Solve 2/50^2+1/30b=6ab+5a^2/15a^3b=

$$\frac { 2 } { 50 ^ { 2 } } + \frac { 1 } { 30 b } = \frac { 6 a b + 5 a ^ { 2 } } { 15 a ^ { 3 } b } =$$

Solve for b

$$b=-\frac{125a\left(a-10\right)}{3\left(a^{2}-500\right)}$$

$$a\neq 10\text{ and }a\neq 0\text{ and }|a|\neq 10\sqrt{5}$$

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 30ba^{3}, the least common multiple of 30b,15a^{3}b.

$$\frac{3}{250}ba^{3}\times 2+a^{3}=2\left(6ab+5a^{2}\right)$$

Multiply \frac{3}{250} and 2 to get \frac{3}{125}.

$$\frac{3}{125}ba^{3}+a^{3}=2\left(6ab+5a^{2}\right)$$

Use the distributive property to multiply 2 by 6ab+5a^{2}.

$$\frac{3}{125}ba^{3}+a^{3}=12ab+10a^{2}$$

Subtract 12ab from both sides.

$$\frac{3}{125}ba^{3}+a^{3}-12ab=10a^{2}$$

Subtract a^{3} from both sides.

$$\frac{3}{125}ba^{3}-12ab=10a^{2}-a^{3}$$

Combine all terms containing b.

$$\left(\frac{3}{125}a^{3}-12a\right)b=10a^{2}-a^{3}$$

The equation is in standard form.

$$\left(\frac{3a^{3}}{125}-12a\right)b=10a^{2}-a^{3}$$

Divide both sides by \frac{3}{125}a^{3}-12a.

$$\frac{\left(\frac{3a^{3}}{125}-12a\right)b}{\frac{3a^{3}}{125}-12a}=\frac{\left(10-a\right)a^{2}}{\frac{3a^{3}}{125}-12a}$$

Dividing by \frac{3}{125}a^{3}-12a undoes the multiplication by \frac{3}{125}a^{3}-12a.

$$b=\frac{\left(10-a\right)a^{2}}{\frac{3a^{3}}{125}-12a}$$

Divide \left(10-a\right)a^{2} by \frac{3}{125}a^{3}-12a.

$$b=\frac{125a\left(10-a\right)}{3\left(a^{2}-500\right)}$$

Variable b cannot be equal to 0.

$$b=\frac{125a\left(10-a\right)}{3\left(a^{2}-500\right)}\text{, }b\neq 0$$

Solve for a

$$\left\{\begin{matrix}a=\frac{5\left(\sqrt{5\left(36b^{2}+1500b+3125\right)}+125\right)}{3b+125}\text{; }a=\frac{5\left(-\sqrt{5\left(36b^{2}+1500b+3125\right)}+125\right)}{3b+125}\text{, }&\left(b\neq 0\text{ and }b\geq \frac{25\sqrt{5}}{3}-\frac{125}{6}\right)\text{ or }\left(b\neq -\frac{125}{3}\text{ and }b\leq -\frac{25\sqrt{5}}{3}-\frac{125}{6}\right)\\a=50\text{, }&b=-\frac{125}{3}\end{matrix}\right.$$