Solve for x 5/(x+2)+(x+2)/(2-x)=(x^2)/(4-x^2)

$\frac{5}{x + 2} + \frac{x + 2}{2 - x} = \frac{x^{2}}{4 - x^{2}}$

Step 1

Factor each term.

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

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

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

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

Step 2

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.

$x + 2, 2 - x, (2 + x) (2 - x)$

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

$1$

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

$(x + 2) = x + 2$

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

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

$(2 - x) = 2 - x$

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

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

$(2 + x) = 2 + x$

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

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

$(2 - x) = 2 - x$

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

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

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

Step 3

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

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

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

Simplify the left side.

Simplify each term.

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

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

Multiply $5$ by $2$ .

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

Multiply $- 1$ by $5$ .

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

Cancel the common factor of $2 - x$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Raise $x + 2$ to the power of $1$ .

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

Raise $x + 2$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

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

Simplify the right side.

Cancel the common factor of $2 - x$ .

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $\frac{x^{2}}{2 + x}$ by $x + 2$ .

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

Cancel the common factor of $x + 2$ and $2 + x$ .

Reorder terms.

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

Cancel the common factor.

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

Divide $x^{2}$ by $1$ .

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

Step 4

Solve the equation.

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

Subtract $x^{2}$ from both sides of the equation.

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

Simplify each term.

Rewrite ${(x + 2)}^{2}$ as $(x + 2) (x + 2)$ .

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

$10 - 5 x + x \cdot x + x \cdot 2 + 2 x + 2 \cdot 2 - x^{2} = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$10 - 5 x + x^{2} + x \cdot 2 + 2 x + 2 \cdot 2 - x^{2} = 0$

Move $2$ to the left of $x$ .

$10 - 5 x + x^{2} + 2 \cdot x + 2 x + 2 \cdot 2 - x^{2} = 0$

Multiply $2$ by $2$ .

$10 - 5 x + x^{2} + 2 x + 2 x + 4 - x^{2} = 0$

Add $2 x$ and $2 x$ .

$10 - 5 x + x^{2} + 4 x + 4 - x^{2} = 0$

Combine the opposite terms in $10 - 5 x + x^{2} + 4 x + 4 - x^{2}$ .

Subtract $x^{2}$ from $x^{2}$ .

$10 - 5 x + 4 x + 4 + 0 = 0$

Add $10 - 5 x + 4 x + 4$ and $0$ .

$10 - 5 x + 4 x + 4 = 0$

Add $10$ and $4$ .

$- 5 x + 4 x + 14 = 0$

Add $- 5 x$ and $4 x$ .

$- x + 14 = 0$

Subtract $14$ from both sides of the equation.

$- x = - 14$

Divide each term in $- x = - 14$ by $- 1$ and simplify.

Divide each term in $- x = - 14$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 14}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 14}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 14}{- 1}$

Simplify the right side.

Divide $- 14$ by $- 1$ .

$x = 14$

Solve for x 5/(x+2)+(x+2)/(2-x)=(x^2)/(4-x^2)

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