Algebraic fractions and rational expressions practice problems.

Problem Analysis

The task involves simplifying and manipulating algebraic fractions. We need to simplify each fraction step by step and identify any patterns or relationships between them. Let's go through each fraction systematically.

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Left Column: Simplifying Fractions

#### 1. First Fraction:
$$
\frac{2(x+3)}{3(x+3)}
$$
- The numerator is \(2(x+3)\).
- The denominator is \(3(x+3)\).
- Since \(x+3\) is a common factor in both the numerator and the denominator, we can cancel it out (assuming \(x \neq -3\)):
$$
\frac{2(x+3)}{3(x+3)} = \frac{2}{3}
$$

#### 2. Second Fraction:
$$
\frac{2(x+3)}{(x+3)}
$$
- The numerator is \(2(x+3)\).
- The denominator is \(x+3\).
- Again, \(x+3\) is a common factor, so we can cancel it out (assuming \(x \neq -3\)):
$$
\frac{2(x+3)}{(x+3)} = 2
$$

#### 3. Third Fraction:
$$
\frac{(x+3)}{2(x+3)}
$$
- The numerator is \(x+3\).
- The denominator is \(2(x+3)\).
- Here, \(x+3\) is a common factor, so we can cancel it out (assuming \(x \neq -3\)):
$$
\frac{(x+3)}{2(x+3)} = \frac{1}{2}
$$

#### 4. Fourth Fraction:
$$
\frac{x+3}{2x+6}
$$
- The numerator is \(x+3\).
- The denominator is \(2x+6\), which can be factored as \(2(x+3)\):
$$
\frac{x+3}{2x+6} = \frac{x+3}{2(x+3)}
$$
- Now, \(x+3\) is a common factor, so we can cancel it out (assuming \(x \neq -3\)):
$$
\frac{x+3}{2(x+3)} = \frac{1}{2}
$$

#### 5. Fifth Fraction:
$$
\frac{(x+3)(x+4)}{2x+6}
$$
- The numerator is \((x+3)(x+4)\).
- The denominator is \(2x+6\), which can be factored as \(2(x+3)\):
$$
\frac{(x+3)(x+4)}{2x+6} = \frac{(x+3)(x+4)}{2(x+3)}
$$
- Here, \(x+3\) is a common factor, so we can cancel it out (assuming \(x \neq -3\)):
$$
\frac{(x+3)(x+4)}{2(x+3)} = \frac{x+4}{2}
$$

#### 6. Sixth Fraction:
$$
\frac{(x+3)(x+4)}{2x+8}
$$
- The numerator is \((x+3)(x+4)\).
- The denominator is \(2x+8\), which can be factored as \(2(x+4)\):
$$
\frac{(x+3)(x+4)}{2x+8} = \frac{(x+3)(x+4)}{2(x+4)}
$$
- Here, \(x+4\) is a common factor, so we can cancel it out (assuming \(x \neq -4\)):
$$
\frac{(x+3)(x+4)}{2(x+4)} = \frac{x+3}{2}
$$

#### 7. Seventh Fraction:
$$
\frac{x^2 + 7x + 12}{2x + 8}
$$
- The numerator is \(x^2 + 7x + 12\), which can be factored as \((x+3)(x+4)\):
$$
x^2 + 7x + 12 = (x+3)(x+4)
$$
- The denominator is \(2x+8\), which can be factored as \(2(x+4)\):
$$
\frac{x^2 + 7x + 12}{2x + 8} = \frac{(x+3)(x+4)}{2(x+4)}
$$
- Here, \(x+4\) is a common factor, so we can cancel it out (assuming \(x \neq -4\)):
$$
\frac{(x+3)(x+4)}{2(x+4)} = \frac{x+3}{2}
$$

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Right Column: Simplifying Fractions

#### 1. First Fraction:
$$
\frac{x^2 + 7x + 12}{(x+4)(x-9)}
$$
- The numerator is \(x^2 + 7x + 12\), which can be factored as \((x+3)(x+4)\):
$$
x^2 + 7x + 12 = (x+3)(x+4)
$$
- The denominator is \((x+4)(x-9)\).
- Here, \(x+4\) is a common factor, so we can cancel it out (assuming \(x \neq -4\)):
$$
\frac{x^2 + 7x + 12}{(x+4)(x-9)} = \frac{(x+3)(x+4)}{(x+4)(x-9)} = \frac{x+3}{x-9}
$$

#### 2. Second Fraction:
$$
\frac{x^2 + 7x + 12}{x^2 - 5x - 36}
$$
- The numerator is \(x^2 + 7x + 12\), which can be factored as \((x+3)(x+4)\):
$$
x^2 + 7x + 12 = (x+3)(x+4)
$$
- The denominator is \(x^2 - 5x - 36\), which can be factored as \((x+4)(x-9)\):
$$
x^2 - 5x - 36 = (x+4)(x-9)
$$
- Here, \(x+4\) is a common factor, so we can cancel it out (assuming \(x \neq -4\)):
$$
\frac{x^2 + 7x + 12}{x^2 - 5x - 36} = \frac{(x+3)(x+4)}{(x+4)(x-9)} = \frac{x+3}{x-9}
$$

#### 3. Third Fraction:
$$
\frac{2x^2 + 14x + 24}{x^2 - 5x - 36}
$$
- The numerator is \(2x^2 + 14x + 24\), which can be factored as \(2(x^2 + 7x + 12)\):
$$
2x^2 + 14x + 24 = 2(x^2 + 7x + 12) = 2(x+3)(x+4)
$$
- The denominator is \(x^2 - 5x - 36\), which can be factored as \((x+4)(x-9)\):
$$
x^2 - 5x - 36 = (x+4)(x-9)
$$
- Here, \(x+4\) is a common factor, so we can cancel it out (assuming \(x \neq -4\)):
$$
\frac{2x^2 + 14x + 24}{x^2 - 5x - 36} = \frac{2(x+3)(x+4)}{(x+4)(x-9)} = \frac{2(x+3)}{x-9}
$$

#### 4. Fourth Fraction:
$$
\frac{2x^2 + 14x + 24}{3x^2 - 15x - 108}
$$
- The numerator is \(2x^2 + 14x + 24\), which can be factored as \(2(x+3)(x+4)\):
$$
2x^2 + 14x + 24 = 2(x+3)(x+4)
$$
- The denominator is \(3x^2 - 15x - 108\), which can be factored as \(3(x^2 - 5x - 36)\):
$$
3x^2 - 15x - 108 = 3(x^2 - 5x - 36) = 3(x+4)(x-9)
$$
- Here, \(x+4\) is a common factor, so we can cancel it out (assuming \(x \neq -4\)):
$$
\frac{2x^2 + 14x + 24}{3x^2 - 15x - 108} = \frac{2(x+3)(x+4)}{3(x+4)(x-9)} = \frac{2(x+3)}{3(x-9)}
$$

#### 5. Fifth Fraction:
$$
\frac{2x^2 + 14x + 24}{3x^2 + 4x - 15}
$$
- The numerator is \(2x^2 + 14x + 24\), which can be factored as \(2(x+3)(x+4)\):
$$
2x^2 + 14x + 24 = 2(x+3)(x+4)
$$
- The denominator is \(3x^2 + 4x - 15\), which can be factored as \((3x-5)(x+3)\):
$$
3x^2 + 4x - 15 = (3x-5)(x+3)
$$
- Here, \(x+3\) is a common factor, so we can cancel it out (assuming \(x \neq -3\)):
$$
\frac{2x^2 + 14x + 24}{3x^2 + 4x - 15} = \frac{2(x+3)(x+4)}{(3x-5)(x+3)} = \frac{2(x+4)}{3x-5}
$$

#### 6. Sixth Fraction:
$$
\frac{14x - 24 - 2x^2}{3x^2 + 4x - 15}
$$
- The numerator is \(14x - 24 - 2x^2\), which can be rewritten as \(-2(x^2 - 7x + 12)\):
$$
14x - 24 - 2x^2 = -2(x^2 - 7x + 12)
$$
- The quadratic \(x^2 - 7x + 12\) can be factored as \((x-3)(x-4)\):
$$
x^2 - 7x + 12 = (x-3)(x-4)
$$
- So, the numerator becomes:
$$
14x - 24 - 2x^2 = -2(x-3)(x-4)
$$
- The denominator is \(3x^2 + 4x - 15\), which can be factored as \((3x-5)(x+3)\):
$$
3x^2 + 4x - 15 = (3x-5)(x+3)
$$
- There are no common factors between the numerator and the denominator, so the fraction remains:
$$
\frac{14x - 24 - 2x^2}{3x^2 + 4x - 15} = \frac{-2(x-3)(x-4)}{(3x-5)(x+3)}
$$

#### 7. Seventh Fraction:
$$
\frac{14x - 24 - 2x^2}{3x^2 - 4x - 15}
$$
- The numerator is \(14x - 24 - 2x^2\), which can be rewritten as \(-2(x^2 - 7x + 12)\):
$$
14x - 24 - 2x^2 = -2(x^2 - 7x + 12)
$$
- The quadratic \(x^2 - 7x + 12\) can be factored as \((x-3)(x-4)\):
$$
x^2 - 7x + 12 = (x-3)(x-4)
$$
- So, the numerator becomes:
$$
14x - 24 - 2x^2 = -2(x-3)(x-4)
$$
- The denominator is \(3x^2 - 4x - 15\), which can be factored as \((3x+5)(x-3)\):
$$
3x^2 - 4x - 15 = (3x+5)(x-3)
$$
- Here, \(x-3\) is a common factor, so we can cancel it out (assuming \(x \neq 3\)):
$$
\frac{14x - 24 - 2x^2}{3x^2 - 4x - 15} = \frac{-2(x-3)(x-4)}{(3x+5)(x-3)} = \frac{-2(x-4)}{3x+5}
$$

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Final Answer

The simplified forms of the fractions are:

Left Column:
1. \(\frac{2}{3}\)
2. \(2\)
3. \(\frac{1}{2}\)
4. \(\frac{1}{2}\)
5. \(\frac{x+4}{2}\)
6. \(\frac{x+3}{2}\)
7. \(\frac{x+3}{2}\)

Right Column:
1. \(\frac{x+3}{x-9}\)
2. \(\frac{x+3}{x-9}\)
3. \(\frac{2(x+3)}{x-9}\)
4. \(\frac{2(x+3)}{3(x-9)}\)
5. \(\frac{2(x+4)}{3x-5}\)
6. \(\frac{-2(x-3)(x-4)}{(3x-5)(x+3)}\)
7. \(\frac{-2(x-4)}{3x+5}\)

$$
\boxed{\text{See above for detailed simplifications.}}
$$