Solve 3x+2-e^x=x(3+2/x-e^x/x)

$$3 x + 2 - e ^ { x } = x ( 3 + \frac { 2 } { x } - \frac { e ^ { x } } { x } )$$

Solve for x

$$x\neq 0$$

Short Solution Steps

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$x\left(3x+2\right)-xe^{x}=x\left(3+\frac{2}{x}-\frac{e^{x}}{x}\right)x$$

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

$$x\left(3x+2\right)-xe^{x}=x^{2}\left(3+\frac{2}{x}-\frac{e^{x}}{x}\right)$$

Use the distributive property to multiply x by 3x+2.

$$3x^{2}+2x-xe^{x}=x^{2}\left(3+\frac{2}{x}-\frac{e^{x}}{x}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x}{x}.

$$3x^{2}+2x-xe^{x}=x^{2}\left(\frac{3x}{x}+\frac{2}{x}-\frac{e^{x}}{x}\right)$$

Since \frac{3x}{x} and \frac{2}{x} have the same denominator, add them by adding their numerators.

$$3x^{2}+2x-xe^{x}=x^{2}\left(\frac{3x+2}{x}-\frac{e^{x}}{x}\right)$$

Since \frac{3x+2}{x} and \frac{e^{x}}{x} have the same denominator, subtract them by subtracting their numerators.

$$3x^{2}+2x-xe^{x}=x^{2}\times \frac{3x+2-e^{x}}{x}$$

Express x^{2}\times \frac{3x+2-e^{x}}{x} as a single fraction.

$$3x^{2}+2x-xe^{x}=\frac{x^{2}\left(3x+2-e^{x}\right)}{x}$$

Use the distributive property to multiply x^{2} by 3x+2-e^{x}.

$$3x^{2}+2x-xe^{x}=\frac{3x^{3}+2x^{2}-x^{2}e^{x}}{x}$$

Subtract \frac{3x^{3}+2x^{2}-x^{2}e^{x}}{x} from both sides.

$$3x^{2}+2x-xe^{x}-\frac{3x^{3}+2x^{2}-x^{2}e^{x}}{x}=0$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$x\left(3x^{2}+2x-xe^{x}\right)-\left(3x^{3}+2x^{2}-x^{2}e^{x}\right)=0$$

Use the distributive property to multiply x by 3x^{2}+2x-xe^{x}.

$$3x^{3}+2x^{2}-e^{x}x^{2}-\left(3x^{3}+2x^{2}-x^{2}e^{x}\right)=0$$

To find the opposite of 3x^{3}+2x^{2}-x^{2}e^{x}, find the opposite of each term.

$$3x^{3}+2x^{2}-e^{x}x^{2}-3x^{3}-2x^{2}+x^{2}e^{x}=0$$

Combine 3x^{3} and -3x^{3} to get 0.

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

Combine 2x^{2} and -2x^{2} to get 0.

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

Combine -e^{x}x^{2} and x^{2}e^{x} to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$

Variable x cannot be equal to 0.

$$x\in \mathrm{R}\setminus 0$$