Solve (a-b)x^2+(b-c)x+(c-a)=0

$$( a - b ) x ^ { 2 } + ( b - c ) x + ( c - a ) = 0$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{bx+c}{x+1}\text{, }&x\neq -1\\a\in \mathrm{C}\text{, }&x=1\text{ or }\left(b=c\text{ and }x=-1\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a-b by x^{2}.

$$ax^{2}-bx^{2}+\left(b-c\right)x+c-a=0$$

Use the distributive property to multiply b-c by x.

$$ax^{2}-bx^{2}+bx-cx+c-a=0$$

Add bx^{2} to both sides. Anything plus zero gives itself.

$$ax^{2}+bx-cx+c-a=bx^{2}$$

Subtract bx from both sides.

$$ax^{2}-cx+c-a=bx^{2}-bx$$

Add cx to both sides.

$$ax^{2}+c-a=bx^{2}-bx+cx$$

Subtract c from both sides.

$$ax^{2}-a=bx^{2}-bx+cx-c$$

Combine all terms containing a.

$$\left(x^{2}-1\right)a=bx^{2}-bx+cx-c$$

The equation is in standard form.

$$\left(x^{2}-1\right)a=bx^{2}+cx-bx-c$$

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

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

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

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

Divide \left(-1+x\right)\left(bx+c\right) by x^{2}-1.

$$a=\frac{bx+c}{x+1}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=\frac{ax+a-c}{x}\text{, }&x\neq 0\\b\in \mathrm{C}\text{, }&x=1\text{ or }\left(a=c\text{ and }x=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a-b by x^{2}.

$$ax^{2}-bx^{2}+\left(b-c\right)x+c-a=0$$

Use the distributive property to multiply b-c by x.

$$ax^{2}-bx^{2}+bx-cx+c-a=0$$

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

$$-bx^{2}+bx-cx+c-a=-ax^{2}$$

Add cx to both sides.

$$-bx^{2}+bx+c-a=-ax^{2}+cx$$

Subtract c from both sides.

$$-bx^{2}+bx-a=-ax^{2}+cx-c$$

Add a to both sides.

$$-bx^{2}+bx=-ax^{2}+cx-c+a$$

Reorder the terms.

$$-bx^{2}+bx=-ax^{2}+cx+a-c$$

Combine all terms containing b.

$$\left(-x^{2}+x\right)b=-ax^{2}+cx+a-c$$

The equation is in standard form.

$$\left(x-x^{2}\right)b=-ax^{2}+cx+a-c$$

Divide both sides by -x^{2}+x.

$$\frac{\left(x-x^{2}\right)b}{x-x^{2}}=\frac{\left(1-x\right)\left(ax+a-c\right)}{x-x^{2}}$$

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

$$b=\frac{\left(1-x\right)\left(ax+a-c\right)}{x-x^{2}}$$

Divide \left(1-x\right)\left(a+ax-c\right) by -x^{2}+x.

$$b=\frac{ax+a-c}{x}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{bx+c}{x+1}\text{, }&x\neq -1\\a\in \mathrm{R}\text{, }&x=1\text{ or }\left(b=c\text{ and }x=-1\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a-b by x^{2}.

$$ax^{2}-bx^{2}+\left(b-c\right)x+c-a=0$$

Use the distributive property to multiply b-c by x.

$$ax^{2}-bx^{2}+bx-cx+c-a=0$$

Add bx^{2} to both sides. Anything plus zero gives itself.

$$ax^{2}+bx-cx+c-a=bx^{2}$$

Subtract bx from both sides.

$$ax^{2}-cx+c-a=bx^{2}-bx$$

Add cx to both sides.

$$ax^{2}+c-a=bx^{2}-bx+cx$$

Subtract c from both sides.

$$ax^{2}-a=bx^{2}-bx+cx-c$$

Combine all terms containing a.

$$\left(x^{2}-1\right)a=bx^{2}-bx+cx-c$$

The equation is in standard form.

$$\left(x^{2}-1\right)a=bx^{2}+cx-bx-c$$

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

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

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

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

Divide \left(-1+x\right)\left(bx+c\right) by x^{2}-1.

$$a=\frac{bx+c}{x+1}$$

Solve for b

$$\left\{\begin{matrix}b=\frac{ax+a-c}{x}\text{, }&x\neq 0\\b\in \mathrm{R}\text{, }&x=1\text{ or }\left(a=c\text{ and }x=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply a-b by x^{2}.

$$ax^{2}-bx^{2}+\left(b-c\right)x+c-a=0$$

Use the distributive property to multiply b-c by x.

$$ax^{2}-bx^{2}+bx-cx+c-a=0$$

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

$$-bx^{2}+bx-cx+c-a=-ax^{2}$$

Add cx to both sides.

$$-bx^{2}+bx+c-a=-ax^{2}+cx$$

Subtract c from both sides.

$$-bx^{2}+bx-a=-ax^{2}+cx-c$$

Add a to both sides.

$$-bx^{2}+bx=-ax^{2}+cx-c+a$$

Reorder the terms.

$$-bx^{2}+bx=-ax^{2}+cx+a-c$$

Combine all terms containing b.

$$\left(-x^{2}+x\right)b=-ax^{2}+cx+a-c$$

The equation is in standard form.

$$\left(x-x^{2}\right)b=-ax^{2}+cx+a-c$$

Divide both sides by -x^{2}+x.

$$\frac{\left(x-x^{2}\right)b}{x-x^{2}}=\frac{\left(1-x\right)\left(ax+a-c\right)}{x-x^{2}}$$

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

$$b=\frac{\left(1-x\right)\left(ax+a-c\right)}{x-x^{2}}$$

Divide \left(1-x\right)\left(a+ax-c\right) by -x^{2}+x.

$$b=\frac{ax+a-c}{x}$$