Solve x/a+y/b+z/c=1

$$\frac { x } { a } + \frac { y } { b } + \frac { z } { c } = 1$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{bcx}{cy+bz-bc}\text{, }&b\neq 0\text{ and }c\neq 0\text{ and }x\neq 0\text{ and }y\neq \frac{b\left(c-z\right)}{c}\\a\neq 0\text{, }&y=\frac{b\left(c-z\right)}{c}\text{ and }c\neq 0\text{ and }x=0\text{ and }b\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by abc, the least common multiple of a,b,c.

$$bcx+acy+abz=abc$$

Subtract abc from both sides.

$$bcx+acy+abz-abc=0$$

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

$$acy+abz-abc=-bcx$$

Combine all terms containing a.

$$\left(cy+bz-bc\right)a=-bcx$$

Divide both sides by yc+zb-cb.

$$\frac{\left(cy+bz-bc\right)a}{cy+bz-bc}=-\frac{bcx}{cy+bz-bc}$$

Dividing by yc+zb-cb undoes the multiplication by yc+zb-cb.

$$a=-\frac{bcx}{cy+bz-bc}$$

Variable a cannot be equal to 0.

$$a=-\frac{bcx}{cy+bz-bc}\text{, }a\neq 0$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{bcx}{cy+bz-bc}\text{, }&b\neq 0\text{ and }x\neq 0\text{ and }c\neq 0\text{ and }y\neq \frac{b\left(c-z\right)}{c}\\a\neq 0\text{, }&y=\frac{b\left(c-z\right)}{c}\text{ and }c\neq 0\text{ and }x=0\text{ and }b\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by abc, the least common multiple of a,b,c.

$$bcx+acy+abz=abc$$

Subtract abc from both sides.

$$bcx+acy+abz-abc=0$$

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

$$acy+abz-abc=-bcx$$

Combine all terms containing a.

$$\left(cy+bz-bc\right)a=-bcx$$

Divide both sides by yc+zb-cb.

$$\frac{\left(cy+bz-bc\right)a}{cy+bz-bc}=-\frac{bcx}{cy+bz-bc}$$

Dividing by yc+zb-cb undoes the multiplication by yc+zb-cb.

$$a=-\frac{bcx}{cy+bz-bc}$$

Variable a cannot be equal to 0.

$$a=-\frac{bcx}{cy+bz-bc}\text{, }a\neq 0$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{acy}{cx+az-ac}\text{, }&a\neq 0\text{ and }c\neq 0\text{ and }y\neq 0\text{ and }x\neq \frac{a\left(c-z\right)}{c}\\b\neq 0\text{, }&x=\frac{a\left(c-z\right)}{c}\text{ and }c\neq 0\text{ and }y=0\text{ and }a\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by abc, the least common multiple of a,b,c.

$$bcx+acy+abz=abc$$

Subtract abc from both sides.

$$bcx+acy+abz-abc=0$$

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

$$bcx+abz-abc=-acy$$

Combine all terms containing b.

$$\left(cx+az-ac\right)b=-acy$$

Divide both sides by cx+az-ca.

$$\frac{\left(cx+az-ac\right)b}{cx+az-ac}=-\frac{acy}{cx+az-ac}$$

Dividing by cx+az-ca undoes the multiplication by cx+az-ca.

$$b=-\frac{acy}{cx+az-ac}$$

Variable b cannot be equal to 0.

$$b=-\frac{acy}{cx+az-ac}\text{, }b\neq 0$$