Solve y=1(2*x^-1+2)^1+3

$$y = 1 ( 2 \cdot x ^ { - 1 } + 2 ) ^ { 1 } + 3$$

Solve for x

$$x=-\frac{2}{5-y}$$

$$y\neq 5$$

Steps for Solving Linear Equation

Calculate 2x^{-1}+2 to the power of 1 and get 2x^{-1}+2.

$$y=1\left(2x^{-1}+2\right)+3$$

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

$$y=2x^{-1}+2+3$$

Add 2 and 3 to get 5.

$$y=2x^{-1}+5$$

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

$$2x^{-1}+5=y$$

Reorder the terms.

$$5+2\times \frac{1}{x}=y$$

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

$$x\times 5+2\times 1=yx$$

Multiply 2 and 1 to get 2.

$$x\times 5+2=yx$$

Subtract yx from both sides.

$$x\times 5+2-yx=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$x\times 5-yx=-2$$

Combine all terms containing x.

$$\left(5-y\right)x=-2$$

Divide both sides by 5-y.

$$\frac{\left(5-y\right)x}{5-y}=-\frac{2}{5-y}$$

Dividing by 5-y undoes the multiplication by 5-y.

$$x=-\frac{2}{5-y}$$

Variable x cannot be equal to 0.

$$x=-\frac{2}{5-y}\text{, }x\neq 0$$

Solve for y

$$y=5+\frac{2}{x}$$

$$x\neq 0$$