Solve e^x_{0}(1-x)-1=2-x

$$e ^ { x _ { 0 } } ( 1 - x ) - 1 = 2 - x$$

Solve for x

$$x=-\frac{e^{x_{0}}-3}{1-e^{x_{0}}}$$

$$x_{0}\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply e^{x_{0}} by 1-x.

$$e^{x_{0}}-e^{x_{0}}x-1=2-x$$

Add x to both sides.

$$e^{x_{0}}-e^{x_{0}}x-1+x=2$$

Subtract e^{x_{0}} from both sides.

$$-e^{x_{0}}x-1+x=2-e^{x_{0}}$$

Add 1 to both sides.

$$-e^{x_{0}}x+x=2-e^{x_{0}}+1$$

Add 2 and 1 to get 3.

$$-e^{x_{0}}x+x=3-e^{x_{0}}$$

Combine all terms containing x.

$$\left(-e^{x_{0}}+1\right)x=3-e^{x_{0}}$$

The equation is in standard form.

$$\left(1-e^{x_{0}}\right)x=3-e^{x_{0}}$$

Divide both sides by -e^{x_{0}}+1.

$$\frac{\left(1-e^{x_{0}}\right)x}{1-e^{x_{0}}}=\frac{3-e^{x_{0}}}{1-e^{x_{0}}}$$

Dividing by -e^{x_{0}}+1 undoes the multiplication by -e^{x_{0}}+1.

$$x=\frac{3-e^{x_{0}}}{1-e^{x_{0}}}$$

Solve for x_0

$$x_{0}=\ln(\frac{3-x}{1-x})$$

$$x<1\text{ or }x>3$$