Solve for x 5/x+5=5/3x+31/6

Subtract ${\displaystyle 5}$ from both sides of the equation.

To write ${\displaystyle -5}$ as a fraction with a common denominator, multiply by ${\displaystyle \frac{6}{6}}$.

Combine ${\displaystyle -5}$ and ${\displaystyle \frac{6}{6}}$.

Combine the numerators over the common denominator.

Simplify the numerator.

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

Since ${\displaystyle x,3,6}$ contain both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part ${\displaystyle 1,3,6}$ then find LCM for the variable part ${\displaystyle x^1}$.

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

Since ${\displaystyle 3}$ has no factors besides ${\displaystyle 1}$ and ${\displaystyle 3}$.

${\displaystyle 3}$ is a prime number

${\displaystyle 6}$ has factors of ${\displaystyle 2}$ and ${\displaystyle 3}$.

Multiply ${\displaystyle 2}$ by ${\displaystyle 3}$.

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 LCM for ${\displaystyle x,3,6}$ is the numeric part ${\displaystyle 6}$ multiplied by the variable part.

Multiply each term in ${\displaystyle \frac{5}{x}=\frac{5x}{3}+\frac{1}{6}}$ by ${\displaystyle 6x}$.

Simplify the left side.

Simplify the right side.

Rewrite the equation as ${\displaystyle 10x^2+x=30}$.

Subtract ${\displaystyle 30}$ from both sides of the equation.

Use the quadratic formula to find the solutions.

Substitute the values ${\displaystyle a=10}$, ${\displaystyle b=1}$, and ${\displaystyle c=-30}$ into the quadratic formula and solve for ${\displaystyle x}$.

Simplify.

The final answer is the combination of both solutions.