Solve 6x^2+(b-2a)x=b^2+ab

$$6 x ^ { 2 } + ( b - 2 a ) x = b ^ { 2 } + a b$$

Solve for a (complex solution)

$$\left\{\begin{matrix}\\a=3x-b\text{, }&\text{unconditionally}\\a\in \mathrm{C}\text{, }&x=-\frac{b}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply b-2a by x.

$$6x^{2}+bx-2ax=b^{2}+ab$$

Subtract ab from both sides.

$$6x^{2}+bx-2ax-ab=b^{2}$$

Subtract 6x^{2} from both sides.

$$bx-2ax-ab=b^{2}-6x^{2}$$

Subtract bx from both sides.

$$-2ax-ab=b^{2}-6x^{2}-bx$$

Combine all terms containing a.

$$\left(-2x-b\right)a=b^{2}-6x^{2}-bx$$

The equation is in standard form.

$$\left(-2x-b\right)a=b^{2}-bx-6x^{2}$$

Divide both sides by -2x-b.

$$\frac{\left(-2x-b\right)a}{-2x-b}=\frac{\left(b-3x\right)\left(2x+b\right)}{-2x-b}$$

Dividing by -2x-b undoes the multiplication by -2x-b.

$$a=\frac{\left(b-3x\right)\left(2x+b\right)}{-2x-b}$$

Divide \left(-3x+b\right)\left(2x+b\right) by -2x-b.

$$a=3x-b$$

Solve for a

$$\left\{\begin{matrix}\\a=3x-b\text{, }&\text{unconditionally}\\a\in \mathrm{R}\text{, }&x=-\frac{b}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply b-2a by x.

$$6x^{2}+bx-2ax=b^{2}+ab$$

Subtract ab from both sides.

$$6x^{2}+bx-2ax-ab=b^{2}$$

Subtract 6x^{2} from both sides.

$$bx-2ax-ab=b^{2}-6x^{2}$$

Subtract bx from both sides.

$$-2ax-ab=b^{2}-6x^{2}-bx$$

Combine all terms containing a.

$$\left(-2x-b\right)a=b^{2}-6x^{2}-bx$$

The equation is in standard form.

$$\left(-2x-b\right)a=b^{2}-bx-6x^{2}$$

Divide both sides by -2x-b.

$$\frac{\left(-2x-b\right)a}{-2x-b}=\frac{\left(b-3x\right)\left(2x+b\right)}{-2x-b}$$

Dividing by -2x-b undoes the multiplication by -2x-b.

$$a=\frac{\left(b-3x\right)\left(2x+b\right)}{-2x-b}$$

Divide \left(-3x+b\right)\left(2x+b\right) by -2x-b.

$$a=3x-b$$

Solve for b

$$b=3x-a$$

$$b=-2x$$