Solve 9x^2-3ax+1=01/c

$$9 x ^ { 2 } - 3 a x + 1 = 0 \frac { 1 } { c }$$

Solve for a

$$a=3x+\frac{1}{3x}$$

$$x\neq 0\text{ and }c\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by c.

$$9x^{2}c-3axc+c=0\times 1$$

Multiply 0 and 1 to get 0.

$$9x^{2}c-3axc+c=0$$

Subtract 9x^{2}c from both sides. Anything subtracted from zero gives its negation.

$$-3axc+c=-9x^{2}c$$

Subtract c from both sides.

$$-3axc=-9x^{2}c-c$$

The equation is in standard form.

$$\left(-3cx\right)a=-9cx^{2}-c$$

Divide both sides by -3xc.

$$\frac{\left(-3cx\right)a}{-3cx}=-\frac{c\left(9x^{2}+1\right)}{-3cx}$$

Dividing by -3xc undoes the multiplication by -3xc.

$$a=-\frac{c\left(9x^{2}+1\right)}{-3cx}$$

Divide -c\left(9x^{2}+1\right) by -3xc.

$$a=3x+\frac{1}{3x}$$

Solve for c

$$c\neq 0$$

$$a=3x+\frac{1}{3x}\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Variable c cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by c.

$$9x^{2}c-3axc+c=0\times 1$$

Multiply 0 and 1 to get 0.

$$9x^{2}c-3axc+c=0$$

Combine all terms containing c.

$$\left(9x^{2}-3ax+1\right)c=0$$

Divide 0 by 9x^{2}-3ax+1.

$$c=0$$

Variable c cannot be equal to 0.

$$c\in \emptyset$$