Solve x^2-axb/2x^2-b=3

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

Solve for a

$$a=\frac{3b-5x^{2}}{bx}$$

$$b\neq 2x^{2}\text{ and }b\neq 0\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2x^{2}-b.

$$x^{2}-axb=3\left(2x^{2}-b\right)$$

Use the distributive property to multiply 3 by 2x^{2}-b.

$$x^{2}-axb=6x^{2}-3b$$

Subtract x^{2} from both sides.

$$-axb=6x^{2}-3b-x^{2}$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$-axb=5x^{2}-3b$$

The equation is in standard form.

$$\left(-bx\right)a=5x^{2}-3b$$

Divide both sides by -xb.

$$\frac{\left(-bx\right)a}{-bx}=\frac{5x^{2}-3b}{-bx}$$

Dividing by -xb undoes the multiplication by -xb.

$$a=\frac{5x^{2}-3b}{-bx}$$

Divide 5x^{2}-3b by -xb.

$$a=-\frac{5x}{b}+\frac{3}{x}$$

Solve for b

$$b=\frac{5x^{2}}{3-ax}$$

$$\left(a=0\text{ or }x\neq \frac{3}{a}\right)\text{ and }\left(a=0\text{ or }x\neq \frac{1}{2a}\right)\text{ and }x\neq 0$$