Rational Expressions Quiz with detailed solutions for factoring, canceling, and simplifying algebraic fractions.
A math worksheet titled "Rational Expressions Quiz" featuring several problems involving factoring, canceling, and simplifying rational expressions, with step-by-step solutions shown.
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Step-by-step solution for: SOLUTION: Simplifying rational expressions worksheet algebra ...
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Step-by-step solution for: SOLUTION: Simplifying rational expressions worksheet algebra ...
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 answers.
---
$$
\frac{x^2 + 11x - 12}{x^2 - 1} \cdot \frac{5x + 5}{3x + 1}
$$
#### 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:
$$
\boxed{\frac{5(x + 12)}{3x + 1}} \quad \text{or} \quad \frac{5x + 60}{3x + 1}
$$
---
$$
(2x + 16) \cdot \frac{5}{2x^2 + 17x + 8}
$$
#### 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:
$$
\boxed{\frac{10}{2x + 1}}
$$
---
$$
\frac{y - 6}{10} \div \frac{6 - y}{5}
$$
#### 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 $ (assuming $ y \ne 6 $):
$$
\frac{1}{10} \cdot \frac{5}{-1} = \frac{5}{-10} = -\frac{1}{2}
$$
✔ Final Answer:
$$
\boxed{-\frac{1}{2}}
$$
---
$$
\frac{2x^2 - 5x - 3}{4x^2 - 12x - 7} \div \frac{4x + 9}{2x - 7}
$$
#### Step 1: Invert and multiply.
$$
\frac{2x^2 - 5x - 3}{4x^2 - 12x - 7} \cdot \frac{2x - 7}{4x + 9}
$$
#### Step 2: Factor all quadratics.
- $ 2x^2 - 5x - 3 = (2x + 1)(x - 3) $
- $ 4x^2 - 12x - 7 = (2x - 7)(2x + 1) $
- $ 4x + 9 $ is irreducible
- $ 2x - 7 $ stays as is
Now:
$$
\frac{(2x + 1)(x - 3)}{(2x - 7)(2x + 1)} \cdot \frac{2x - 7}{4x + 9}
$$
#### Step 3: Cancel common factors.
- $ (2x + 1) $ cancels
- $ (2x - 7) $ cancels
Remaining:
$$
\frac{x - 3}{1} \cdot \frac{1}{4x + 9} = \frac{x - 3}{4x + 9}
$$
✔ Final Answer:
$$
\boxed{\frac{x - 3}{4x + 9}}
$$
---
$$
\frac{4s^2 + 8s - 12}{s^2 - 6s + 5} \div \frac{2s^2 + 7s + 3}{4s^2 - 8s - 5}
$$
#### 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 all expressions.
- $ 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
Remaining:
$$
\frac{4}{s - 5} \cdot (2s - 5) = \frac{4(2s - 5)}{s - 5}
$$
✔ Final Answer:
$$
\boxed{\frac{4(2s - 5)}{s - 5}}
$$
---
$$
\frac{\frac{x^2 + 3x + 2}{x^2 - 1}}{\frac{4x^2 + 8x}{5(x - 1)}}
$$
This is a complex fraction — division of two fractions.
#### 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 everything.
- $ x^2 + 3x + 2 = (x + 2)(x + 1) $
- $ x^2 - 1 = (x + 1)(x - 1) $
- $ 4x^2 + 8x = 4x(x + 2) $
- $ 5(x - 1) $ stays
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
Remaining:
$$
\frac{1}{1} \cdot \frac{5}{4x} = \frac{5}{4x}
$$
✔ Final Answer:
$$
\boxed{\frac{5}{4x}}
$$
---
| Problem | Simplified Answer |
|--------|-------------------|
| 1 | $ \frac{5(x + 12)}{3x + 1} $ |
| 2 | $ \frac{10}{2x + 1} $ |
| 3 | $ -\frac{1}{2} $ |
| 4 | $ \frac{x - 3}{4x + 9} $ |
| 5 | $ \frac{4(2s - 5)}{s - 5} $ |
| 6 | $ \frac{5}{4x} $ |
All solutions in the image are correct and well-explained.
---
- Factoring quadratics and special forms (difference of squares).
- Inverting and multiplying when dividing rational expressions.
- Canceling common factors in numerator and denominator.
- Using $-1$ as a factor to handle sign changes (e.g., $6 - y = -(y - 6)$).
Let me know if you'd like these problems explained in video format or want practice problems!
---
🔷 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:
$$
\boxed{\frac{5(x + 12)}{3x + 1}} \quad \text{or} \quad \frac{5x + 60}{3x + 1}
$$
---
🔷 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:
$$
\boxed{\frac{10}{2x + 1}}
$$
---
🔷 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 $ (assuming $ y \ne 6 $):
$$
\frac{1}{10} \cdot \frac{5}{-1} = \frac{5}{-10} = -\frac{1}{2}
$$
✔ Final Answer:
$$
\boxed{-\frac{1}{2}}
$$
---
🔷 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 all quadratics.
- $ 2x^2 - 5x - 3 = (2x + 1)(x - 3) $
- $ 4x^2 - 12x - 7 = (2x - 7)(2x + 1) $
- $ 4x + 9 $ is irreducible
- $ 2x - 7 $ stays as is
Now:
$$
\frac{(2x + 1)(x - 3)}{(2x - 7)(2x + 1)} \cdot \frac{2x - 7}{4x + 9}
$$
#### Step 3: Cancel common factors.
- $ (2x + 1) $ cancels
- $ (2x - 7) $ cancels
Remaining:
$$
\frac{x - 3}{1} \cdot \frac{1}{4x + 9} = \frac{x - 3}{4x + 9}
$$
✔ Final Answer:
$$
\boxed{\frac{x - 3}{4x + 9}}
$$
---
🔷 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 all expressions.
- $ 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
Remaining:
$$
\frac{4}{s - 5} \cdot (2s - 5) = \frac{4(2s - 5)}{s - 5}
$$
✔ Final Answer:
$$
\boxed{\frac{4(2s - 5)}{s - 5}}
$$
---
🔷 Problem 6:
$$
\frac{\frac{x^2 + 3x + 2}{x^2 - 1}}{\frac{4x^2 + 8x}{5(x - 1)}}
$$
This is a complex fraction — division of two fractions.
#### 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 everything.
- $ x^2 + 3x + 2 = (x + 2)(x + 1) $
- $ x^2 - 1 = (x + 1)(x - 1) $
- $ 4x^2 + 8x = 4x(x + 2) $
- $ 5(x - 1) $ stays
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
Remaining:
$$
\frac{1}{1} \cdot \frac{5}{4x} = \frac{5}{4x}
$$
✔ Final Answer:
$$
\boxed{\frac{5}{4x}}
$$
---
✔ Summary of All Answers:
| Problem | Simplified Answer |
|--------|-------------------|
| 1 | $ \frac{5(x + 12)}{3x + 1} $ |
| 2 | $ \frac{10}{2x + 1} $ |
| 3 | $ -\frac{1}{2} $ |
| 4 | $ \frac{x - 3}{4x + 9} $ |
| 5 | $ \frac{4(2s - 5)}{s - 5} $ |
| 6 | $ \frac{5}{4x} $ |
All solutions in the image are correct and well-explained.
---
📌 Key Concepts Used:
- Factoring quadratics and special forms (difference of squares).
- Inverting and multiplying when dividing rational expressions.
- Canceling common factors in numerator and denominator.
- Using $-1$ as a factor to handle sign changes (e.g., $6 - y = -(y - 6)$).
Let me know if you'd like these problems explained in video format or want practice problems!
Parent Tip: Review the logic above to help your child master the concept of simplifying rational expressions worksheet answer key.