Solve Using the Quadratic Formula 1/x+1/(x+4)=1/3

$\frac{1}{x} + \frac{1}{x + 4} = \frac{1}{3}$

Move all terms to the left side of the equation and simplify.

Subtract $\frac{1}{3}$ from both sides of the equation.

$\frac{1}{x} + \frac{1}{x + 4} - \frac{1}{3} = 0$

Simplify $\frac{1}{x} + \frac{1}{x + 4} - \frac{1}{3}$ .

Find the common denominator.

Multiply $\frac{1}{x}$ by $\frac{(x + 4) \cdot 3}{(x + 4) \cdot 3}$ .

$\frac{1}{x} \cdot \frac{(x + 4) \cdot 3}{(x + 4) \cdot 3} + \frac{1}{x + 4} - \frac{1}{3} = 0$

Multiply $\frac{1}{x}$ by $\frac{(x + 4) \cdot 3}{(x + 4) \cdot 3}$ .

$\frac{(x + 4) \cdot 3}{x ((x + 4) \cdot 3)} + \frac{1}{x + 4} - \frac{1}{3} = 0$

Multiply $\frac{1}{x + 4}$ by $\frac{x \cdot 3}{x \cdot 3}$ .

$\frac{(x + 4) \cdot 3}{x ((x + 4) \cdot 3)} + \frac{1}{x + 4} \cdot \frac{x \cdot 3}{x \cdot 3} - \frac{1}{3} = 0$

Multiply $\frac{1}{x + 4}$ by $\frac{x \cdot 3}{x \cdot 3}$ .

$\frac{(x + 4) \cdot 3}{x ((x + 4) \cdot 3)} + \frac{x \cdot 3}{(x + 4) (x \cdot 3)} - \frac{1}{3} = 0$

Multiply $\frac{1}{3}$ by $\frac{x (x + 4)}{x (x + 4)}$ .

$\frac{(x + 4) \cdot 3}{x ((x + 4) \cdot 3)} + \frac{x \cdot 3}{(x + 4) (x \cdot 3)} - (\frac{1}{3} \cdot \frac{x (x + 4)}{x (x + 4)}) = 0$

Multiply $\frac{1}{3}$ by $\frac{x (x + 4)}{x (x + 4)}$ .

$\frac{(x + 4) \cdot 3}{x ((x + 4) \cdot 3)} + \frac{x \cdot 3}{(x + 4) (x \cdot 3)} - \frac{x (x + 4)}{3 (x (x + 4))} = 0$

Reorder the factors of $x ((x + 4) \cdot 3)$ .

$\frac{(x + 4) \cdot 3}{3 x (x + 4)} + \frac{x \cdot 3}{(x + 4) (x \cdot 3)} - \frac{x (x + 4)}{3 (x (x + 4))} = 0$

Reorder the factors of $(x + 4) (x \cdot 3)$ .

$\frac{(x + 4) \cdot 3}{3 x (x + 4)} + \frac{x \cdot 3}{3 x (x + 4)} - \frac{x (x + 4)}{3 (x (x + 4))} = 0$

Reorder the factors of $3 (x (x + 4))$ .

$\frac{(x + 4) \cdot 3}{3 x (x + 4)} + \frac{x \cdot 3}{3 x (x + 4)} - \frac{x (x + 4)}{3 x (x + 4)} = 0$

Combine the numerators over the common denominator.

$\frac{(x + 4) \cdot 3 + x \cdot 3 - x (x + 4)}{3 x (x + 4)} = 0$

Simplify each term.

Apply the distributive property.

$\frac{x \cdot 3 + 4 \cdot 3 + x \cdot 3 - x (x + 4)}{3 x (x + 4)} = 0$

Move $3$ to the left of $x$ .

$\frac{3 \cdot x + 4 \cdot 3 + x \cdot 3 - x (x + 4)}{3 x (x + 4)} = 0$

Multiply $4$ by $3$ .

$\frac{3 \cdot x + 12 + x \cdot 3 - x (x + 4)}{3 x (x + 4)} = 0$

Move $3$ to the left of $x$ .

$\frac{3 x + 12 + 3 \cdot x - x (x + 4)}{3 x (x + 4)} = 0$

Apply the distributive property.

$\frac{3 x + 12 + 3 x - x \cdot x - x \cdot 4}{3 x (x + 4)} = 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$\frac{3 x + 12 + 3 x - (x \cdot x) - x \cdot 4}{3 x (x + 4)} = 0$

Multiply $x$ by $x$ .

$\frac{3 x + 12 + 3 x - x^{2} - x \cdot 4}{3 x (x + 4)} = 0$

Multiply $4$ by $- 1$ .

$\frac{3 x + 12 + 3 x - x^{2} - 4 x}{3 x (x + 4)} = 0$

Add $3 x$ and $3 x$ .

$\frac{6 x + 12 - x^{2} - 4 x}{3 x (x + 4)} = 0$

Subtract $4 x$ from $6 x$ .

$\frac{2 x + 12 - x^{2}}{3 x (x + 4)} = 0$

Reorder terms.

$\frac{- x^{2} + 2 x + 12}{3 x (x + 4)} = 0$

Split the fraction $\frac{- x^{2} + 2 x + 12}{3 x (x + 4)}$ into two fractions.

$\frac{- x^{2} + 2 x}{3 x (x + 4)} + \frac{12}{3 x (x + 4)} = 0$

Simplify each term.

Factor $x$ out of $- x^{2} + 2 x$ .

Factor $x$ out of $- x^{2}$ .

$\frac{x (- x) + 2 x}{3 x (x + 4)} + \frac{12}{3 x (x + 4)} = 0$

Factor $x$ out of $2 x$ .

$\frac{x (- x) + x \cdot 2}{3 x (x + 4)} + \frac{12}{3 x (x + 4)} = 0$

Factor $x$ out of $x (- x) + x \cdot 2$ .

$\frac{x (- x + 2)}{3 x (x + 4)} + \frac{12}{3 x (x + 4)} = 0$

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{x (- x + 2)}{3 x (x + 4)} + \frac{12}{3 x (x + 4)} = 0$

Rewrite the expression.

$\frac{- x + 2}{3 (x + 4)} + \frac{12}{3 x (x + 4)} = 0$

Split the fraction $\frac{- x + 2}{3 (x + 4)}$ into two fractions.

$\frac{- x}{3 (x + 4)} + \frac{2}{3 (x + 4)} + \frac{12}{3 x (x + 4)} = 0$

Move the negative in front of the fraction.

$- \frac{x}{3 (x + 4)} + \frac{2}{3 (x + 4)} + \frac{12}{3 x (x + 4)} = 0$

Cancel the common factor of $12$ and $3$ .

Factor $3$ out of $12$ .

$- \frac{x}{3 (x + 4)} + \frac{2}{3 (x + 4)} + \frac{3 \cdot 4}{3 x (x + 4)} = 0$

Cancel the common factors.

Factor $3$ out of $3 x (x + 4)$ .

$- \frac{x}{3 (x + 4)} + \frac{2}{3 (x + 4)} + \frac{3 \cdot 4}{3 (x (x + 4))} = 0$

Cancel the common factor.

$- \frac{x}{3 (x + 4)} + \frac{2}{3 (x + 4)} + \frac{3 \cdot 4}{3 (x (x + 4))} = 0$

Rewrite the expression.

$- \frac{x}{3 (x + 4)} + \frac{2}{3 (x + 4)} + \frac{4}{x (x + 4)} = 0$

Find the LCD of the terms in the equation.

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

$3 (x + 4), 3 (x + 4), x (x + 4), 1$

Since $3 (x + 4), 3 (x + 4), x (x + 4), 1$ 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 $3 (x + 4), 3 (x + 4), x (x + 4), 1$ are:

1. Find the LCM for the numeric part $3, 3, 1, 1$ .

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

3. Find the LCM for the compound variable part $x + 4, x + 4, x + 4$ .

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.

Since $3$ has no factors besides $1$ and $3$ .

$3$ is a prime number

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

Not prime

The LCM of $3, 3, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$3$

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

$x^{1} = x$

$x$ occurs $1$ time.

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

$x$

The factor for $x + 4$ is $x + 4$ itself.

$(x + 4) = x + 4$

$(x + 4)$ occurs $1$ time.

The LCM of $x + 4, x + 4, x + 4$ is the result of multiplying all factors the greatest number of times they occur in either term.

$x + 4$

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$3 x (x + 4)$

Multiply each term in $- \frac{x}{3 (x + 4)} + \frac{2}{3 (x + 4)} + \frac{4}{x (x + 4)} = 0$ by $3 x (x + 4)$ to eliminate the fractions.

Multiply each term in $- \frac{x}{3 (x + 4)} + \frac{2}{3 (x + 4)} + \frac{4}{x (x + 4)} = 0$ by $3 x (x + 4)$ .

$- \frac{x}{3 (x + 4)} (3 x (x + 4)) + \frac{2}{3 (x + 4)} (3 x (x + 4)) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Simplify the left side.

Simplify each term.

Cancel the common factor of $3 (x + 4)$ .

Move the leading negative in $- \frac{x}{3 (x + 4)}$ into the numerator.

$\frac{- x}{3 (x + 4)} (3 x (x + 4)) + \frac{2}{3 (x + 4)} (3 x (x + 4)) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Factor $3 (x + 4)$ out of $3 x (x + 4)$ .

$\frac{- x}{3 (x + 4)} (3 (x + 4) (x)) + \frac{2}{3 (x + 4)} (3 x (x + 4)) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Cancel the common factor.

$\frac{- x}{3 (x + 4)} (3 (x + 4) x) + \frac{2}{3 (x + 4)} (3 x (x + 4)) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Rewrite the expression.

$- x \cdot x + \frac{2}{3 (x + 4)} (3 x (x + 4)) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Raise $x$ to the power of $1$ .

$- (x^{1} x) + \frac{2}{3 (x + 4)} (3 x (x + 4)) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Raise $x$ to the power of $1$ .

$- (x^{1} x^{1}) + \frac{2}{3 (x + 4)} (3 x (x + 4)) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- x^{1 + 1} + \frac{2}{3 (x + 4)} (3 x (x + 4)) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Add $1$ and $1$ .

$- x^{2} + \frac{2}{3 (x + 4)} (3 x (x + 4)) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Rewrite using the commutative property of multiplication.

$- x^{2} + 3 \frac{2}{3 (x + 4)} (x (x + 4)) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Cancel the common factor of $3$ .

Cancel the common factor.

$- x^{2} + 3 \frac{2}{3 (x + 4)} (x (x + 4)) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Rewrite the expression.

$- x^{2} + \frac{2}{x + 4} (x (x + 4)) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Cancel the common factor of $x + 4$ .

Factor $x + 4$ out of $x (x + 4)$ .

$- x^{2} + \frac{2}{x + 4} ((x + 4) x) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Cancel the common factor.

$- x^{2} + \frac{2}{x + 4} ((x + 4) x) + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Rewrite the expression.

$- x^{2} + 2 x + \frac{4}{x (x + 4)} (3 x (x + 4)) = 0 (3 x (x + 4))$

Rewrite using the commutative property of multiplication.

$- x^{2} + 2 x + 3 \frac{4}{x (x + 4)} (x (x + 4)) = 0 (3 x (x + 4))$

Multiply $3 \frac{4}{x (x + 4)}$ .

Combine $3$ and $\frac{4}{x (x + 4)}$ .

$- x^{2} + 2 x + \frac{3 \cdot 4}{x (x + 4)} (x (x + 4)) = 0 (3 x (x + 4))$

Multiply $3$ by $4$ .

$- x^{2} + 2 x + \frac{12}{x (x + 4)} (x (x + 4)) = 0 (3 x (x + 4))$

Cancel the common factor of $x (x + 4)$ .

Cancel the common factor.

$- x^{2} + 2 x + \frac{12}{x (x + 4)} (x (x + 4)) = 0 (3 x (x + 4))$

Rewrite the expression.

$- x^{2} + 2 x + 12 = 0 (3 x (x + 4))$

Simplify the right side.

Apply the distributive property.

$- x^{2} + 2 x + 12 = 0 (3 x \cdot x + 3 x \cdot 4)$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$- x^{2} + 2 x + 12 = 0 (3 (x \cdot x) + 3 x \cdot 4)$

Multiply $x$ by $x$ .

$- x^{2} + 2 x + 12 = 0 (3 x^{2} + 3 x \cdot 4)$

Multiply $4$ by $3$ .

$- x^{2} + 2 x + 12 = 0 (3 x^{2} + 12 x)$

Multiply $0$ by $3 x^{2} + 12 x$ .

$- x^{2} + 2 x + 12 = 0$

Solve the equation.

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = - 1$ , $b = 2$ , and $c = 12$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (- 1 \cdot 12)}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1 \cdot 12}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4 \cdot 12}}{2 \cdot - 1}$

Multiply $4$ by $12$ .

$x = \frac{- 2 \pm \sqrt{4 + 48}}{2 \cdot - 1}$

Add $4$ and $48$ .

$x = \frac{- 2 \pm \sqrt{52}}{2 \cdot - 1}$

Rewrite $52$ as $2^{2} \cdot 13$ .

Factor $4$ out of $52$ .

$x = \frac{- 2 \pm \sqrt{4 (13)}}{2 \cdot - 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 13}}{2 \cdot - 1}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{13}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- 2 \pm 2 \sqrt{13}}{- 2}$

Simplify $\frac{- 2 \pm 2 \sqrt{13}}{- 2}$ .

$x = 1 \pm \sqrt{13}$

The final answer is the combination of both solutions.

$x = 1 + \sqrt{13}, 1 - \sqrt{13}$

$x = 1 \pm \sqrt{13}$

The result can be shown in multiple forms.

Exact Form:

$x = 1 \pm \sqrt{13}$

Decimal Form:

$x = 4.60555127 \dots, - 2.60555127 \dots$

Solve Using the Quadratic Formula 1/x+1/(x+4)=1/3

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