Solve x=1+1/p+1^2/2!+1^3/2!

$$x = 1 + \frac { 1 } { p } + \frac { 1 ^ { 2 } } { 2 ! } + \frac { 1 ^ { 3 } } { 2 ! }$$

Solve for p

$$p=-\frac{1}{2-x}$$

$$x\neq 2$$

Steps for Solving Linear Equation

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

$$xp=p+1+\frac{1}{2}p\times 1^{2}+\frac{1}{2}p\times 1^{3}$$

Calculate 1 to the power of 2 and get 1.

$$xp=p+1+\frac{1}{2}p\times 1+\frac{1}{2}p\times 1^{3}$$

Multiply \frac{1}{2} and 1 to get \frac{1}{2}.

$$xp=p+1+\frac{1}{2}p+\frac{1}{2}p\times 1^{3}$$

Combine p and \frac{1}{2}p to get \frac{3}{2}p.

$$xp=\frac{3}{2}p+1+\frac{1}{2}p\times 1^{3}$$

Calculate 1 to the power of 3 and get 1.

$$xp=\frac{3}{2}p+1+\frac{1}{2}p\times 1$$

Multiply \frac{1}{2} and 1 to get \frac{1}{2}.

$$xp=\frac{3}{2}p+1+\frac{1}{2}p$$

Combine \frac{3}{2}p and \frac{1}{2}p to get 2p.

$$xp=2p+1$$

Subtract 2p from both sides.

$$xp-2p=1$$

Combine all terms containing p.

$$\left(x-2\right)p=1$$

Divide both sides by x-2.

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

Dividing by x-2 undoes the multiplication by x-2.

$$p=\frac{1}{x-2}$$

Variable p cannot be equal to 0.

$$p=\frac{1}{x-2}\text{, }p\neq 0$$

Solve for x

$$x=2+\frac{1}{p}$$

$$p\neq 0$$