Question

each expression.
answer on the blank provided.

1. \\(\frac{x^2 - 9}{x^2 - 4x - 21}\\)
2. \\(\frac{x + 7}{3x - 12} cdot \frac{x - 4}{x^2 + 2x - 35}\\)
3. \\(\frac{-7x + 5}{x^2 - 9} - \frac{-5x + 11}{x^2 - 9}\\)
4. \\(\frac{8x}{8x^2 - 16x} div \frac{3}{12x - 24}\\)
5. \\(\frac{x^2 - 25}{3x + 12} cdot \frac{x + 4}{x - 5}\\)
6. \\(\frac{x}{x^2 - 3x - 10} + \frac{5x + 1}{x^2 - 3x - 10}\\)
7. \\(\frac{2x + 8}{x^2 + 7x + 12}\\)
8. \\(\frac{x^2 + 2x}{x^2 - 49} div \frac{x^2 + 5x + 6}{x^2 + 10x + 21}\\)

Explanation:

Step1: Factor numerator and denominator

Factor $x^2-9=(x-3)(x+3)$, $x^2-4x-21=(x-7)(x+3)$
$\frac{(x-3)(x+3)}{(x-7)(x+3)}$

Step2: Cancel common factors

Cancel $(x+3)$ (where $x
eq-3$)
$\frac{x-3}{x-7}$

---

Step1: Factor all terms

Factor $3x-12=3(x-4)$, $x^2+2x-35=(x+7)(x-5)$
$\frac{x+7}{3(x-4)} \cdot \frac{x-4}{(x+7)(x-5)}$

Step2: Cancel common factors

Cancel $(x+7)$ and $(x-4)$ (where $x
eq-7,4$)
$\frac{1}{3(x-5)} = \frac{1}{3x-15}$

---

Step1: Combine fractions over common denominator

Same denominator, subtract numerators
$\frac{(-7x+5)-(-5x+11)}{x^2-9}$

Step2: Simplify numerator

$\frac{-7x+5+5x-11}{x^2-9} = \frac{-2x-6}{x^2-9}$

Step3: Factor and cancel

Factor numerator $-2(x+3)$, denominator $(x-3)(x+3)$; cancel $(x+3)$ (where $x
eq-3$)
$\frac{-2}{x-3} = \frac{2}{3-x}$

---

Step1: Rewrite division as multiplication

Multiply by reciprocal of second fraction
$\frac{8x}{8x^2-16x} \cdot \frac{12x-24}{3}$

Step2: Factor all terms

Factor $8x^2-16x=8x(x-2)$, $12x-24=12(x-2)$
$\frac{8x}{8x(x-2)} \cdot \frac{12(x-2)}{3}$

Step3: Cancel common factors

Cancel $8x$ and $(x-2)$ (where $x
eq0,2$)
$\frac{12}{3}=4$

---

Step1: Factor all terms

Factor $x^2-25=(x-5)(x+5)$, $3x+12=3(x+4)$
$\frac{(x-5)(x+5)}{3(x+4)} \cdot \frac{x+4}{x-5}$

Step2: Cancel common factors

Cancel $(x-5)$ and $(x+4)$ (where $x
eq5,-4$)
$\frac{x+5}{3}$

---

Step1: Combine fractions over common denominator

Same denominator, add numerators
$\frac{x+(5x+1)}{x^2-3x-10}$

Step2: Simplify numerator and factor denominator

$\frac{6x+1}{(x-5)(x+2)}$ (no common factors to cancel)

---

Step1: Factor numerator and denominator

Factor $2x+8=2(x+4)$, $x^2+7x+12=(x+3)(x+4)$
$\frac{2(x+4)}{(x+3)(x+4)}$

Step2: Cancel common factors

Cancel $(x+4)$ (where $x
eq-4$)
$\frac{2}{x+3}$

---

Step1: Rewrite division as multiplication

Multiply by reciprocal of second fraction
$\frac{x^2+2x}{x^2-49} \cdot \frac{x^2+10x+21}{x^2+5x+6}$

Step2: Factor all terms

Factor $x^2+2x=x(x+2)$, $x^2-49=(x-7)(x+7)$, $x^2+10x+21=(x+3)(x+7)$, $x^2+5x+6=(x+2)(x+3)$
$\frac{x(x+2)}{(x-7)(x+7)} \cdot \frac{(x+3)(x+7)}{(x+2)(x+3)}$

Step3: Cancel common factors

Cancel $(x+2)$, $(x+7)$, $(x+3)$ (where $x
eq-2,-7,-3$)
$\frac{x}{x-7}$

Answer:

1. $\boldsymbol{\frac{x-3}{x-7}}$
2. $\boldsymbol{\frac{1}{3x-15}}$
3. $\boldsymbol{\frac{2}{3-x}}$
4. $\boldsymbol{4}$
5. $\boldsymbol{\frac{x+5}{3}}$
6. $\boldsymbol{\frac{6x+1}{(x-5)(x+2)}}$
7. $\boldsymbol{\frac{2}{x+3}}$
8. $\boldsymbol{\frac{x}{x-7}}$