Solve 10ab+b^2=3a

$$10 a b + b ^ { 2 } = 3 a$$

Solve for a

$$a=-\frac{b^{2}}{10b-3}$$

$$b\neq \frac{3}{10}$$

Steps for Solving Linear Equation

Subtract 3a from both sides.

$$10ab+b^{2}-3a=0$$

Subtract b^{2} from both sides. Anything subtracted from zero gives its negation.

$$10ab-3a=-b^{2}$$

Combine all terms containing a.

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

Divide both sides by 10b-3.

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

Dividing by 10b-3 undoes the multiplication by 10b-3.

$$a=-\frac{b^{2}}{10b-3}$$

Solve for b (complex solution)

$$b=\sqrt{a\left(25a+3\right)}-5a$$

$$b=-\sqrt{a\left(25a+3\right)}-5a$$

Solve for b

$$b=\sqrt{a\left(25a+3\right)}-5a$$

$$b=-\sqrt{a\left(25a+3\right)}-5a\text{, }a\leq -\frac{3}{25}\text{ or }a\geq 0$$