Question

1. solve \\(\frac{x^2 - 2x - 24}{x^2 - 16} \div \frac{x^2 + 11x + 30}{x^2 + 9x + 20}\\). completely simplify your answer and state any restrictions on the variable.

\\(\bigcirc \frac{x+5}{x+5}, x \
eq -4, x \
eq -5, x \
eq 3, x \
eq -6\\)
\\(\bigcirc \frac{x+6}{x+6}, x \
eq -4, x \
eq -5, x \
eq 3, x \
eq -6\\)
\\(\bigcirc \frac{x+5}{x+6}, x \
eq -4, x \
eq -5, x \
eq 3, x \
eq -6\\)
\\(\bigcirc \frac{x+6}{x+5}, x \
eq -4, x \
eq -5, x \
eq 3, x \
eq -6\\)

Explanation:

⚡ Using what you learned: Solving Rational Equations

Step 1: Identify the equation and find restrictions

The equation from the image is:

$$ \frac{x - 2}{x^2 - 16} = \frac{x - 3}{x^2 + 10x + 24} $$

Factor the denominators to find the values of \( x \) that make any denominator zero:

$$ x^2 - 16 = (x - 4)(x + 4) $$

$$ x^2 + 10x + 24 = (x + 4)(x + 6) $$

The restricted values are:

$$ x eq 4, \quad x eq -4, \quad x eq -6 $$

Step 2: Solve the equation

Multiply both sides by the least common denominator, \( (x - 4)(x + 4)(x + 6) \):

$$ (x - 2)(x + 6) = (x - 3)(x - 4) $$

Expand both sides:

$$ x^2 + 4x - 12 = x^2 - 7x + 12 $$

Subtract \( x^2 \) from both sides:

$$ 4x - 12 = -7x + 12 $$

Add \( 7x \) to both sides:

$$ 11x - 12 = 12 $$

Add \( 12 \) to both sides:

$$ 11x = 24 $$

$$ x = \frac{24}{11} $$

Since \( \frac{24}{11} \) is not one of the restricted values, it is the valid solution.

Answer:

$$ x = \frac{24}{11}, \quad x eq -4, \quad x eq -6, \quad x eq 4 $$