Solve v=-x^2/2+frac{sqrt{3}}{b}x+2

$$v = - \frac { x ^ { 2 } } { 2 } + \frac { \sqrt { 3 } } { b } x + 2$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{2\sqrt{3}x}{4-2v-x^{2}}\text{, }&x\neq 0\text{ and }v\neq -\frac{x^{2}}{2}+2\\b\neq 0\text{, }&v=2\text{ and }x=0\end{matrix}\right.$$

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 2b, the least common multiple of 2,b.

$$v\times 2b=-bx^{2}+2\sqrt{3}x+2b\times 2$$

Multiply 2 and 2 to get 4.

$$v\times 2b=-bx^{2}+2\sqrt{3}x+4b$$

Add bx^{2} to both sides.

$$v\times 2b+bx^{2}=2\sqrt{3}x+4b$$

Subtract 4b from both sides.

$$v\times 2b+bx^{2}-4b=2\sqrt{3}x$$

Combine all terms containing b.

$$\left(v\times 2+x^{2}-4\right)b=2\sqrt{3}x$$

The equation is in standard form.

$$\left(x^{2}+2v-4\right)b=2\sqrt{3}x$$

Divide both sides by x^{2}+2v-4.

$$\frac{\left(x^{2}+2v-4\right)b}{x^{2}+2v-4}=\frac{2\sqrt{3}x}{x^{2}+2v-4}$$

Dividing by x^{2}+2v-4 undoes the multiplication by x^{2}+2v-4.

$$b=\frac{2\sqrt{3}x}{x^{2}+2v-4}$$

Variable b cannot be equal to 0.

$$b=\frac{2\sqrt{3}x}{x^{2}+2v-4}\text{, }b\neq 0$$

Solve for v

$$v=-\frac{x^{2}}{2}+\frac{\sqrt{3}x}{b}+2$$

$$b\neq 0$$