Solve for x 1/(x-2)-1/(x+3)=1/5

$\frac{1}{x - 2} - \frac{1}{x + 3} = \frac{1}{5}$

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.

$x - 2, x + 3, 5$

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

Not prime

Since $5$ has no factors besides $1$ and $5$ .

$5$ is a prime number

The LCM of $1, 1, 5$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$5$

The factor for $x - 2$ is $x - 2$ itself.

$(x - 2) = x - 2$

$(x - 2)$ occurs $1$ time.

The factor for $x + 3$ is $x + 3$ itself.

$(x + 3) = x + 3$

$(x + 3)$ occurs $1$ time.

The LCM of $x - 2, x + 3$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x - 2) (x + 3)$

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

$5 (x - 2) (x + 3)$

Multiply each term in $\frac{1}{x - 2} - \frac{1}{x + 3} = \frac{1}{5}$ by $5 (x - 2) (x + 3)$ to eliminate the fractions.

Multiply each term in $\frac{1}{x - 2} - \frac{1}{x + 3} = \frac{1}{5}$ by $5 (x - 2) (x + 3)$ .

$\frac{1}{x - 2} (5 (x - 2) (x + 3)) - \frac{1}{x + 3} (5 (x - 2) (x + 3)) = \frac{1}{5} (5 (x - 2) (x + 3))$

Simplify the left side.

Simplify each term.

Rewrite using the commutative property of multiplication.

$5 \frac{1}{x - 2} ((x - 2) (x + 3)) - \frac{1}{x + 3} (5 (x - 2) (x + 3)) = \frac{1}{5} (5 (x - 2) (x + 3))$

Combine $5$ and $\frac{1}{x - 2}$ .

$\frac{5}{x - 2} ((x - 2) (x + 3)) - \frac{1}{x + 3} (5 (x - 2) (x + 3)) = \frac{1}{5} (5 (x - 2) (x + 3))$

Cancel the common factor of $x - 2$ .

Cancel the common factor.

$\frac{5}{x - 2} ((x - 2) (x + 3)) - \frac{1}{x + 3} (5 (x - 2) (x + 3)) = \frac{1}{5} (5 (x - 2) (x + 3))$

Rewrite the expression.

$5 (x + 3) - \frac{1}{x + 3} (5 (x - 2) (x + 3)) = \frac{1}{5} (5 (x - 2) (x + 3))$

Apply the distributive property.

$5 x + 5 \cdot 3 - \frac{1}{x + 3} (5 (x - 2) (x + 3)) = \frac{1}{5} (5 (x - 2) (x + 3))$

Multiply $5$ by $3$ .

$5 x + 15 - \frac{1}{x + 3} (5 (x - 2) (x + 3)) = \frac{1}{5} (5 (x - 2) (x + 3))$

Cancel the common factor of $x + 3$ .

Move the leading negative in $- \frac{1}{x + 3}$ into the numerator.

$5 x + 15 + \frac{- 1}{x + 3} (5 (x - 2) (x + 3)) = \frac{1}{5} (5 (x - 2) (x + 3))$

Factor $x + 3$ out of $5 (x - 2) (x + 3)$ .

$5 x + 15 + \frac{- 1}{x + 3} ((x + 3) (5 (x - 2))) = \frac{1}{5} (5 (x - 2) (x + 3))$

Cancel the common factor.

$5 x + 15 + \frac{- 1}{x + 3} ((x + 3) (5 (x - 2))) = \frac{1}{5} (5 (x - 2) (x + 3))$

Rewrite the expression.

$5 x + 15 - (5 (x - 2)) = \frac{1}{5} (5 (x - 2) (x + 3))$

Multiply $5$ by $- 1$ .

$5 x + 15 - 5 (x - 2) = \frac{1}{5} (5 (x - 2) (x + 3))$

Apply the distributive property.

$5 x + 15 - 5 x - 5 \cdot - 2 = \frac{1}{5} (5 (x - 2) (x + 3))$

Multiply $- 5$ by $- 2$ .

$5 x + 15 - 5 x + 10 = \frac{1}{5} (5 (x - 2) (x + 3))$

Simplify by adding terms.

Combine the opposite terms in $5 x + 15 - 5 x + 10$ .

Subtract $5 x$ from $5 x$ .

$0 + 15 + 10 = \frac{1}{5} (5 (x - 2) (x + 3))$

Add $0$ and $15$ .

$15 + 10 = \frac{1}{5} (5 (x - 2) (x + 3))$

Add $15$ and $10$ .

$25 = \frac{1}{5} (5 (x - 2) (x + 3))$

Simplify the right side.

Cancel the common factor of $5$ .

Factor $5$ out of $5 (x - 2) (x + 3)$ .

$25 = \frac{1}{5} (5 ((x - 2) (x + 3)))$

Cancel the common factor.

$25 = \frac{1}{5} (5 ((x - 2) (x + 3)))$

Rewrite the expression.

$25 = (x - 2) (x + 3)$

Expand $(x - 2) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$25 = x (x + 3) - 2 (x + 3)$

Apply the distributive property.

$25 = x \cdot x + x \cdot 3 - 2 (x + 3)$

Apply the distributive property.

$25 = x \cdot x + x \cdot 3 - 2 x - 2 \cdot 3$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$25 = x^{2} + x \cdot 3 - 2 x - 2 \cdot 3$

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

$25 = x^{2} + 3 \cdot x - 2 x - 2 \cdot 3$

Multiply $- 2$ by $3$ .

$25 = x^{2} + 3 x - 2 x - 6$

Subtract $2 x$ from $3 x$ .

$25 = x^{2} + x - 6$

Solve the equation.

Rewrite the equation as $x^{2} + x - 6 = 25$ .

$x^{2} + x - 6 = 25$

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

Subtract $25$ from both sides of the equation.

$x^{2} + x - 6 - 25 = 0$

Subtract $25$ from $- 6$ .

$x^{2} + x - 31 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 1$ , and $c = - 31$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot - 31)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 31}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 31}}{2 \cdot 1}$

Multiply $- 4$ by $- 31$ .

$x = \frac{- 1 \pm \sqrt{1 + 124}}{2 \cdot 1}$

Add $1$ and $124$ .

$x = \frac{- 1 \pm \sqrt{125}}{2 \cdot 1}$

Rewrite $125$ as $5^{2} \cdot 5$ .

Factor $25$ out of $125$ .

$x = \frac{- 1 \pm \sqrt{25 (5)}}{2 \cdot 1}$

Rewrite $25$ as $5^{2}$ .

$x = \frac{- 1 \pm \sqrt{5^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 1 \pm 5 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm 5 \sqrt{5}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{1 - 5 \sqrt{5}}{2}, - \frac{1 + 5 \sqrt{5}}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1 - 5 \sqrt{5}}{2}, - \frac{1 + 5 \sqrt{5}}{2}$

Decimal Form:

$x = 5.09016994 \dots, - 6.09016994 \dots$

Solve for x 1/(x-2)-1/(x+3)=1/5

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