Solve for x 1+2/(x-5)=(3x+5)/(x^2-25)

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

Move all terms not containing $x$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

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

Simplify each term.

Simplify the denominator.

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

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

Split the fraction $\frac{3 x + 5}{(x + 5) (x - 5)}$ into two fractions.

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

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.

$x - 5, (x + 5) (x - 5), (x + 5) (x - 5), 1$

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

Not prime

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

$1$

The factor for $x - 5$ is $x - 5$ itself.

$(x - 5) = x - 5$

$(x - 5)$ occurs $1$ time.

The factor for $x + 5$ is $x + 5$ itself.

$(x + 5) = x + 5$

$(x + 5)$ occurs $1$ time.

The factor for $x - 5$ is $x - 5$ itself.

$(x - 5) = x - 5$

$(x - 5)$ occurs $1$ time.

The factor for $x + 5$ is $x + 5$ itself.

$(x + 5) = x + 5$

$(x + 5)$ occurs $1$ time.

The factor for $x - 5$ is $x - 5$ itself.

$(x - 5) = x - 5$

$(x - 5)$ occurs $1$ time.

The LCM of $x - 5, x + 5, x - 5, x + 5, x - 5$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x - 5) (x + 5)$

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

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

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

Simplify the left side.

Cancel the common factor of $x - 5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

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

Multiply $2$ by $5$ .

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

Simplify the right side.

Simplify each term.

Cancel the common factor of $(x - 5) (x + 5)$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $(x - 5) (x + 5)$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

$2 x + 10 = 3 x + 5 - ((x - 5) (x + 5))$

Expand $(x - 5) (x + 5)$ using the FOIL Method.

Apply the distributive property.

$2 x + 10 = 3 x + 5 - (x (x + 5) - 5 (x + 5))$

Apply the distributive property.

$2 x + 10 = 3 x + 5 - (x \cdot x + x \cdot 5 - 5 (x + 5))$

Apply the distributive property.

$2 x + 10 = 3 x + 5 - (x \cdot x + x \cdot 5 - 5 x - 5 \cdot 5)$

Combine the opposite terms in $x \cdot x + x \cdot 5 - 5 x - 5 \cdot 5$ .

Reorder the factors in the terms $x \cdot 5$ and $- 5 x$ .

$2 x + 10 = 3 x + 5 - (x \cdot x + 5 x - 5 x - 5 \cdot 5)$

Subtract $5 x$ from $5 x$ .

$2 x + 10 = 3 x + 5 - (x \cdot x + 0 - 5 \cdot 5)$

Add $x \cdot x$ and $0$ .

$2 x + 10 = 3 x + 5 - (x \cdot x - 5 \cdot 5)$

Simplify each term.

Multiply $x$ by $x$ .

$2 x + 10 = 3 x + 5 - (x^{2} - 5 \cdot 5)$

Multiply $- 5$ by $5$ .

$2 x + 10 = 3 x + 5 - (x^{2} - 25)$

Apply the distributive property.

$2 x + 10 = 3 x + 5 - x^{2} - - 25$

Multiply $- 1$ by $- 25$ .

$2 x + 10 = 3 x + 5 - x^{2} + 25$

Add $5$ and $25$ .

$2 x + 10 = 3 x - x^{2} + 30$

Solve the equation.

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$3 x - x^{2} + 30 = 2 x + 10$

Move all terms containing $x$ to the left side of the equation.

Subtract $2 x$ from both sides of the equation.

$3 x - x^{2} + 30 - 2 x = 10$

Subtract $2 x$ from $3 x$ .

$x - x^{2} + 30 = 10$

Subtract $10$ from both sides of the equation.

$x - x^{2} + 30 - 10 = 0$

Subtract $10$ from $30$ .

$x - x^{2} + 20 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $x - x^{2} + 20$ .

Reorder $x$ and $- x^{2}$ .

$- x^{2} + x + 20 = 0$

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + x + 20 = 0$

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 20 = 0$

Rewrite $20$ as $- 1 (- 20)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 20 = 0$

Factor $- 1$ out of $- (x^{2}) - 1 (- x)$ .

$- (x^{2} - x) - 1 \cdot - 20 = 0$

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 20)$ .

$- (x^{2} - x - 20) = 0$

Factor.

Factor $x^{2} - x - 20$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Write the factored form using these integers.

$- ((x - 5) (x + 4)) = 0$

Remove unnecessary parentheses.

$- (x - 5) (x + 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 5 = 0$

$x + 4 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $- (x - 5) (x + 4) = 0$ true.

$x = 5, - 4$

Exclude the solutions that do not make $1 + \frac{2}{x - 5} = \frac{3 x + 5}{x^{2} - 25}$ true.

$x = - 4$

Solve for x 1+2/(x-5)=(3x+5)/(x^2-25)

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