Handwritten solutions to a math worksheet on multiplying and dividing rational expressions.

Let’s solve each problem one by one. We’ll multiply or divide rational expressions, factor when needed, and simplify.

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Problem 1:

$$
\frac{2x + 6}{5x + 10} \cdot \frac{x + 2}{x^2 + 4x + 3}
$$

Step 1: Factor everything.

- $2x + 6 = 2(x + 3)$
- $5x + 10 = 5(x + 2)$
- $x^2 + 4x + 3 = (x + 1)(x + 3)$

So now:

$$
\frac{2(x + 3)}{5(x + 2)} \cdot \frac{x + 2}{(x + 1)(x + 3)}
$$

Step 2: Cancel common factors.

- $(x + 3)$ cancels top and bottom.
- $(x + 2)$ cancels top and bottom.

Left with:

$$
\frac{2}{5(x + 1)}
$$

✔ Final Answer for #1: $\boxed{\dfrac{2}{5(x + 1)}}$

---

Problem 2:

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

Dividing means flip the second fraction and multiply:

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

Factor:

- $x^2 - x - 12 = (x - 4)(x + 3)$
- $3x - 9 = 3(x - 3)$

Now:

$$
\frac{(x - 4)(x + 3)}{3(x - 3)} \cdot \frac{12}{x - 4}
$$

Cancel $(x - 4)$ top and bottom.

Multiply numerators: $(x + 3) \cdot 12 = 12(x + 3)$

Denominator: $3(x - 3)$

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

✔ Final Answer for #2: $\boxed{\dfrac{4(x + 3)}{x - 3}}$

---

Problem 3:

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

Flip and multiply:

$$
\frac{x^2 - 5x + 4}{x^2} \cdot \frac{x}{x - 4}
$$

Factor numerator: $x^2 - 5x + 4 = (x - 1)(x - 4)$

Now:

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

Cancel $(x - 4)$ and one $x$:

Left with: $\frac{(x - 1)}{x}$

✔ Final Answer for #3: $\boxed{\dfrac{x - 1}{x}}$

---

Problem 4:

$$
\frac{6}{x^2 + 9x + 20} \cdot \frac{8x + 40}{6x - 12}
$$

Factor:

- $x^2 + 9x + 20 = (x + 4)(x + 5)$
- $8x + 40 = 8(x + 5)$
- $6x - 12 = 6(x - 2)$

Now:

$$
\frac{6}{(x + 4)(x + 5)} \cdot \frac{8(x + 5)}{6(x - 2)}
$$

Cancel 6 top and bottom.

Cancel $(x + 5)$ top and bottom.

Left with:

$$
\frac{8}{(x + 4)(x - 2)}
$$

Wait — let me check again.

After canceling 6 and (x+5):

Numerator: 8
Denominator: (x + 4)(x - 2)

Yes.

But wait — original has 6 in numerator and denominator? Let's write it clearly:

Original after factoring:

$$
\frac{6}{(x+4)(x+5)} \cdot \frac{8(x+5)}{6(x-2)}
= \frac{6 \cdot 8(x+5)}{(x+4)(x+5) \cdot 6(x-2)}
$$

Cancel 6, cancel (x+5):

→ $\frac{8}{(x+4)(x-2)}$

✔ Final Answer for #4: $\boxed{\dfrac{8}{(x + 4)(x - 2)}}$

---

Problem 5:

$$
\frac{5x - 15}{4x^2} \cdot \frac{x^3}{6x - 18}
$$

Factor:

- $5x - 15 = 5(x - 3)$
- $6x - 18 = 6(x - 3)$

Now:

$$
\frac{5(x - 3)}{4x^2} \cdot \frac{x^3}{6(x - 3)}
$$

Cancel $(x - 3)$

Cancel $x^2$ from denominator with $x^3$ → leaves $x$ on top

So:

Numerator: $5 \cdot x = 5x$
Denominator: $4 \cdot 6 = 24$

Answer: $\frac{5x}{24}$

Wait — let’s do step by step:

After canceling (x-3):

$$
\frac{5}{4x^2} \cdot \frac{x^3}{6} = \frac{5 \cdot x^3}{4x^2 \cdot 6} = \frac{5x}{24}
$$

✔ Final Answer for #5: $\boxed{\dfrac{5x}{24}}$

---

Problem 6:

$$
\frac{7x^2}{12x} \cdot \frac{14x^3}{48y^3} \cdot \frac{48y^3}{14x^3}
$$

Look closely — last two fractions are reciprocals!

$\frac{14x^3}{48y^3} \cdot \frac{48y^3}{14x^3} = 1$

So whole thing becomes:

$$
\frac{7x^2}{12x} \cdot 1 = \frac{7x^2}{12x} = \frac{7x}{12}
$$

Wait — but looking at student work, they got $\frac{24y^3}{x}$ — that doesn’t match.

Wait — maybe I misread.

Actually, looking back:

The expression is:

$$
\frac{7x^2}{12x} \cdot \frac{14x^3}{48y^3} \cdot \frac{48y^3}{14x^3}
$$

Yes — middle and last cancel to 1.

First term: $\frac{7x^2}{12x} = \frac{7x}{12}$

But student wrote answer as $\frac{24y^3}{x}$ — which is wrong.

Wait — perhaps typo in problem? Or maybe I need to recompute.

Alternatively, maybe the third fraction is different? In image, it looks like:

Student wrote: “48y³ / 14x³” — yes.

So multiplication:

Numerator: $7x^2 \cdot 14x^3 \cdot 48y^3$

Denominator: $12x \cdot 48y^3 \cdot 14x^3$

Everything cancels except:

From first fraction: 7x² / 12x → 7x/12

Others cancel completely.

So answer should be $\frac{7x}{12}$

But student has boxed $\frac{24y^3}{x}$ — incorrect.

Wait — unless the problem was written differently?

Looking again at user image — Problem 6:

It says:

$$
\frac{7x^2}{12x} \cdot \frac{14x^3}{48y^3} \cdot \frac{48y^3}{14x^3}
$$

Yes — so indeed, last two multiply to 1.

So final answer is $\frac{7x}{12}$

But student has $\frac{24y^3}{x}$ — probably made a mistake.

However, since we’re solving correctly, we go with math.

✔ Final Answer for #6: $\boxed{\dfrac{7x}{12}}$

Wait — hold on! Maybe I misread the problem.

In the image, problem 6 might be:

Actually, looking carefully — the student wrote:

“7x² / 12x · 14x³ / 48y³ · 48y³ / 14x³”

But then their answer is “24y³ / x” — which suggests they may have multiplied wrong.

Perhaps the third fraction is not reciprocal? No — it is.

Another possibility: maybe the problem is:

$$
\frac{7x^2}{12x} \cdot \frac{14x^3}{48y^3} \div \frac{14x^3}{48y^3}
$$

No — it’s written as multiplication.

I think student made error. Correct answer is $\frac{7x}{12}$

But let me double-check with numbers.

Let x=2, y=1

First fraction: 7*(4)/(12*2) = 28/24 = 7/6

Second: 14*8 / 48*1 = 112/48 = 7/3

Third: 48*1 / 14*8 = 48/112 = 3/7

Multiply all: (7/6) * (7/3) * (3/7) = (7/6)*(1) = 7/6

And 7x/12 = 14/12 = 7/6 — matches.

So correct.

✔ Final Answer for #6: $\boxed{\dfrac{7x}{12}}$

---

Problem 7:

$$
\frac{x^2 + 5x - 24}{2x + 2} \cdot \frac{3x + 24}{x^2 - 8x - 9}
$$

Factor:

- $x^2 + 5x - 24 = (x + 8)(x - 3)$
- $2x + 2 = 2(x + 1)$
- $3x + 24 = 3(x + 8)$
- $x^2 - 8x - 9 = (x - 9)(x + 1)$

Now:

$$
\frac{(x + 8)(x - 3)}{2(x + 1)} \cdot \frac{3(x + 8)}{(x - 9)(x + 1)}
$$

Multiply numerators: $(x + 8)(x - 3) \cdot 3(x + 8) = 3(x + 8)^2 (x - 3)$

Denominators: $2(x + 1) \cdot (x - 9)(x + 1) = 2(x + 1)^2 (x - 9)$

No common factors to cancel? Wait — no (x+8) in denominator, etc.

So answer is:

$$
\frac{3(x + 8)^2 (x - 3)}{2(x + 1)^2 (x - 9)}
$$

But student has boxed something else — let’s see what they did.

They wrote: “x² -12 +27 over 6” — which is messy.

Actually, perhaps they tried to combine incorrectly.

We must leave factored unless instructed otherwise.

But let me check if any cancellation possible — no.

So this is simplified.

But maybe we can write as:

$$
\frac{3(x + 8)^2 (x - 3)}{2(x + 1)^2 (x - 9)}
$$

✔ Final Answer for #7: $\boxed{\dfrac{3(x + 8)^2 (x - 3)}{2(x + 1)^2 (x - 9)}}$

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Problem 8:

$$
\frac{30x^4}{24x^3} \cdot \frac{12x}{50x} \cdot \frac{8x^2}{9x}
$$

Simplify each fraction first or multiply all together.

Numerator: $30x^4 \cdot 12x \cdot 8x^2 = 30 \cdot 12 \cdot 8 \cdot x^{4+1+2} = 2880 x^7$

Denominator: $24x^3 \cdot 50x \cdot 9x = 24 \cdot 50 \cdot 9 \cdot x^{3+1+1} = 10800 x^5$

So:

$$
\frac{2880 x^7}{10800 x^5} = \frac{2880}{10800} x^{2}
$$

Simplify fraction: divide numerator and denominator by 1440? Let's find GCD.

2880 ÷ 1440 = 2, 10800 ÷ 1440 = 7.5 — not integer.

Divide by 240: 2880÷240=12, 10800÷240=45 → 12/45 = 4/15

Or step by step:

2880 / 10800 = 288 / 1080 = 144 / 540 = 72 / 270 = 36 / 135 = 12 / 45 = 4 / 15

So $\frac{4}{15} x^2$

But student has $\frac{9x}{5x}$ — which is 9/5 — wrong.

Wait — perhaps I miscalculated coefficients.

30 * 12 * 8 = 30*96 = 2880 — ok

24 * 50 * 9 = 24*450 = 10800 — ok

Exponents: x^4 * x * x^2 = x^7; x^3 * x * x = x^5 — ok

2880 / 10800 = reduce by dividing by 240: 2880÷240=12, 10800÷240=45 → 12/45=4/15

Yes.

So $\frac{4x^2}{15}$

✔ Final Answer for #8: $\boxed{\dfrac{4x^2}{15}}$

---

Problem 9:

$$
\frac{4x}{8x + 8} \cdot \frac{x^2 + 8x + 7}{8x^2}
$$

Factor:

- $8x + 8 = 8(x + 1)$
- $x^2 + 8x + 7 = (x + 1)(x + 7)$

Now:

$$
\frac{4x}{8(x + 1)} \cdot \frac{(x + 1)(x + 7)}{8x^2}
$$

Cancel (x+1)

Cancel x: 4x / 8x^2 = 4/(8x) = 1/(2x) after canceling x

Better:

Numerator: 4x * (x+1)(x+7)

Denominator: 8(x+1) * 8x^2 = 64 x^2 (x+1)

Cancel (x+1) and one x:

Left: numerator: 4(x+7), denominator: 64x

So: $\frac{4(x+7)}{64x} = \frac{x+7}{16x}$

✔ Final Answer for #9: $\boxed{\dfrac{x + 7}{16x}}$

---

Problem 10:

$$
\frac{6x - 12}{x^2 - 9x + 18} \cdot \frac{7x - 21}{5x - 10}
$$

Factor:

- $6x - 12 = 6(x - 2)$
- $x^2 - 9x + 18 = (x - 3)(x - 6)$
- $7x - 21 = 7(x - 3)$
- $5x - 10 = 5(x - 2)$

Now:

$$
\frac{6(x - 2)}{(x - 3)(x - 6)} \cdot \frac{7(x - 3)}{5(x - 2)}
$$

Cancel (x-2) and (x-3)

Left with:

Numerator: 6 * 7 = 42

Denominator: (x - 6) * 5 = 5(x - 6)

So: $\frac{42}{5(x - 6)}$

✔ Final Answer for #10: $\boxed{\dfrac{42}{5(x - 6)}}$

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## Final Answers Summary:

1. $\boxed{\dfrac{2}{5(x + 1)}}$
2. $\boxed{\dfrac{4(x + 3)}{x - 3}}$
3. $\boxed{\dfrac{x - 1}{x}}$
4. $\boxed{\dfrac{8}{(x + 4)(x - 2)}}$
5. $\boxed{\dfrac{5x}{24}}$
6. $\boxed{\dfrac{7x}{12}}$
7. $\boxed{\dfrac{3(x + 8)^2 (x - 3)}{2(x + 1)^2 (x - 9)}}$
8. $\boxed{\dfrac{4x^2}{15}}$
9. $\boxed{\dfrac{x + 7}{16x}}$
10. $\boxed{\dfrac{42}{5(x - 6)}}$