Solve a+b+c+abc=25

$$a+b+c+abc=25$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{b+c-25}{bc+1}\text{, }&c=0\text{ or }b\neq -\frac{1}{c}\\a\in \mathrm{R}\text{, }&\left(b=\frac{\sqrt{629}+25}{2}\text{ and }c=\frac{25-\sqrt{629}}{2}\right)\text{ or }\left(b=\frac{25-\sqrt{629}}{2}\text{ and }c=\frac{\sqrt{629}+25}{2}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract b from both sides.

$$a+c+abc=25-b$$

Subtract c from both sides.

$$a+abc=25-b-c$$

Combine all terms containing a.

$$\left(1+bc\right)a=25-b-c$$

The equation is in standard form.

$$\left(bc+1\right)a=25-c-b$$

Divide both sides by 1+bc.

$$\frac{\left(bc+1\right)a}{bc+1}=\frac{25-c-b}{bc+1}$$

Dividing by 1+bc undoes the multiplication by 1+bc.

$$a=\frac{25-c-b}{bc+1}$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{a+c-25}{ac+1}\text{, }&c=0\text{ or }a\neq -\frac{1}{c}\\b\in \mathrm{R}\text{, }&\left(a=\frac{\sqrt{629}+25}{2}\text{ and }c=\frac{25-\sqrt{629}}{2}\right)\text{ or }\left(a=\frac{25-\sqrt{629}}{2}\text{ and }c=\frac{\sqrt{629}+25}{2}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract a from both sides.

$$b+c+abc=25-a$$

Subtract c from both sides.

$$b+abc=25-a-c$$

Combine all terms containing b.

$$\left(1+ac\right)b=25-a-c$$

The equation is in standard form.

$$\left(ac+1\right)b=25-c-a$$

Divide both sides by 1+ac.

$$\frac{\left(ac+1\right)b}{ac+1}=\frac{25-c-a}{ac+1}$$

Dividing by 1+ac undoes the multiplication by 1+ac.

$$b=\frac{25-c-a}{ac+1}$$