Solve for f
f=-\frac{x}{-2x^{2}+5x-1}
x\neq 0\text{ and }x\neq \frac{\sqrt{17}+5}{4}\text{ and }x\neq \frac{5-\sqrt{17}}{4}
Solve for x (complex solution)
x=-\frac{\sqrt{17f^{2}+10f+1}-5f-1}{4f}
x=\frac{\sqrt{17f^{2}+10f+1}+5f+1}{4f}\text{, }f\neq 0
Solve for x
x=-\frac{\sqrt{17f^{2}+10f+1}-5f-1}{4f}
x=\frac{\sqrt{17f^{2}+10f+1}+5f+1}{4f}\text{, }f\leq \frac{-2\sqrt{2}-5}{17}\text{ or }\left(f\neq 0\text{ and }f\geq \frac{2\sqrt{2}-5}{17}\right)
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\frac{1}{f}x=2x^{2}-5x+1
Reorder the terms.
1x=2x^{2}f-5xf+f
Variable f cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by f.
2x^{2}f-5xf+f=1x
Swap sides so that all variable terms are on the left hand side.
2fx^{2}-5fx+f=x
Reorder the terms.
\left(2x^{2}-5x+1\right)f=x
Combine all terms containing f.
\frac{\left(2x^{2}-5x+1\right)f}{2x^{2}-5x+1}=\frac{x}{2x^{2}-5x+1}
Divide both sides by 2x^{2}-5x+1.
f=\frac{x}{2x^{2}-5x+1}
Dividing by 2x^{2}-5x+1 undoes the multiplication by 2x^{2}-5x+1.
f=\frac{x}{2x^{2}-5x+1}\text{, }f\neq 0
Variable f cannot be equal to 0.
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