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.

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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 $x$ .

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

Add $- 20$ and $5$ .

$\frac{- x - 15}{x + 5} = 0$

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

$\frac{- (x) - 15}{x + 5} = 0$

Rewrite $- 15$ as $- 1 (15)$ .

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

Factor $- 1$ out of $- (x) - 1 (15)$ .

$\frac{- (x + 15)}{x + 5} = 0$

Rewrite $- (x + 15)$ as $- 1 (x + 15)$ .

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

Move the negative in front of the fraction.

$- \frac{x + 15}{x + 5} = 0$

Step 3

Set the numerator equal to zero.

$x + 15 = 0$

Step 4

Subtract $15$ from both sides of the equation.

$x = - 15$

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

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