{1}-x_{2}x_{1}-x_{1}\right)x_{2}$$

Use the distributive property to multiply -2x_{2}^{2}x_{1}-x_{2}x_{1}-x_{1} by x_{2}.

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{3}x_{1}=2x_{1}^{2}x_{2}^{2}+x_{1}x_{2}^{2}+x_{2}^{2}-2x_{1}x_{2}^{3}-x_{1}x_{2}^{2}-x_{1}x_{2}$$

Combine x_{1}x_{2}^{2} and -x_{1}x_{2}^{2} to get 0.

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{3}x_{1}=2x_{1}^{2}x_{2}^{2}+x_{2}^{2}-2x_{1}x_{2}^{3}-x_{1}x_{2}$$

Combine all terms containing f.

$$\left(x_{1}^{2}\times 2x_{2}^{2}-2x_{2}^{3}x_{1}\right)f=2x_{1}^{2}x_{2}^{2}+x_{2}^{2}-2x_{1}x_{2}^{3}-x_{1}x_{2}$$

The equation is in standard form.

$$\left(2x_{1}^{2}x_{2}^{2}-2x_{1}x_{2}^{3}\right)f=2x_{1}^{2}x_{2}^{2}-2x_{1}x_{2}^{3}-x_{1}x_{2}+x_{2}^{2}$$

Divide both sides by 2x_{1}^{2}x_{2}^{2}-2x_{2}^{3}x_{1}.

$$\frac{\left(2x_{1}^{2}x_{2}^{2}-2x_{1}x_{2}^{3}\right)f}{2x_{1}^{2}x_{2}^{2}-2x_{1}x_{2}^{3}}=\frac{x_{2}\left(x_{1}-x_{2}\right)\left(2x_{1}x_{2}-1\right)}{2x_{1}^{2}x_{2}^{2}-2x_{1}x_{2}^{3}}$$

Dividing by 2x_{1}^{2}x_{2}^{2}-2x_{2}^{3}x_{1} undoes the multiplication by 2x_{1}^{2}x_{2}^{2}-2x_{2}^{3}x_{1}.

$$f=\frac{x_{2}\left(x_{1}-x_{2}\right)\left(2x_{1}x_{2}-1\right)}{2x_{1}^{2}x_{2}^{2}-2x_{1}x_{2}^{3}}$$

Divide x_{2}\left(-1+2x_{1}x_{2}\right)\left(x_{1}-x_{2}\right) by 2x_{1}^{2}x_{2}^{2}-2x_{2}^{3}x_{1}.

$$f=1-\frac{1}{2x_{1}x_{2}}$$

Solve for f

$$\left\{\begin{matrix}f=\frac{2x_{1}x_{2}-1}{2x_{1}x_{2}}\text{, }&x_{2}\neq 0\text{ and }x_{1}\neq 0\\f\in \mathrm{R}\text{, }&x_{1}=x_{2}\text{ and }x_{2}\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2x_{1}x_{2}, the least common multiple of 2x_{1},2,2x_{2}.

$$fx_{1}\times 2x_{1}x_{2}-fx_{2}\times 2x_{1}x_{2}=2x_{1}x_{2}\left(x_{1}+\frac{1}{2x_{1}}+\frac{1}{2}\right)-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

Multiply x_{1} and x_{1} to get x_{1}^{2}.

$$fx_{1}^{2}\times 2x_{2}-fx_{2}\times 2x_{1}x_{2}=2x_{1}x_{2}\left(x_{1}+\frac{1}{2x_{1}}+\frac{1}{2}\right)-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

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

$$fx_{1}^{2}\times 2x_{2}-fx_{2}^{2}\times 2x_{1}=2x_{1}x_{2}\left(x_{1}+\frac{1}{2x_{1}}+\frac{1}{2}\right)-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

Multiply -1 and 2 to get -2.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{1}x_{2}\left(x_{1}+\frac{1}{2x_{1}}+\frac{1}{2}\right)-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2x_{1} and 2 is 2x_{1}. Multiply \frac{1}{2} times \frac{x_{1}}{x_{1}}.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{1}x_{2}\left(x_{1}+\frac{1}{2x_{1}}+\frac{x_{1}}{2x_{1}}\right)-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

Since \frac{1}{2x_{1}} and \frac{x_{1}}{2x_{1}} have the same denominator, add them by adding their numerators.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{1}x_{2}\left(x_{1}+\frac{1+x_{1}}{2x_{1}}\right)-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

Use the distributive property to multiply 2x_{1}x_{2} by x_{1}+\frac{1+x_{1}}{2x_{1}}.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+2x_{1}x_{2}\times \frac{1+x_{1}}{2x_{1}}-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

Express 2\times \frac{1+x_{1}}{2x_{1}} as a single fraction.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+\frac{2\left(1+x_{1}\right)}{2x_{1}}x_{1}x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

Cancel out 2 in both numerator and denominator.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+\frac{x_{1}+1}{x_{1}}x_{1}x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{1}{2}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2x_{2} and 2 is 2x_{2}. Multiply \frac{1}{2} times \frac{x_{2}}{x_{2}}.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+\frac{x_{1}+1}{x_{1}}x_{1}x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1}{2x_{2}}+\frac{x_{2}}{2x_{2}}\right)$$

Since \frac{1}{2x_{2}} and \frac{x_{2}}{2x_{2}} have the same denominator, add them by adding their numerators.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+\frac{x_{1}+1}{x_{1}}x_{1}x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1+x_{2}}{2x_{2}}\right)$$

Express \frac{x_{1}+1}{x_{1}}x_{1} as a single fraction.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+\frac{\left(x_{1}+1\right)x_{1}}{x_{1}}x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1+x_{2}}{2x_{2}}\right)$$

Cancel out x_{1} in both numerator and denominator.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+\left(x_{1}+1\right)x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1+x_{2}}{2x_{2}}\right)$$

Use the distributive property to multiply x_{1}+1 by x_{2}.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}-2x_{1}x_{2}\left(x_{2}+\frac{1+x_{2}}{2x_{2}}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x_{2} times \frac{2x_{2}}{2x_{2}}.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}-2x_{1}x_{2}\left(\frac{x_{2}\times 2x_{2}}{2x_{2}}+\frac{1+x_{2}}{2x_{2}}\right)$$

Since \frac{x_{2}\times 2x_{2}}{2x_{2}} and \frac{1+x_{2}}{2x_{2}} have the same denominator, add them by adding their numerators.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}-2x_{1}x_{2}\times \frac{x_{2}\times 2x_{2}+1+x_{2}}{2x_{2}}$$

Do the multiplications in x_{2}\times 2x_{2}+1+x_{2}.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}-2x_{1}x_{2}\times \frac{2x_{2}^{2}+1+x_{2}}{2x_{2}}$$

Express 2\times \frac{2x_{2}^{2}+1+x_{2}}{2x_{2}} as a single fraction.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}-\frac{2\left(2x_{2}^{2}+1+x_{2}\right)}{2x_{2}}x_{1}x_{2}$$

Cancel out 2 in both numerator and denominator.

$$fx_{1}^{2}\times 2x_{2}-2fx_{2}^{2}x_{1}=2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}-\frac{2x_{2}^{2}+x_{2}+1}{x_{2}}x_{1}x_{2}$$

Multiply both sides of the equation by x_{2}.

$$fx_{1}^{2}\times 2x_{2}x_{2}-2fx_{2}^{2}x_{1}x_{2}=x_{2}\left(2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}\right)-\left(2x_{2}^{2}+x_{2}+1\right)x_{1}x_{2}$$

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

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{2}x_{1}x_{2}=x_{2}\left(2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}\right)-\left(2x_{2}^{2}+x_{2}+1\right)x_{1}x_{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{3}x_{1}=x_{2}\left(2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}\right)-\left(2x_{2}^{2}+x_{2}+1\right)x_{1}x_{2}$$

Use the distributive property to multiply x_{2} by 2x_{2}x_{1}^{2}+x_{1}x_{2}+x_{2}.

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{3}x_{1}=2x_{1}^{2}x_{2}^{2}+x_{1}x_{2}^{2}+x_{2}^{2}-\left(2x_{2}^{2}+x_{2}+1\right)x_{1}x_{2}$$

Use the distributive property to multiply -1 by 2x_{2}^{2}+x_{2}+1.

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{3}x_{1}=2x_{1}^{2}x_{2}^{2}+x_{1}x_{2}^{2}+x_{2}^{2}+\left(-2x_{2}^{2}-x_{2}-1\right)x_{1}x_{2}$$

Use the distributive property to multiply -2x_{2}^{2}-x_{2}-1 by x_{1}.

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{3}x_{1}=2x_{1}^{2}x_{2}^{2}+x_{1}x_{2}^{2}+x_{2}^{2}+\left(-2x_{2}^{2}x_{1}-x_{2}x_{1}-x_{1}\right)x_{2}$$

Use the distributive property to multiply -2x_{2}^{2}x_{1}-x_{2}x_{1}-x_{1} by x_{2}.

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{3}x_{1}=2x_{1}^{2}x_{2}^{2}+x_{1}x_{2}^{2}+x_{2}^{2}-2x_{1}x_{2}^{3}-x_{1}x_{2}^{2}-x_{1}x_{2}$$

Combine x_{1}x_{2}^{2} and -x_{1}x_{2}^{2} to get 0.

$$fx_{1}^{2}\times 2x_{2}^{2}-2fx_{2}^{3}x_{1}=2x_{1}^{2}x_{2}^{2}+x_{2}^{2}-2x_{1}x_{2}^{3}-x_{1}x_{2}$$

Combine all terms containing f.

$$\left(x_{1}^{2}\times 2x_{2}^{2}-2x_{2}^{3}x_{1}\right)f=2x_{1}^{2}x_{2}^{2}+x_{2}^{2}-2x_{1}x_{2}^{3}-x_{1}x_{2}$$

The equation is in standard form.

$$\left(2x_{1}^{2}x_{2}^{2}-2x_{1}x_{2}^{3}\right)f=2x_{1}^{2}x_{2}^{2}-2x_{1}x_{2}^{3}-x_{1}x_{2}+x_{2}^{2}$$

Divide both sides by 2x_{1}^{2}x_{2}^{2}-2x_{2}^{3}x_{1}.

$$\frac{\left(2x_{1}^{2}x_{2}^{2}-2x_{1}x_{2}^{3}\right)f}{2x_{1}^{2}x_{2}^{2}-2x_{1}x_{2}^{3}}=\frac{x_{2}\left(x_{1}-x_{2}\right)\left(2x_{1}x_{2}-1\right)}{2x_{1}^{2}x_{2}^{2}-2x_{1}x_{2}^{3}}$$

Dividing by 2x_{1}^{2}x_{2}^{2}-2x_{2}^{3}x_{1} undoes the multiplication by 2x_{1}^{2}x_{2}^{2}-2x_{2}^{3}x_{1}.

$$f=\frac{x_{2}\left(x_{1}-x_{2}\right)\left(2x_{1}x_{2}-1\right)}{2x_{1}^{2}x_{2}^{2}-2x_{1}x_{2}^{3}}$$

Divide x_{2}\left(-1+2x_{1}x_{2}\right)\left(x_{1}-x_{2}\right) by 2x_{1}^{2}x_{2}^{2}-2x_{2}^{3}x_{1}.

$$f=1-\frac{1}{2x_{1}x_{2}}$$

Solve for x_1

$$\left\{\begin{matrix}x_{1}=-\frac{1}{2x_{2}\left(f-1\right)}\text{, }&f\neq 1\text{ and }x_{2}\neq 0\\x_{1}=x_{2}\text{, }&x_{2}\neq 0\end{matrix}\right.$$