Solve 4/ax+4=2/3x

$$\frac { 4 } { a x } + 4 = \frac { 2 } { 3 } x$$

Solve for a

$$a=-\frac{6}{x\left(6-x\right)}$$

$$x\neq 6\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3ax, the least common multiple of ax,3.

$$3\times 4+3ax\times 4=\frac{2}{3}x\times 3ax$$

Multiply 3 and 4 to get 12.

$$12+3ax\times 4=\frac{2}{3}x\times 3ax$$

Multiply 3 and 4 to get 12.

$$12+12ax=\frac{2}{3}x\times 3ax$$

Multiply x and x to get x^{2}.

$$12+12ax=\frac{2}{3}x^{2}\times 3a$$

Multiply \frac{2}{3} and 3 to get 2.

$$12+12ax=2x^{2}a$$

Subtract 2x^{2}a from both sides.

$$12+12ax-2x^{2}a=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$12ax-2x^{2}a=-12$$

Combine all terms containing a.

$$\left(12x-2x^{2}\right)a=-12$$

Divide both sides by 12x-2x^{2}.

$$\frac{\left(12x-2x^{2}\right)a}{12x-2x^{2}}=-\frac{12}{12x-2x^{2}}$$

Dividing by 12x-2x^{2} undoes the multiplication by 12x-2x^{2}.

$$a=-\frac{12}{12x-2x^{2}}$$

Divide -12 by 12x-2x^{2}.

$$a=-\frac{6}{x\left(6-x\right)}$$

Variable a cannot be equal to 0.

$$a=-\frac{6}{x\left(6-x\right)}\text{, }a\neq 0$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{9a^{2}+6a}}{a}+3$$

$$x=\frac{\sqrt{9a^{2}+6a}}{a}+3\text{, }a\neq 0$$

Solve for x

$$x=-\frac{\sqrt{9a^{2}+6a}}{a}+3$$

$$x=\frac{\sqrt{9a^{2}+6a}}{a}+3\text{, }a\leq -\frac{2}{3}\text{ or }a>0$$