Question

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\\\frac{\frac{x^2-4}{x-2} + \frac{x^2-9}{x-3}}{\frac{x^2-1}{x-1} - \frac{x^2-16}{x-4}} \quad (x \
eq 2, 3, 1, 4)\\

Explanation:

⚡ Using what you learned: Simplifying Rational Expressions

Step 1: Simplify the numerator terms

Factor the difference of squares in each numerator of the top expression:

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

$$ \frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3} = x+3 $$

Sum of the simplified numerator terms:

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

Step 2: Simplify the denominator terms

Factor the difference of squares in each numerator of the bottom expression:

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

$$ \frac{x^2-16}{x-4} = \frac{(x-4)(x+4)}{x-4} = x+4 $$

Difference of the simplified denominator terms:

$$ (x+1) - (x+4) = x + 1 - x - 4 = -3 $$

Step 3: Combine and write the final expression

Combine the simplified numerator and denominator:

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

Answer:

$$ -\frac{2x+5}{3} $$