Solve 2/frac{6{b}}+y/b=1

$$\frac { 2 } { \frac { 6 } { b } } + \frac { y } { b } = 1$$

Solve for y

$$y=-\frac{b\left(b-3\right)}{3}$$

$$b\neq 0$$

Solution Steps

Multiply both sides of the equation by b.

$$\frac{1}{6}b^{2}\times 2+y=b$$

Multiply \frac{1}{6} and 2 to get \frac{1}{3}.

$$\frac{1}{3}b^{2}+y=b$$

Subtract \frac{1}{3}b^{2} from both sides.

$$y=b-\frac{1}{3}b^{2}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=\frac{-\sqrt{9-12y}+3}{2}\text{; }b=\frac{\sqrt{9-12y}+3}{2}\text{, }&y\neq 0\\b=3\text{, }&y=0\end{matrix}\right.$$

Solve for b

$$\left\{\begin{matrix}b=\frac{-\sqrt{9-12y}+3}{2}\text{; }b=\frac{\sqrt{9-12y}+3}{2}\text{, }&y\neq 0\text{ and }y\leq \frac{3}{4}\\b=3\text{, }&y=0\end{matrix}\right.$$