Solve B=2ax^2-5ax^2+ax

$$B = 2 a x ^ { 2 } - 5 a x ^ { 2 } + a x$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{B}{x\left(1-3x\right)}\text{, }&x\neq \frac{1}{3}\text{ and }x\neq 0\\a\in \mathrm{C}\text{, }&\left(x=0\text{ or }x=\frac{1}{3}\right)\text{ and }B=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Combine 2ax^{2} and -5ax^{2} to get -3ax^{2}.

$$B=-3ax^{2}+ax$$

Swap sides so that all variable terms are on the left hand side.

$$-3ax^{2}+ax=B$$

Combine all terms containing a.

$$\left(-3x^{2}+x\right)a=B$$

The equation is in standard form.

$$\left(x-3x^{2}\right)a=B$$

Divide both sides by -3x^{2}+x.

$$\frac{\left(x-3x^{2}\right)a}{x-3x^{2}}=\frac{B}{x-3x^{2}}$$

Dividing by -3x^{2}+x undoes the multiplication by -3x^{2}+x.

$$a=\frac{B}{x-3x^{2}}$$

Divide B by -3x^{2}+x.

$$a=\frac{B}{x\left(1-3x\right)}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{B}{x\left(1-3x\right)}\text{, }&x\neq \frac{1}{3}\text{ and }x\neq 0\\a\in \mathrm{R}\text{, }&\left(x=0\text{ or }x=\frac{1}{3}\right)\text{ and }B=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Combine 2ax^{2} and -5ax^{2} to get -3ax^{2}.

$$B=-3ax^{2}+ax$$

Swap sides so that all variable terms are on the left hand side.

$$-3ax^{2}+ax=B$$

Combine all terms containing a.

$$\left(-3x^{2}+x\right)a=B$$

The equation is in standard form.

$$\left(x-3x^{2}\right)a=B$$

Divide both sides by -3x^{2}+x.

$$\frac{\left(x-3x^{2}\right)a}{x-3x^{2}}=\frac{B}{x-3x^{2}}$$

Dividing by -3x^{2}+x undoes the multiplication by -3x^{2}+x.

$$a=\frac{B}{x-3x^{2}}$$

Divide B by -3x^{2}+x.

$$a=\frac{B}{x\left(1-3x\right)}$$

Solve for B

$$B=ax\left(1-3x\right)$$