Solve the Rational Equation for x (x-3)/(x+1)=(4x-6)/((x+1)(x+2))

Factor ${\displaystyle 2}$ out of ${\displaystyle 4x}$.

Factor ${\displaystyle 2}$ out of ${\displaystyle -6}$.

Factor ${\displaystyle 2}$ out of ${\displaystyle 2\left(2x\right)+2(-3)}$.

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

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 ${\displaystyle 1,1}$ 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 ${\displaystyle x+1}$ itself.

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

The factor for ${\displaystyle x+2}$ is ${\displaystyle x+2}$ itself.

${\displaystyle (x+2)}$ occurs ${\displaystyle 1}$ time.

The LCM of ${\displaystyle x+1,x+1,x+2}$ is the result of multiplying all factors the greatest number of times they occur in either term.

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

Simplify the left side.

Simplify the right side.

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

Add ${\displaystyle 6}$ to both sides of the equation.

Combine the opposite terms in ${\displaystyle x^2-5x-6+6}$.

Factor ${\displaystyle x}$ out of ${\displaystyle x^2-5x}$.

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 ${\displaystyle x}$ equal to ${\displaystyle 0}$.

Set ${\displaystyle x-5}$ equal to ${\displaystyle 0}$ and solve for ${\displaystyle x}$.

The final solution is all the values that make ${\displaystyle x(x-5)=0}$ true.