Rational Expressions Quiz worksheet with step-by-step solutions for simplifying algebraic fractions.

The image you've uploaded is a "Rational Expressions Quiz" that includes several problems involving multiplying and dividing rational expressions, with step-by-step solutions provided. I'll walk through each problem, explain the steps, and verify the solutions.

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🔷 Problem 1:

$$
\frac{x^2 + 11x - 12}{x^2 - 1} \cdot \frac{5x + 5}{3x + 1}
$$

#### Step 1: Factor all expressions.
- $ x^2 + 11x - 12 = (x - 1)(x + 12) $
- $ x^2 - 1 = (x + 1)(x - 1) $ → difference of squares
- $ 5x + 5 = 5(x + 1) $

So the expression becomes:
$$
\frac{(x - 1)(x + 12)}{(x + 1)(x - 1)} \cdot \frac{5(x + 1)}{3x + 1}
$$

#### Step 2: Cancel common factors.
- $ (x - 1) $ cancels from numerator and denominator.
- $ (x + 1) $ cancels from numerator and denominator.

Remaining:
$$
\frac{(x + 12)}{1} \cdot \frac{5}{3x + 1} = \frac{5(x + 12)}{3x + 1}
$$

✔ Final Answer:
$$
\boxed{\frac{5(x + 12)}{3x + 1}} \quad \text{or} \quad \frac{5x + 60}{3x + 1}
$$

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🔷 Problem 2:

$$
(2x + 16) \cdot \frac{5}{2x^2 + 17x + 8}
$$

#### Step 1: Factor.
- $ 2x + 16 = 2(x + 8) $
- $ 2x^2 + 17x + 8 = (2x + 1)(x + 8) $ → factor quadratic

Now:
$$
\frac{2(x + 8)}{1} \cdot \frac{5}{(2x + 1)(x + 8)}
$$

#### Step 2: Cancel common factors.
- $ (x + 8) $ cancels.

Remaining:
$$
\frac{2}{1} \cdot \frac{5}{2x + 1} = \frac{10}{2x + 1}
$$

✔ Final Answer:
$$
\boxed{\frac{10}{2x + 1}}
$$

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🔷 Problem 3:

$$
\frac{y - 6}{10} \div \frac{6 - y}{5}
$$

#### Step 1: Invert and multiply.
$$
\frac{y - 6}{10} \cdot \frac{5}{6 - y}
$$

Note: $ 6 - y = -(y - 6) $, so write:
$$
\frac{y - 6}{10} \cdot \frac{5}{-(y - 6)}
$$

#### Step 2: Cancel $ (y - 6) $
$$
\frac{1}{10} \cdot \frac{5}{-1} = \frac{5}{-10} = -\frac{1}{2}
$$

✔ Final Answer:
$$
\boxed{-\frac{1}{2}}
$$

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🔷 Problem 4:

$$
\frac{2x^2 - 5x - 3}{4x^2 - 12x - 7} \div \frac{4x + 9}{2x - 7}
$$

#### Step 1: Invert and multiply.
$$
\frac{2x^2 - 5x - 3}{4x^2 - 12x - 7} \cdot \frac{2x - 7}{4x + 9}
$$

#### Step 2: Factor numerators and denominators.
- $ 2x^2 - 5x - 3 = (2x + 1)(x - 3) $
- $ 4x^2 - 12x - 7 = (2x - 7)(2x + 1) $

So:
$$
\frac{(2x + 1)(x - 3)}{(2x - 7)(2x + 1)} \cdot \frac{2x - 7}{4x + 9}
$$

#### Step 3: Cancel common factors.
- $ (2x + 1) $ cancels
- $ (2x - 7) $ cancels

Left with:
$$
\frac{x - 3}{1} \cdot \frac{1}{4x + 9} = \frac{x - 3}{4x + 9}
$$

✔ Final Answer:
$$
\boxed{\frac{x - 3}{4x + 9}}
$$

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🔷 Problem 5:

$$
\frac{4s^2 + 8s - 12}{s^2 - 6s + 5} \div \frac{2s^2 + 7s + 3}{4s^2 - 8s - 5}
$$

#### Step 1: Invert and multiply.
$$
\frac{4s^2 + 8s - 12}{s^2 - 6s + 5} \cdot \frac{4s^2 - 8s - 5}{2s^2 + 7s + 3}
$$

#### Step 2: Factor everything.
- $ 4s^2 + 8s - 12 = 4(s^2 + 2s - 3) = 4(s + 3)(s - 1) $
- $ s^2 - 6s + 5 = (s - 1)(s - 5) $
- $ 4s^2 - 8s - 5 = (2s + 1)(2s - 5) $
- $ 2s^2 + 7s + 3 = (2s + 1)(s + 3) $

Now plug in:
$$
\frac{4(s + 3)(s - 1)}{(s - 1)(s - 5)} \cdot \frac{(2s + 1)(2s - 5)}{(2s + 1)(s + 3)}
$$

#### Step 3: Cancel common factors.
- $ (s - 1) $ cancels
- $ (s + 3) $ cancels
- $ (2s + 1) $ cancels

Left:
$$
\frac{4}{(s - 5)} \cdot (2s - 5) = \frac{4(2s - 5)}{s - 5}
$$

✔ Final Answer:
$$
\boxed{\frac{4(2s - 5)}{s - 5}}
$$

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🔷 Problem 6:

$$
\frac{\frac{x^2 + 3x + 2}{x^2 - 1}}{\frac{4x^2 + 8x}{5(x - 1)}}
$$

This is a complex fraction — divide the two rational expressions.

#### Step 1: Rewrite as multiplication by reciprocal.
$$
\frac{x^2 + 3x + 2}{x^2 - 1} \cdot \frac{5(x - 1)}{4x^2 + 8x}
$$

#### Step 2: Factor.
- $ x^2 + 3x + 2 = (x + 2)(x + 1) $
- $ x^2 - 1 = (x + 1)(x - 1) $
- $ 4x^2 + 8x = 4x(x + 2) $

Now:
$$
\frac{(x + 2)(x + 1)}{(x + 1)(x - 1)} \cdot \frac{5(x - 1)}{4x(x + 2)}
$$

#### Step 3: Cancel common factors.
- $ (x + 1) $ cancels
- $ (x - 1) $ cancels
- $ (x + 2) $ cancels

Left:
$$
\frac{1}{1} \cdot \frac{5}{4x} = \frac{5}{4x}
$$

✔ Final Answer:
$$
\boxed{\frac{5}{4x}}
$$

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✔ Summary of Answers:

| Problem | Simplified Answer |
|--------|-------------------|
| 1 | $ \dfrac{5(x + 12)}{3x + 1} $ |
| 2 | $ \dfrac{10}{2x + 1} $ |
| 3 | $ -\dfrac{1}{2} $ |
| 4 | $ \dfrac{x - 3}{4x + 9} $ |
| 5 | $ \dfrac{4(2s - 5)}{s - 5} $ |
| 6 | $ \dfrac{5}{4x} $ |

All solutions in the image are correct.

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📌 Key Concepts Used:

- Factoring quadratics
- Canceling common factors (only when they're identical)
- Inverting and multiplying for division
- Recognizing opposites: $ 6 - y = -(y - 6) $
- Simplifying complex fractions

Let me know if you'd like a video-style explanation or practice problems!