Solve by Factoring 100/(x^2-25)-1=(2x)/(x-5)

$\frac{100}{x^{2} - 25} - 1 = \frac{2 x}{x - 5}$

Step 1

Subtract $\frac{2 x}{x - 5}$ from both sides of the equation.

$\frac{100}{x^{2} - 25} - 1 - \frac{2 x}{x - 5} = 0$

Step 2

Simplify $\frac{100}{x^{2} - 25} - 1 - \frac{2 x}{x - 5}$ .

Simplify the denominator.

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

$\frac{100}{x^{2} - 5^{2}} - 1 - \frac{2 x}{x - 5} = 0$

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 $b = 5$ .

$\frac{100}{(x + 5) (x - 5)} - 1 - \frac{2 x}{x - 5} = 0$

To write $- \frac{2 x}{x - 5}$ as a fraction with a common denominator, multiply by $\frac{x + 5}{x + 5}$ .

$\frac{100}{(x + 5) (x - 5)} - \frac{2 x}{x - 5} \cdot \frac{x + 5}{x + 5} - 1 = 0$

Write each expression with a common denominator of $(x + 5) (x - 5)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2 x}{x - 5}$ by $\frac{x + 5}{x + 5}$ .

$\frac{100}{(x + 5) (x - 5)} - \frac{2 x (x + 5)}{(x - 5) (x + 5)} - 1 = 0$

Reorder the factors of $(x - 5) (x + 5)$ .

$\frac{100}{(x + 5) (x - 5)} - \frac{2 x (x + 5)}{(x + 5) (x - 5)} - 1 = 0$

Combine the numerators over the common denominator.

$\frac{100 - 2 x (x + 5)}{(x + 5) (x - 5)} - 1 = 0$

Simplify each term.

Simplify the numerator.

Factor $2$ out of $100 - 2 x (x + 5)$ .

Factor $2$ out of $100$ .

$\frac{2 (50) - 2 x (x + 5)}{(x + 5) (x - 5)} - 1 = 0$

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

$\frac{2 (50) + 2 (- x (x + 5))}{(x + 5) (x - 5)} - 1 = 0$

Factor $2$ out of $2 (50) + 2 (- x (x + 5))$ .

$\frac{2 (50 - x (x + 5))}{(x + 5) (x - 5)} - 1 = 0$

Apply the distributive property.

$\frac{2 (50 - x \cdot x - x \cdot 5)}{(x + 5) (x - 5)} - 1 = 0$

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

Move $x$ .

$\frac{2 (50 - (x \cdot x) - x \cdot 5)}{(x + 5) (x - 5)} - 1 = 0$

Multiply $x$ by $x$ .

$\frac{2 (50 - x^{2} - x \cdot 5)}{(x + 5) (x - 5)} - 1 = 0$

Multiply $5$ by $- 1$ .

$\frac{2 (50 - x^{2} - 5 x)}{(x + 5) (x - 5)} - 1 = 0$

Reorder terms.

$\frac{2 (- x^{2} - 5 x + 50)}{(x + 5) (x - 5)} - 1 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 50 = - 50$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

$\frac{2 (- x^{2} - 5 (x) + 50)}{(x + 5) (x - 5)} - 1 = 0$

Rewrite $- 5$ as $5$ plus $- 10$

$\frac{2 (- x^{2} + (5 - 10) x + 50)}{(x + 5) (x - 5)} - 1 = 0$

Apply the distributive property.

$\frac{2 (- x^{2} + 5 x - 10 x + 50)}{(x + 5) (x - 5)} - 1 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{2 ((- x^{2} + 5 x) - 10 x + 50)}{(x + 5) (x - 5)} - 1 = 0$

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $- x + 5$ .

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

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

Cancel the common factor of $- x + 5$ and $x - 5$ .

Factor $- 1$ out of $- x$ .

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

Rewrite $5$ as $- 1 (- 5)$ .

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

Factor $- 1$ out of $- (x) - 1 (- 5)$ .

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

Rewrite $- (x - 5)$ as $- 1 (x - 5)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $2$ by $- 1$ .

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

Move the negative in front of the fraction.

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{x + 5}{x + 5}$ .

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Factor $- 1$ out of $- 2 (x + 10) - (x + 5)$ .

Factor $- 1$ out of $- 2 (x + 10)$ .

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

Factor $- 1$ out of $- (2 (x + 10)) - (x + 5)$ .

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

Apply the distributive property.

$\frac{- (2 x + 2 \cdot 10 + x + 5)}{x + 5} = 0$

Multiply $2$ by $10$ .

$\frac{- (2 x + 20 + x + 5)}{x + 5} = 0$

Add $2 x$ and $x$ .

$\frac{- (3 x + 20 + 5)}{x + 5} = 0$

Add $20$ and $5$ .

$\frac{- (3 x + 25)}{x + 5} = 0$

Move the negative in front of the fraction.

$- \frac{3 x + 25}{x + 5} = 0$

Step 3

Set the numerator equal to zero.

$3 x + 25 = 0$

Step 4

Solve the equation for $x$ .

Subtract $25$ from both sides of the equation.

$3 x = - 25$

Divide each term in $3 x = - 25$ by $3$ and simplify.

Divide each term in $3 x = - 25$ by $3$ .

$\frac{3 x}{3} = \frac{- 25}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 25}{3}$

Divide $x$ by $1$ .

$x = \frac{- 25}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{25}{3}$

Solve by Factoring 100/(x^2-25)-1=(2x)/(x-5)

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