Solve x-a/3=x+3/a-2

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

Solve for x (complex solution)

$$\left\{\begin{matrix}x=a-3\text{, }&a\neq 0\\x\in \mathrm{C}\text{, }&a=3\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 3a, the least common multiple of 3,a.

$$a\left(x-a\right)=3\left(x+3\right)+3a\left(-2\right)$$

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

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

Use the distributive property to multiply 3 by x+3.

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

Multiply 3 and -2 to get -6.

$$ax-a^{2}=3x+9-6a$$

Subtract 3x from both sides.

$$ax-a^{2}-3x=9-6a$$

Add a^{2} to both sides.

$$ax-3x=9-6a+a^{2}$$

Combine all terms containing x.

$$\left(a-3\right)x=9-6a+a^{2}$$

The equation is in standard form.

$$\left(a-3\right)x=a^{2}-6a+9$$

Divide both sides by -3+a.

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

Dividing by -3+a undoes the multiplication by -3+a.

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

Divide \left(a-3\right)^{2} by -3+a.

$$x=a-3$$

Solve for x

$$\left\{\begin{matrix}x=a-3\text{, }&a\neq 0\\x\in \mathrm{R}\text{, }&a=3\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 3a, the least common multiple of 3,a.

$$a\left(x-a\right)=3\left(x+3\right)+3a\left(-2\right)$$

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

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

Use the distributive property to multiply 3 by x+3.

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

Multiply 3 and -2 to get -6.

$$ax-a^{2}=3x+9-6a$$

Subtract 3x from both sides.

$$ax-a^{2}-3x=9-6a$$

Add a^{2} to both sides.

$$ax-3x=9-6a+a^{2}$$

Combine all terms containing x.

$$\left(a-3\right)x=9-6a+a^{2}$$

The equation is in standard form.

$$\left(a-3\right)x=a^{2}-6a+9$$

Divide both sides by -3+a.

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

Dividing by -3+a undoes the multiplication by -3+a.

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

Divide \left(a-3\right)^{2} by -3+a.

$$x=a-3$$

Solve for a

$$\left\{\begin{matrix}\\a=3\text{, }&\text{unconditionally}\\a=x+3\text{, }&x\neq -3\end{matrix}\right.$$