Solve f(y^2)+(2x-1)y_{1}=0

$$f ( y ^ { 2 } ) + ( 2 x - 1 ) y _ { 1 } = 0$$

Solve for f (complex solution)

$$\left\{\begin{matrix}f=-\frac{y_{1}\left(2x-1\right)}{y^{2}}\text{, }&y\neq 0\\f\in \mathrm{C}\text{, }&\left(x=\frac{1}{2}\text{ or }y_{1}=0\right)\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2x-1 by y_{1}.

$$fy^{2}+2xy_{1}-y_{1}=0$$

Subtract 2xy_{1} from both sides. Anything subtracted from zero gives its negation.

$$fy^{2}-y_{1}=-2xy_{1}$$

Add y_{1} to both sides.

$$fy^{2}=-2xy_{1}+y_{1}$$

The equation is in standard form.

$$y^{2}f=y_{1}-2xy_{1}$$

Divide both sides by y^{2}.

$$\frac{y^{2}f}{y^{2}}=\frac{y_{1}-2xy_{1}}{y^{2}}$$

Dividing by y^{2} undoes the multiplication by y^{2}.

$$f=\frac{y_{1}-2xy_{1}}{y^{2}}$$

Divide -2y_{1}x+y_{1} by y^{2}.

$$f=\frac{y_{1}\left(1-2x\right)}{y^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{fy^{2}}{2y_{1}}+\frac{1}{2}\text{, }&y_{1}\neq 0\\x\in \mathrm{C}\text{, }&\left(f=0\text{ or }y=0\right)\text{ and }y_{1}=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2x-1 by y_{1}.

$$fy^{2}+2xy_{1}-y_{1}=0$$

Subtract fy^{2} from both sides. Anything subtracted from zero gives its negation.

$$2xy_{1}-y_{1}=-fy^{2}$$

Add y_{1} to both sides.

$$2xy_{1}=-fy^{2}+y_{1}$$

The equation is in standard form.

$$2y_{1}x=y_{1}-fy^{2}$$

Divide both sides by 2y_{1}.

$$\frac{2y_{1}x}{2y_{1}}=\frac{y_{1}-fy^{2}}{2y_{1}}$$

Dividing by 2y_{1} undoes the multiplication by 2y_{1}.

$$x=\frac{y_{1}-fy^{2}}{2y_{1}}$$

Divide y_{1}-fy^{2} by 2y_{1}.

$$x=-\frac{fy^{2}}{2y_{1}}+\frac{1}{2}$$

Solve for f

$$\left\{\begin{matrix}f=-\frac{y_{1}\left(2x-1\right)}{y^{2}}\text{, }&y\neq 0\\f\in \mathrm{R}\text{, }&\left(x=\frac{1}{2}\text{ or }y_{1}=0\right)\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2x-1 by y_{1}.

$$fy^{2}+2xy_{1}-y_{1}=0$$

Subtract 2xy_{1} from both sides. Anything subtracted from zero gives its negation.

$$fy^{2}-y_{1}=-2xy_{1}$$

Add y_{1} to both sides.

$$fy^{2}=-2xy_{1}+y_{1}$$

The equation is in standard form.

$$y^{2}f=y_{1}-2xy_{1}$$

Divide both sides by y^{2}.

$$\frac{y^{2}f}{y^{2}}=\frac{y_{1}-2xy_{1}}{y^{2}}$$

Dividing by y^{2} undoes the multiplication by y^{2}.

$$f=\frac{y_{1}-2xy_{1}}{y^{2}}$$

Divide -2y_{1}x+y_{1} by y^{2}.

$$f=\frac{y_{1}\left(1-2x\right)}{y^{2}}$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{fy^{2}}{2y_{1}}+\frac{1}{2}\text{, }&y_{1}\neq 0\\x\in \mathrm{R}\text{, }&\left(f=0\text{ or }y=0\right)\text{ and }y_{1}=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2x-1 by y_{1}.

$$fy^{2}+2xy_{1}-y_{1}=0$$

Subtract fy^{2} from both sides. Anything subtracted from zero gives its negation.

$$2xy_{1}-y_{1}=-fy^{2}$$

Add y_{1} to both sides.

$$2xy_{1}=-fy^{2}+y_{1}$$

The equation is in standard form.

$$2y_{1}x=y_{1}-fy^{2}$$

Divide both sides by 2y_{1}.

$$\frac{2y_{1}x}{2y_{1}}=\frac{y_{1}-fy^{2}}{2y_{1}}$$

Dividing by 2y_{1} undoes the multiplication by 2y_{1}.

$$x=\frac{y_{1}-fy^{2}}{2y_{1}}$$

Divide y_{1}-fy^{2} by 2y_{1}.

$$x=-\frac{fy^{2}}{2y_{1}}+\frac{1}{2}$$