Teaching Algebraic Fractions - Free Printable

To solve the given problems, we will simplify each algebraic fraction step by step. Let's go through each problem:

---

Problem 1:

$$
\frac{x^2 - 64}{x - 8}
$$

#### Solution:
- The numerator $ x^2 - 64 $ is a difference of squares, which can be factored as:
$$
x^2 - 64 = (x - 8)(x + 8)
$$
- Substitute this into the fraction:
$$
\frac{x^2 - 64}{x - 8} = \frac{(x - 8)(x + 8)}{x - 8}
$$
- Cancel the common factor $ x - 8 $ (assuming $ x \neq 8 $):
$$
\frac{(x - 8)(x + 8)}{x - 8} = x + 8
$$

#### Final Answer:
$$
\boxed{x + 8}
$$

---

Problem 2:

$$
\frac{3x^2 + 3x}{9x^2 + 9x}
$$

#### Solution:
- Factor out the greatest common factor (GCF) in both the numerator and the denominator:
$$
3x^2 + 3x = 3x(x + 1)
$$
$$
9x^2 + 9x = 9x(x + 1)
$$
- Substitute these into the fraction:
$$
\frac{3x^2 + 3x}{9x^2 + 9x} = \frac{3x(x + 1)}{9x(x + 1)}
$$
- Cancel the common factors $ 3x $ and $ x + 1 $ (assuming $ x \neq 0 $ and $ x \neq -1 $):
$$
\frac{3x(x + 1)}{9x(x + 1)} = \frac{3}{9} = \frac{1}{3}
$$

#### Final Answer:
$$
\boxed{\frac{1}{3}}
$$

---

Problem 3:

$$
\frac{6x^2 + 9x}{4x + 6}
$$

#### Solution:
- Factor out the GCF in both the numerator and the denominator:
$$
6x^2 + 9x = 3x(2x + 3)
$$
$$
4x + 6 = 2(2x + 3)
$$
- Substitute these into the fraction:
$$
\frac{6x^2 + 9x}{4x + 6} = \frac{3x(2x + 3)}{2(2x + 3)}
$$
- Cancel the common factor $ 2x + 3 $ (assuming $ x \neq -\frac{3}{2} $):
$$
\frac{3x(2x + 3)}{2(2x + 3)} = \frac{3x}{2}
$$

#### Final Answer:
$$
\boxed{\frac{3x}{2}}
$$

---

Problem 4:

$$
\frac{6y}{xy + y^2}
$$

#### Solution:
- Factor out the GCF in the denominator:
$$
xy + y^2 = y(x + y)
$$
- Substitute this into the fraction:
$$
\frac{6y}{xy + y^2} = \frac{6y}{y(x + y)}
$$
- Cancel the common factor $ y $ (assuming $ y \neq 0 $):
$$
\frac{6y}{y(x + y)} = \frac{6}{x + y}
$$

#### Final Answer:
$$
\boxed{\frac{6}{x + y}}
$$

---

Problem 5:

$$
\frac{a + b}{5a + 5b}
$$

#### Solution:
- Factor out the GCF in the denominator:
$$
5a + 5b = 5(a + b)
$$
- Substitute this into the fraction:
$$
\frac{a + b}{5a + 5b} = \frac{a + b}{5(a + b)}
$$
- Cancel the common factor $ a + b $ (assuming $ a + b \neq 0 $):
$$
\frac{a + b}{5(a + b)} = \frac{1}{5}
$$

#### Final Answer:
$$
\boxed{\frac{1}{5}}
$$

---

Problem 6:

$$
\frac{(x - 3)(x + 4)}{2x - 6}
$$

#### Solution:
- Factor out the GCF in the denominator:
$$
2x - 6 = 2(x - 3)
$$
- Substitute this into the fraction:
$$
\frac{(x - 3)(x + 4)}{2x - 6} = \frac{(x - 3)(x + 4)}{2(x - 3)}
$$
- Cancel the common factor $ x - 3 $ (assuming $ x \neq 3 $):
$$
\frac{(x - 3)(x + 4)}{2(x - 3)} = \frac{x + 4}{2}
$$

#### Final Answer:
$$
\boxed{\frac{x + 4}{2}}
$$

---

Problem 7:

$$
\frac{4x}{7} - \frac{x - 1}{9}
$$

#### Solution:
- Find a common denominator for the fractions. The least common denominator (LCD) of 7 and 9 is 63.
- Rewrite each fraction with the common denominator:
$$
\frac{4x}{7} = \frac{4x \cdot 9}{7 \cdot 9} = \frac{36x}{63}
$$
$$
\frac{x - 1}{9} = \frac{(x - 1) \cdot 7}{9 \cdot 7} = \frac{7(x - 1)}{63} = \frac{7x - 7}{63}
$$
- Subtract the fractions:
$$
\frac{4x}{7} - \frac{x - 1}{9} = \frac{36x}{63} - \frac{7x - 7}{63} = \frac{36x - (7x - 7)}{63}
$$
- Simplify the numerator:
$$
36x - (7x - 7) = 36x - 7x + 7 = 29x + 7
$$
- Write the final fraction:
$$
\frac{29x + 7}{63}
$$

#### Final Answer:
$$
\boxed{\frac{29x + 7}{63}}
$$

---

Problem 8:

$$
\frac{p^2 - 4}{p - 2}
$$

#### Solution:
- The numerator $ p^2 - 4 $ is a difference of squares, which can be factored as:
$$
p^2 - 4 = (p - 2)(p + 2)
$$
- Substitute this into the fraction:
$$
\frac{p^2 - 4}{p - 2} = \frac{(p - 2)(p + 2)}{p - 2}
$$
- Cancel the common factor $ p - 2 $ (assuming $ p \neq 2 $):
$$
\frac{(p - 2)(p + 2)}{p - 2} = p + 2
$$

#### Final Answer:
$$
\boxed{p + 2}
$$

---

Problem 9:

$$
\frac{3a - ab}{3a - ab^2}
$$

#### Solution:
- Factor out the GCF in both the numerator and the denominator:
$$
3a - ab = a(3 - b)
$$
$$
3a - ab^2 = a(3 - b^2)
$$
- Substitute these into the fraction:
$$
\frac{3a - ab}{3a - ab^2} = \frac{a(3 - b)}{a(3 - b^2)}
$$
- Cancel the common factor $ a $ (assuming $ a \neq 0 $):
$$
\frac{a(3 - b)}{a(3 - b^2)} = \frac{3 - b}{3 - b^2}
$$
- The denominator $ 3 - b^2 $ is a difference of squares, which can be factored as:
$$
3 - b^2 = ( \sqrt{3} - b )( \sqrt{3} + b )
$$
- However, no further simplification is possible without additional context.

#### Final Answer:
$$
\boxed{\frac{3 - b}{3 - b^2}}
$$

---

Problem 10:

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

#### Solution:
- Factor the numerator and the denominator:
$$
x^2 + 3x = x(x + 3)
$$
$$
x^2 - 9 = (x - 3)(x + 3)
$$
- Substitute these into the fraction:
$$
\frac{x^2 + 3x}{x^2 - 9} = \frac{x(x + 3)}{(x - 3)(x + 3)}
$$
- Cancel the common factor $ x + 3 $ (assuming $ x \neq -3 $):
$$
\frac{x(x + 3)}{(x - 3)(x + 3)} = \frac{x}{x - 3}
$$

#### Final Answer:
$$
\boxed{\frac{x}{x - 3}}
$$

---

Problem 11:

$$
\frac{1 + x}{1 - x^2}
$$

#### Solution:
- The denominator $ 1 - x^2 $ is a difference of squares, which can be factored as:
$$
1 - x^2 = (1 - x)(1 + x)
$$
- Substitute this into the fraction:
$$
\frac{1 + x}{1 - x^2} = \frac{1 + x}{(1 - x)(1 + x)}
$$
- Cancel the common factor $ 1 + x $ (assuming $ x \neq -1 $):
$$
\frac{1 + x}{(1 - x)(1 + x)} = \frac{1}{1 - x}
$$

#### Final Answer:
$$
\boxed{\frac{1}{1 - x}}
$$

---

Problem 12:

$$
\frac{4x - 12}{x^2 - 3x}
$$

#### Solution:
- Factor out the GCF in the numerator and the denominator:
$$
4x - 12 = 4(x - 3)
$$
$$
x^2 - 3x = x(x - 3)
$$
- Substitute these into the fraction:
$$
\frac{4x - 12}{x^2 - 3x} = \frac{4(x - 3)}{x(x - 3)}
$$
- Cancel the common factor $ x - 3 $ (assuming $ x \neq 3 $):
$$
\frac{4(x - 3)}{x(x - 3)} = \frac{4}{x}
$$

#### Final Answer:
$$
\boxed{\frac{4}{x}}
$$

---

Problem 13:

$$
\frac{7x}{y(x + y)}
$$

#### Solution:
- This fraction is already in its simplest form, as there are no common factors to cancel.

#### Final Answer:
$$
\boxed{\frac{7x}{y(x + y)}}
$$

---

Problem 14:

$$
\frac{1}{x(x^2 + 3)}
$$

#### Solution:
- This fraction is already in its simplest form, as there are no common factors to cancel.

#### Final Answer:
$$
\boxed{\frac{1}{x(x^2 + 3)}}
$$

---

Final Answers:

1. $\boxed{x + 8}$
2. $\boxed{\frac{1}{3}}$
3. $\boxed{\frac{3x}{2}}$
4. $\boxed{\frac{6}{x + y}}$
5. $\boxed{\frac{1}{5}}$
6. $\boxed{\frac{x + 4}{2}}$
7. $\boxed{\frac{29x + 7}{63}}$
8. $\boxed{p + 2}$
9. $\boxed{\frac{3 - b}{3 - b^2}}$
10. $\boxed{\frac{x}{x - 3}}$
11. $\boxed{\frac{1}{1 - x}}$
12. $\boxed{\frac{4}{x}}$
13. $\boxed{\frac{7x}{y(x + y)}}$
14. $\boxed{\frac{1}{x(x^2 + 3)}}$