Solve +(2k-1)x+k^2-1=0

$$+ ( 2 k - 1 ) x + k ^ { 2 } - 1 = 0$$

Solve for x

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

$$k\neq \frac{1}{2}$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2k-1 by x.

$$2kx-x+k^{2}-1=0$$

Subtract k^{2} from both sides. Anything subtracted from zero gives its negation.

$$2kx-x-1=-k^{2}$$

Add 1 to both sides.

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

Combine all terms containing x.

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

The equation is in standard form.

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

Divide both sides by 2k-1.

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

Dividing by 2k-1 undoes the multiplication by 2k-1.

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

Solve for k

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

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