Solve 2x+1/x-1=-x+k

$$\frac { 2 x + 1 } { x - 1 } = - x + k$$

Solve for k

$$k=-\frac{x^{2}+x+1}{1-x}$$

$$x\neq 1$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x-1.

$$2x+1=-\left(x-1\right)x+\left(x-1\right)k$$

Use the distributive property to multiply x-1 by x.

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

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

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

Use the distributive property to multiply x-1 by k.

$$2x+1=-x^{2}+x+xk-k$$

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

$$-x^{2}+x+xk-k=2x+1$$

Add x^{2} to both sides.

$$x+xk-k=2x+1+x^{2}$$

Subtract x from both sides.

$$xk-k=2x+1+x^{2}-x$$

Combine 2x and -x to get x.

$$xk-k=x+1+x^{2}$$

Combine all terms containing k.

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

The equation is in standard form.

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

Divide both sides by x-1.

$$\frac{\left(x-1\right)k}{x-1}=\frac{x^{2}+x+1}{x-1}$$

Dividing by x-1 undoes the multiplication by x-1.

$$k=\frac{x^{2}+x+1}{x-1}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{k^{2}-6k-3}+k-1}{2}$$

$$x=\frac{-\sqrt{k^{2}-6k-3}+k-1}{2}$$

Solve for x

$$x=\frac{\sqrt{k^{2}-6k-3}+k-1}{2}$$

$$x=\frac{-\sqrt{k^{2}-6k-3}+k-1}{2}\text{, }k\geq 2\sqrt{3}+3\text{ or }k\leq 3-2\sqrt{3}$$