Solve 3kx^2-(2k-3+1)x+1-k=0

$$3 k x ^ { 2 } - ( 2 k - 3 + 1 ) x + 1 - k = 0$$

Solve for k

$$k=-\frac{2x+1}{\left(x-1\right)\left(3x+1\right)}$$

$$x\neq -\frac{1}{3}\text{ and }x\neq 1$$

Steps for Solving Linear Equation

Add -3 and 1 to get -2.

$$3kx^{2}-\left(2k-2\right)x+1-k=0$$

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

$$3kx^{2}-\left(2kx-2x\right)+1-k=0$$

To find the opposite of 2kx-2x, find the opposite of each term.

$$3kx^{2}-2kx+2x+1-k=0$$

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

$$3kx^{2}-2kx+1-k=-2x$$

Subtract 1 from both sides.

$$3kx^{2}-2kx-k=-2x-1$$

Combine all terms containing k.

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

Divide both sides by 3x^{2}-2x-1.

$$\frac{\left(3x^{2}-2x-1\right)k}{3x^{2}-2x-1}=\frac{-2x-1}{3x^{2}-2x-1}$$

Dividing by 3x^{2}-2x-1 undoes the multiplication by 3x^{2}-2x-1.

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

Divide -2x-1 by 3x^{2}-2x-1.

$$k=-\frac{2x+1}{\left(x-1\right)\left(3x+1\right)}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{\left(k-1\right)\left(4k-1\right)}+k-1}{3k}\text{; }x=\frac{-\sqrt{\left(k-1\right)\left(4k-1\right)}+k-1}{3k}\text{, }&k\neq 0\\x=-\frac{1}{2}\text{, }&k=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{\left(k-1\right)\left(4k-1\right)}+k-1}{3k}\text{; }x=\frac{-\sqrt{\left(k-1\right)\left(4k-1\right)}+k-1}{3k}\text{, }&k\geq 1\text{ or }\left(k\neq 0\text{ and }k\leq \frac{1}{4}\right)\\x=-\frac{1}{2}\text{, }&k=0\end{matrix}\right.$$