Solve a+2ab+3ab+0b=0

$$a + 2 a b + 3 a b + 0 b = 0$$

Solve for a

$$\left\{\begin{matrix}\\a=0\text{, }&\text{unconditionally}\\a\in \mathrm{R}\text{, }&b=-\frac{1}{5}\end{matrix}\right.$$

Steps for Solving Linear Equation

Combine 2ab and 3ab to get 5ab.

$$a+5ab+0b=0$$

Anything times zero gives zero.

$$a+5ab+0=0$$

Anything plus zero gives itself.

$$a+5ab=0$$

Combine all terms containing a.

$$\left(1+5b\right)a=0$$

The equation is in standard form.

$$\left(5b+1\right)a=0$$

Divide 0 by 1+5b.

$$a=0$$

Solve for b

$$\left\{\begin{matrix}\\b=-\frac{1}{5}\text{, }&\text{unconditionally}\\b\in \mathrm{R}\text{, }&a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Combine 2ab and 3ab to get 5ab.

$$a+5ab+0b=0$$

Anything times zero gives zero.

$$a+5ab+0=0$$

Anything plus zero gives itself.

$$a+5ab=0$$

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

$$5ab=-a$$

Divide both sides by 5a.

$$\frac{5ab}{5a}=-\frac{a}{5a}$$

Dividing by 5a undoes the multiplication by 5a.

$$b=-\frac{a}{5a}$$

Divide -a by 5a.

$$b=-\frac{1}{5}$$