Multiplying and Dividing Rational Expressions Worksheets - Math Monks - Free Printable

Let's solve each of these problems step by step. We are multiplying and dividing rational expressions. The key steps include:

1. Factor all numerators and denominators.
2. Simplify by canceling common factors.
3. For division, multiply by the reciprocal of the divisor.

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Problem 1:

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

Step 1: Factor
- $2x + 4 = 2(x + 2)$
- $3x + 6 = 3(x + 2)$

So:
$$
= 2(x + 2) \cdot \frac{3x}{3(x + 2)}
$$

Step 2: Cancel common factors
- $3$ in numerator and denominator → cancel
- $(x + 2)$ in numerator and denominator → cancel

$$
= 2 \cdot \frac{x}{1} = 2x
$$

✔ Answer: $ \boxed{2x} $

---

Problem 2:

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

Step 1: Factor
- $3x + 6 = 3(x + 2)$
- $2x + 4 = 2(x + 2)$

So:
$$
= 3(x + 2) \cdot \frac{2x}{2(x + 2)}
$$

Step 2: Cancel
- $2$ in numerator and denominator → cancel
- $(x + 2)$ → cancel

$$
= 3 \cdot x = 3x
$$

✔ Answer: $ \boxed{3x} $

---

Problem 3:

$$
\frac{x^2 - 4}{x^2 - 1} \cdot \frac{x + 1}{x + 2}
$$

Step 1: Factor
- $x^2 - 4 = (x - 2)(x + 2)$
- $x^2 - 1 = (x - 1)(x + 1)$

So:
$$
= \frac{(x - 2)(x + 2)}{(x - 1)(x + 1)} \cdot \frac{x + 1}{x + 2}
$$

Step 2: Cancel
- $(x + 2)$ → cancel
- $(x + 1)$ → cancel

$$
= \frac{x - 2}{x - 1}
$$

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

---

Problem 4:

$$
\frac{x^2 - 1}{x - 3} \cdot \frac{x^2 - 9}{x + 1}
$$

Step 1: Factor
- $x^2 - 1 = (x - 1)(x + 1)$
- $x^2 - 9 = (x - 3)(x + 3)$

So:
$$
= \frac{(x - 1)(x + 1)}{x - 3} \cdot \frac{(x - 3)(x + 3)}{x + 1}
$$

Step 2: Cancel
- $(x + 1)$ → cancel
- $(x - 3)$ → cancel

$$
= (x - 1)(x + 3)
$$

✔ Answer: $ \boxed{(x - 1)(x + 3)} $ or $ \boxed{x^2 + 2x - 3} $

---

Problem 5:

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

Step 1: Factor
- $x^2 - x - 12 = (x - 4)(x + 3)$
- $3x - 9 = 3(x - 3)$

So:
$$
= \frac{(x - 4)(x + 3)}{3(x - 3)} \div \frac{x - 4}{12}
$$

Step 2: Multiply by reciprocal
$$
= \frac{(x - 4)(x + 3)}{3(x - 3)} \cdot \frac{12}{x - 4}
$$

Step 3: Cancel
- $(x - 4)$ → cancel
- $12 / 3 = 4$

$$
= \frac{(x + 3)}{(x - 3)} \cdot 4 = \frac{4(x + 3)}{x - 3}
$$

✔ Answer: $ \boxed{\frac{4(x + 3)}{x - 3}} $

---

Problem 6:

$$
\frac{x^2}{6 - 3x} \div \frac{x}{8 - 4x}
$$

Step 1: Factor
- $6 - 3x = 3(2 - x) = -3(x - 2)$
- $8 - 4x = 4(2 - x) = -4(x - 2)$

So:
$$
= \frac{x^2}{-3(x - 2)} \div \frac{x}{-4(x - 2)}
$$

Step 2: Multiply by reciprocal
$$
= \frac{x^2}{-3(x - 2)} \cdot \frac{-4(x - 2)}{x}
$$

Step 3: Cancel
- $x$ → cancel one from $x^2$
- $(x - 2)$ → cancel
- $-3$ and $-4$: negatives cancel → positive; $4/3$

$$
= \frac{x}{1} \cdot \frac{4}{3} = \frac{4x}{3}
$$

✔ Answer: $ \boxed{\frac{4x}{3}} $

---

Problem 7:

$$
\frac{x - 4}{x^2 - 5x + 6} \div \frac{x - 3}{x^2 - 6x + 9}
$$

Step 1: Factor
- $x^2 - 5x + 6 = (x - 2)(x - 3)$
- $x^2 - 6x + 9 = (x - 3)^2$

So:
$$
= \frac{x - 4}{(x - 2)(x - 3)} \div \frac{x - 3}{(x - 3)^2}
$$

Note: $\frac{x - 3}{(x - 3)^2} = \frac{1}{x - 3}$

But let's do it properly:

$$
= \frac{x - 4}{(x - 2)(x - 3)} \cdot \frac{(x - 3)^2}{x - 3}
$$

Wait — actually, reciprocal of $\frac{x - 3}{(x - 3)^2}$ is $\frac{(x - 3)^2}{x - 3} = x - 3$

So:
$$
= \frac{x - 4}{(x - 2)(x - 3)} \cdot (x - 3)
$$

Cancel $(x - 3)$:
$$
= \frac{x - 4}{x - 2}
$$

✔ Answer: $ \boxed{\frac{x - 4}{x - 2}} $

---

Problem 8:

$$
\frac{x^2 - 9}{6x} \div \frac{x^2 - 9}{18x^3}
$$

Step 1: Factor
- $x^2 - 9 = (x - 3)(x + 3)$

So:
$$
= \frac{(x - 3)(x + 3)}{6x} \div \frac{(x - 3)(x + 3)}{18x^3}
$$

Step 2: Multiply by reciprocal
$$
= \frac{(x - 3)(x + 3)}{6x} \cdot \frac{18x^3}{(x - 3)(x + 3)}
$$

Step 3: Cancel
- $(x - 3)(x + 3)$ → cancel
- $18 / 6 = 3$
- $x^3 / x = x^2$

$$
= 3x^2
$$

✔ Answer: $ \boxed{3x^2} $

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✔ Final Answers:

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

Let me know if you'd like a visual explanation or help with restrictions on variables!