Find All Complex Solutions 2/(x-3)+1/x=(x-1)/(x-3)

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

Since ${\displaystyle x-3,x,x-3}$ contain both numbers and variables, there are four steps to find the LCM. Find LCM for the numeric, variable, and compound variable parts. Then, multiply them all together.

Steps to find the LCM for ${\displaystyle x-3,x,x-3}$ are:

1. Find the LCM for the numeric part ${\displaystyle 1,1,1}$.

2. Find the LCM for the variable part ${\displaystyle x^1}$.

3. Find the LCM for the compound variable part ${\displaystyle x-3,x-3}$.

4. Multiply each LCM together.

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number ${\displaystyle 1}$ is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of ${\displaystyle 1,1,1}$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

The factor for ${\displaystyle x^1}$ is ${\displaystyle x}$ itself.

${\displaystyle x}$ occurs ${\displaystyle 1}$ time.

The LCM of ${\displaystyle x^1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

The factor for ${\displaystyle x-3}$ is ${\displaystyle x-3}$ itself.

${\displaystyle (x-3)}$ occurs ${\displaystyle 1}$ time.

The LCM of ${\displaystyle x-3,x-3}$ is the result of multiplying all factors the greatest number of times they occur in either term.

The Least Common Multiple ${\displaystyle \mathrm{LCM}}$ of some numbers is the smallest number that the numbers are factors of.

Multiply each term in ${\displaystyle \frac{2}{x-3}+\frac{1}{x}=\frac{x-1}{x-3}}$ by ${\displaystyle x(x-3)}$.

Simplify the left side.

Simplify the right side.

Since ${\displaystyle x}$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

Move all terms containing ${\displaystyle x}$ to the left side of the equation.

Add ${\displaystyle 3}$ to both sides of the equation.

Factor ${\displaystyle x^2-4x+3}$ using the AC method.

If any individual factor on the left side of the equation is equal to ${\displaystyle 0}$, the entire expression will be equal to ${\displaystyle 0}$.

Set ${\displaystyle x-3}$ equal to ${\displaystyle 0}$ and solve for ${\displaystyle x}$.

Set ${\displaystyle x-1}$ equal to ${\displaystyle 0}$ and solve for ${\displaystyle x}$.

The final solution is all the values that make ${\displaystyle (x-3)(x-1)=0}$ true.