Solve (4)a*b=a+b+a/b

$$( 4 ) a * b = a + b + \frac { a } { b }$$

Solve for a

$$a=\frac{b^{2}}{4b^{2}-b-1}$$

$$b\neq \frac{\sqrt{17}+1}{8}\text{ and }b\neq \frac{1-\sqrt{17}}{8}\text{ and }b\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by b.

$$4abb=ba+bb+a$$

Multiply b and b to get b^{2}.

$$4ab^{2}=ba+bb+a$$

Multiply b and b to get b^{2}.

$$4ab^{2}=ba+b^{2}+a$$

Subtract ba from both sides.

$$4ab^{2}-ba=b^{2}+a$$

Subtract a from both sides.

$$4ab^{2}-ba-a=b^{2}$$

Combine all terms containing a.

$$\left(4b^{2}-b-1\right)a=b^{2}$$

Divide both sides by 4b^{2}-b-1.

$$\frac{\left(4b^{2}-b-1\right)a}{4b^{2}-b-1}=\frac{b^{2}}{4b^{2}-b-1}$$

Dividing by 4b^{2}-b-1 undoes the multiplication by 4b^{2}-b-1.

$$a=\frac{b^{2}}{4b^{2}-b-1}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=\frac{\sqrt{a\left(17a-4\right)}+a}{2\left(4a-1\right)}\text{; }b=\frac{-\sqrt{a\left(17a-4\right)}+a}{2\left(4a-1\right)}\text{, }&a\neq \frac{1}{4}\text{ and }a\neq 0\\b=-1\text{, }&a=\frac{1}{4}\end{matrix}\right.$$

Solve for b

$$\left\{\begin{matrix}b=\frac{\sqrt{a\left(17a-4\right)}+a}{2\left(4a-1\right)}\text{; }b=\frac{-\sqrt{a\left(17a-4\right)}+a}{2\left(4a-1\right)}\text{, }&a<0\text{ or }\left(a\neq \frac{1}{4}\text{ and }a\geq \frac{4}{17}\right)\\b=-1\text{, }&a=\frac{1}{4}\end{matrix}\right.$$