Solve for x (x^2-29)/(x^2-10x+21)=6/(x-7)+5/(x-3)

Consider the form ${\displaystyle x^2+bx+c}$. Find a pair of integers whose product is ${\displaystyle c}$ and whose sum is ${\displaystyle b}$. In this case, whose product is ${\displaystyle 21}$ and whose sum is ${\displaystyle -10}$.

Write the factored form using these integers.

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

The factor for ${\displaystyle x-7}$ is ${\displaystyle x-7}$ itself.

${\displaystyle (x-7)}$ occurs ${\displaystyle 1}$ time.

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

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

The factor for ${\displaystyle x-7}$ is ${\displaystyle x-7}$ itself.

${\displaystyle (x-7)}$ occurs ${\displaystyle 1}$ time.

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

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

The LCM of ${\displaystyle x-7,x-3,x-7,x-3}$ is the result of multiplying all factors the greatest number of times they occur in either term.

Multiply each term in ${\displaystyle \frac{x^2-29}{(x-7)(x-3)}=\frac{6}{x-7}+\frac{5}{x-3}}$ by ${\displaystyle (x-7)(x-3)}$.

Simplify the left side.

Simplify the right side.

Subtract ${\displaystyle 11x}$ from both sides of the equation.

Add ${\displaystyle 53}$ to both sides of the equation.

Add ${\displaystyle -29}$ and ${\displaystyle 53}$.

Factor ${\displaystyle x^2-11x+24}$ 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 ${\displaystyle x-8}$ equal to ${\displaystyle 0}$ and solve for ${\displaystyle x}$.

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

The final solution is all the values that make ${\displaystyle (x-8)(x-3)=0}$ true.