Solve y=2-1/2f(x+1)

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

Solve for f (complex solution)

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

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

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

Use the distributive property to multiply -\frac{1}{2}f by x+1.

$$2-\frac{1}{2}fx-\frac{1}{2}f=y$$

Subtract 2 from both sides.

$$-\frac{1}{2}fx-\frac{1}{2}f=y-2$$

Combine all terms containing f.

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

The equation is in standard form.

$$\frac{-x-1}{2}f=y-2$$

Divide both sides by -\frac{1}{2}x-\frac{1}{2}.

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

Dividing by -\frac{1}{2}x-\frac{1}{2} undoes the multiplication by -\frac{1}{2}x-\frac{1}{2}.

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

Divide y-2 by -\frac{1}{2}x-\frac{1}{2}.

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

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{4-f-2y}{f}\text{, }&f\neq 0\\x\in \mathrm{C}\text{, }&y=2\text{ and }f=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

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

Use the distributive property to multiply -\frac{1}{2}f by x+1.

$$2-\frac{1}{2}fx-\frac{1}{2}f=y$$

Subtract 2 from both sides.

$$-\frac{1}{2}fx-\frac{1}{2}f=y-2$$

Add \frac{1}{2}f to both sides.

$$-\frac{1}{2}fx=y-2+\frac{1}{2}f$$

The equation is in standard form.

$$\left(-\frac{f}{2}\right)x=\frac{f}{2}+y-2$$

Divide both sides by -\frac{1}{2}f.

$$\frac{\left(-\frac{f}{2}\right)x}{-\frac{f}{2}}=\frac{\frac{f}{2}+y-2}{-\frac{f}{2}}$$

Dividing by -\frac{1}{2}f undoes the multiplication by -\frac{1}{2}f.

$$x=\frac{\frac{f}{2}+y-2}{-\frac{f}{2}}$$

Divide y-2+\frac{f}{2} by -\frac{1}{2}f.

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

Solve for f

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

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

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

Use the distributive property to multiply -\frac{1}{2}f by x+1.

$$2-\frac{1}{2}fx-\frac{1}{2}f=y$$

Subtract 2 from both sides.

$$-\frac{1}{2}fx-\frac{1}{2}f=y-2$$

Combine all terms containing f.

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

The equation is in standard form.

$$\frac{-x-1}{2}f=y-2$$

Divide both sides by -\frac{1}{2}x-\frac{1}{2}.

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

Dividing by -\frac{1}{2}x-\frac{1}{2} undoes the multiplication by -\frac{1}{2}x-\frac{1}{2}.

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

Divide y-2 by -\frac{1}{2}x-\frac{1}{2}.

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

Solve for x

$$\left\{\begin{matrix}x=\frac{4-f-2y}{f}\text{, }&f\neq 0\\x\in \mathrm{R}\text{, }&y=2\text{ and }f=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

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

Use the distributive property to multiply -\frac{1}{2}f by x+1.

$$2-\frac{1}{2}fx-\frac{1}{2}f=y$$

Subtract 2 from both sides.

$$-\frac{1}{2}fx-\frac{1}{2}f=y-2$$

Add \frac{1}{2}f to both sides.

$$-\frac{1}{2}fx=y-2+\frac{1}{2}f$$

The equation is in standard form.

$$\left(-\frac{f}{2}\right)x=\frac{f}{2}+y-2$$

Divide both sides by -\frac{1}{2}f.

$$\frac{\left(-\frac{f}{2}\right)x}{-\frac{f}{2}}=\frac{\frac{f}{2}+y-2}{-\frac{f}{2}}$$

Dividing by -\frac{1}{2}f undoes the multiplication by -\frac{1}{2}f.

$$x=\frac{\frac{f}{2}+y-2}{-\frac{f}{2}}$$

Divide y-2+\frac{f}{2} by -\frac{1}{2}f.

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