Simplifying Fractions Worksheets - Math Monks - Free Printable

Here is the complete solution for the Simplifying Algebraic Fractions Worksheet, with step-by-step explanations for each problem.

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① $\frac{x^2 - 64}{x - 8}$

- Step 1: Recognize that $x^2 - 64$ is a difference of squares:
$x^2 - 64 = (x - 8)(x + 8)$
- Step 2: Rewrite the fraction:
$\frac{(x - 8)(x + 8)}{x - 8}$
- Step 3: Cancel common factor $(x - 8)$, assuming $x \ne 8$:
$= x + 8$

✔ Answer: $x + 8$

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② $\frac{3x^2 + 3x}{9x^2 + 9x}$

- Step 1: Factor numerator and denominator:
Numerator: $3x(x + 1)$
Denominator: $9x(x + 1)$
- Step 2: Rewrite:
$\frac{3x(x + 1)}{9x(x + 1)}$
- Step 3: Cancel common factors $3x(x + 1)$, assuming $x \ne 0$ and $x \ne -1$:
$= \frac{1}{3}$

✔ Answer: $\frac{1}{3}$

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③ $\frac{6x^2 + 9x}{4x + 6}$

- Step 1: Factor numerator and denominator:
Numerator: $3x(2x + 3)$
Denominator: $2(2x + 3)$
- Step 2: Rewrite:
$\frac{3x(2x + 3)}{2(2x + 3)}$
- Step 3: Cancel common factor $(2x + 3)$, assuming $x \ne -\frac{3}{2}$:
$= \frac{3x}{2}$

✔ Answer: $\frac{3x}{2}$

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④ $\frac{6y}{xy + y^2}$

- Step 1: Factor denominator:
$xy + y^2 = y(x + y)$
- Step 2: Rewrite:
$\frac{6y}{y(x + y)}$
- Step 3: Cancel common factor $y$, assuming $y \ne 0$:
$= \frac{6}{x + y}$

✔ Answer: $\frac{6}{x + y}$

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⑤ $\frac{a + b}{5a + 5b}$

- Step 1: Factor denominator:
$5a + 5b = 5(a + b)$
- Step 2: Rewrite:
$\frac{a + b}{5(a + b)}$
- Step 3: Cancel common factor $(a + b)$, assuming $a + b \ne 0$:
$= \frac{1}{5}$

✔ Answer: $\frac{1}{5}$

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⑥ $\frac{(x - 3)(x + 4)}{2x - 6}$

- Step 1: Factor denominator:
$2x - 6 = 2(x - 3)$
- Step 2: Rewrite:
$\frac{(x - 3)(x + 4)}{2(x - 3)}$
- Step 3: Cancel common factor $(x - 3)$, assuming $x \ne 3$:
$= \frac{x + 4}{2}$

✔ Answer: $\frac{x + 4}{2}$

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⑦ $\frac{4x}{7} - \frac{x - 1}{9}$

- Step 1: Find common denominator → LCM of 7 and 9 is 63.
- Step 2: Rewrite both fractions:
$\frac{4x}{7} = \frac{4x \cdot 9}{63} = \frac{36x}{63}$
$\frac{x - 1}{9} = \frac{(x - 1) \cdot 7}{63} = \frac{7x - 7}{63}$
- Step 3: Subtract:
$\frac{36x - (7x - 7)}{63} = \frac{36x - 7x + 7}{63} = \frac{29x + 7}{63}$

✔ Answer: $\frac{29x + 7}{63}$

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⑧ $\frac{p^2 - 4}{p - 2}$

- Step 1: Factor numerator as difference of squares:
$p^2 - 4 = (p - 2)(p + 2)$
- Step 2: Rewrite:
$\frac{(p - 2)(p + 2)}{p - 2}$
- Step 3: Cancel $(p - 2)$, assuming $p \ne 2$:
$= p + 2$

✔ Answer: $p + 2$

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⑨ $\frac{3a - ab}{3a - ab^2}$

- Step 1: Factor numerator and denominator:
Numerator: $a(3 - b)$
Denominator: $a(3 - b^2)$ → Wait! Let’s check:
Actually, $3a - ab^2 = a(3 - b^2)$ — but that doesn’t match numerator. Let's re-factor carefully.

Actually:

Numerator: $3a - ab = a(3 - b)$
Denominator: $3a - ab^2 = a(3 - b^2)$ — still not matching.

Wait — perhaps we can factor differently? Let’s look again.

Actually, let’s factor out common terms:

Numerator: $a(3 - b)$
Denominator: $a(3 - b^2)$ — but $3 - b^2$ is not the same as $3 - b$. So unless we factor denominator as $a(3 - b^2)$, which doesn't cancel with numerator.

Wait — maybe there's a typo in the problem? Or perhaps it's meant to be factored as:

Let me try factoring denominator:
$3a - ab^2 = a(3 - b^2)$ — still no cancellation.

But notice: if we write numerator: $a(3 - b)$
Denominator: $a(3 - b^2) = a(3 - b)(3 + b)$? No — that’s incorrect. $3 - b^2$ is not factorable over integers unless written as difference of squares: $-(b^2 - 3)$, which doesn’t help.

Wait — perhaps I misread the denominator. Is it $3a - ab^2$ or $3a - ab$? The problem says $3a - ab^2$.

So:

$\frac{a(3 - b)}{a(3 - b^2)} = \frac{3 - b}{3 - b^2}$

Now, note that $3 - b^2 = - (b^2 - 3)$, but that doesn't help much. However, we can write:

$\frac{3 - b}{3 - b^2} = \frac{-(b - 3)}{-(b^2 - 3)} = \frac{b - 3}{b^2 - 3}$ — still not simplifying.

Wait — perhaps the problem was intended to be $\frac{3a - ab}{3a - ab}$? But it’s written as $3a - ab^2$.

Alternatively, maybe factor denominator as $a(3 - b^2)$, and leave it.

But actually, let’s check if numerator and denominator have a common factor.

Numerator: $a(3 - b)$
Denominator: $a(3 - b^2)$ — no common binomial factor unless $b = 1$ or something.

Wait — perhaps we can write:

$\frac{3 - b}{3 - b^2} = \frac{3 - b}{( \sqrt{3} - b)(\sqrt{3} + b)}$ — not helpful for simplification at this level.

I think there might be a typo in the worksheet. Perhaps it was meant to be $\frac{3a - ab}{3a - ab}$ or $\frac{3a - ab}{3a - a b}$ — same thing.

Wait — let me double-check the original image. Since I can’t see it, but based on standard worksheets, perhaps it’s $\frac{3a - ab}{3a - ab}$ — but that would be 1.

Alternatively, perhaps it’s $\frac{3a - ab}{3a - a b^2}$ — which is what we have.

Another possibility: factor denominator as $a(3 - b^2)$, and numerator as $a(3 - b)$, so:

$\frac{a(3 - b)}{a(3 - b^2)} = \frac{3 - b}{3 - b^2}$

And since $3 - b^2 = - (b^2 - 3)$, and $3 - b = - (b - 3)$, so:

$\frac{ -(b - 3) }{ -(b^2 - 3) } = \frac{b - 3}{b^2 - 3}$ — still not simpler.

Perhaps leave as $\frac{3 - b}{3 - b^2}$, but that’s not simplified.

Wait — maybe the denominator is $3a - a b^2 = a(3 - b^2)$, and if we assume $b^2$ is a typo and should be $b$, then it would be $\frac{a(3 - b)}{a(3 - b)} = 1$.

Given the context of the worksheet, it’s likely a typo, and it should be $\frac{3a - ab}{3a - ab} = 1$, or perhaps $\frac{3a - ab}{3a - ab^2}$ is meant to be simplified as is.

But let’s proceed with correct algebra:

$\frac{3a - ab}{3a - ab^2} = \frac{a(3 - b)}{a(3 - b^2)} = \frac{3 - b}{3 - b^2}$

We can factor denominator as difference of squares only if it were $b^2 - 9$ or something. Here, $3 - b^2$ doesn't factor nicely over rationals.

So perhaps the answer is $\frac{3 - b}{3 - b^2}$, but that’s not simplified further.

Alternatively, write it as $\frac{ -(b - 3) }{ -(b^2 - 3) } = \frac{b - 3}{b^2 - 3}$ — still not better.

I think for the purpose of this worksheet, it might be expected to leave it as $\frac{3 - b}{3 - b^2}$, or perhaps there’s a mistake.

Let me check online or standard problems — often, such problems have $3a - ab$ in both, or denominator is $3a - a b^2$ and they expect to factor out a.

Perhaps the intended answer is $\frac{3 - b}{3 - b^2}$, but I’ll note that it doesn’t simplify further.

Wait — another thought: if we write denominator as $3 - b^2 = (\sqrt{3})^2 - b^2 = (\sqrt{3} - b)(\sqrt{3} + b)$, but that’s not standard for this level.

Given the context, I suspect a typo, and it should be $\frac{3a - ab}{3a - ab} = 1$, or perhaps $\frac{3a - ab}{3a - a b^2} = \frac{a(3 - b)}{a(3 - b^2)} = \frac{3 - b}{3 - b^2}$.

But to be safe, I'll box the expression as simplified as possible.

✔ Answer: $\frac{3 - b}{3 - b^2}$ (or if typo, should be 1)

But let’s assume no typo and proceed.

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⑩ $\frac{x^2 + 3x}{x^2 - 9}$

- Step 1: Factor numerator and denominator:
Numerator: $x(x + 3)$
Denominator: $(x - 3)(x + 3)$ (difference of squares)
- Step 2: Rewrite:
$\frac{x(x + 3)}{(x - 3)(x + 3)}$
- Step 3: Cancel $(x + 3)$, assuming $x \ne -3$:
$= \frac{x}{x - 3}$

✔ Answer: $\frac{x}{x - 3}$

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⑪ $\frac{1 + x}{1 - x^2}$

- Step 1: Factor denominator:
$1 - x^2 = (1 - x)(1 + x)$
- Step 2: Rewrite:
$\frac{1 + x}{(1 - x)(1 + x)}$
- Step 3: Cancel $(1 + x)$, assuming $x \ne -1$:
$= \frac{1}{1 - x}$

✔ Answer: $\frac{1}{1 - x}$

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⑫ $\frac{4x - 12}{x^2 - 3x}$

- Step 1: Factor numerator and denominator:
Numerator: $4(x - 3)$
Denominator: $x(x - 3)$
- Step 2: Rewrite:
$\frac{4(x - 3)}{x(x - 3)}$
- Step 3: Cancel $(x - 3)$, assuming $x \ne 3$:
$= \frac{4}{x}$

✔ Answer: $\frac{4}{x}$

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⑬ $\frac{7x}{y(x + y)}$

- This is already a single fraction.
- No common factors to cancel between numerator and denominator.
- So it is already simplified.

✔ Answer: $\frac{7x}{y(x + y)}$

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⑭ $\frac{1}{x(x^2 + 3)}$

- Already a single fraction.
- Denominator cannot be factored further over real numbers (since $x^2 + 3$ has no real roots).
- No common factors with numerator.

✔ Answer: $\frac{1}{x(x^2 + 3)}$

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## ✔ Final Answers Summary:

1. $x + 8$
2. $\frac{1}{3}$
3. $\frac{3x}{2}$
4. $\frac{6}{x + y}$
5. $\frac{1}{5}$
6. $\frac{x + 4}{2}$
7. $\frac{29x + 7}{63}$
8. $p + 2$
9. $\frac{3 - b}{3 - b^2}$ *(Note: May be a typo; if denominator is $3a - ab$, answer is 1)*
10. $\frac{x}{x - 3}$
11. $\frac{1}{1 - x}$
12. $\frac{4}{x}$
13. $\frac{7x}{y(x + y)}$
14. $\frac{1}{x(x^2 + 3)}$

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Let me know if you’d like to see any step explained in more detail!