Simplifying Algebraic Fractions Worksheet for students to practice reducing expressions to single fractions.

Let’s solve each problem one by one. We’ll simplify each algebraic fraction to its simplest form — meaning we factor where possible and cancel common terms in numerator and denominator.

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Problem ①:
$\frac{x^2 - 64}{x - 8}$

Notice: $x^2 - 64$ is a difference of squares: $(x - 8)(x + 8)$
So, $\frac{(x - 8)(x + 8)}{x - 8} = x + 8$ (as long as $x ≠ 8$, but we just simplify)

✔ Answer: $x + 8$

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Problem ②:
$\frac{3x^2 + 3x}{9x^2 + 9x}$

Factor numerator: $3x(x + 1)$
Factor denominator: $9x(x + 1)$
Now: $\frac{3x(x + 1)}{9x(x + 1)} = \frac{3}{9} = \frac{1}{3}$ (cancel $x$ and $x+1$, assuming they’re not zero)

✔ Answer: $\frac{1}{3}$

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Problem ③:
$\frac{6x^2 + 9x}{4x + 6}$

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

✔ Answer: $\frac{3x}{2}$

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Problem ④:
$\frac{6y}{xy + y^2}$

Denominator: $y(x + y)$
So: $\frac{6y}{y(x + y)} = \frac{6}{x + y}$ (cancel $y$)

✔ Answer: $\frac{6}{x + y}$

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Problem ⑤:
$\frac{a + b}{5a + 5b}$

Denominator: $5(a + b)$
So: $\frac{a + b}{5(a + b)} = \frac{1}{5}$

✔ Answer: $\frac{1}{5}$

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Problem ⑥:
$\frac{(x - 3)(x + 4)}{2x - 6}$

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

✔ Answer: $\frac{x + 4}{2}$

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Problem ⑦:
$\frac{4x}{7} - \frac{x - 1}{9}$

We need a common denominator: LCM of 7 and 9 is 63.

Convert:
- $\frac{4x}{7} = \frac{4x \cdot 9}{63} = \frac{36x}{63}$
- $\frac{x - 1}{9} = \frac{(x - 1) \cdot 7}{63} = \frac{7x - 7}{63}$

Now subtract:
$\frac{36x - (7x - 7)}{63} = \frac{36x - 7x + 7}{63} = \frac{29x + 7}{63}$

✔ Answer: $\frac{29x + 7}{63}$

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Problem ⑧:
$\frac{p^2 - 4}{p - 2}$

$p^2 - 4 = (p - 2)(p + 2)$
So: $\frac{(p - 2)(p + 2)}{p - 2} = p + 2$

✔ Answer: $p + 2$

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Problem ⑨:
$\frac{3a - ab}{3a - ab^2}$

Factor numerator: $a(3 - b)$
Factor denominator: $a(3 - b^2)$ → Wait! Actually, let's check:

Wait — denominator: $3a - ab^2 = a(3 - b^2)$? But that doesn’t match numerator.

Actually, let’s factor carefully:

Numerator: $a(3 - b)$
Denominator: $a(3 - b^2)$ — but $3 - b^2$ is NOT same as $3 - b$. So maybe I made a mistake?

Wait — perhaps it’s written as $3a - ab^2 = a(3 - b^2)$, and numerator is $a(3 - b)$. These don’t cancel unless $b=1$ or something — so maybe this can’t be simplified further? Let me double-check the original problem.

Original: $\frac{3a - ab}{3a - ab^2}$

Yes — factor out ‘a’ from both:

Numerator: $a(3 - b)$
Denominator: $a(3 - b^2)$

But $3 - b^2 = (\sqrt{3} - b)(\sqrt{3} + b)$ — not helpful. And no common factors with $3 - b$ unless specified.

Wait — maybe typo? Or perhaps it’s meant to be $3a - ab$ over $3a - ab$? No.

Alternatively — perhaps factor differently? Let’s try plugging numbers to test.

Suppose a=1, b=2:

Numerator: 3(1) - 1*2 = 3 - 2 = 1
Denominator: 3(1) - 1*(4) = 3 - 4 = -1 → result = -1

If we leave as $\frac{3 - b}{3 - b^2}$, for b=2: (3-2)/(3-4) = 1/-1 = -1 → correct.

But can we simplify $\frac{3 - b}{3 - b^2}$? Note that $3 - b^2 = -(b^2 - 3)$, and $3 - b = -(b - 3)$ — still no common factor.

Unless... wait — is it possible the denominator was meant to be $3a - ab$? Then it would be 1. But as written, it’s $ab^2$.

Perhaps the problem is designed to show that sometimes you can’t simplify much? But let’s look again.

Wait — maybe factor denominator as $a(3 - b^2)$ and numerator $a(3 - b)$, so cancel ‘a’, get $\frac{3 - b}{3 - b^2}$

And since $3 - b^2 = ( \sqrt{3} - b )( \sqrt{3} + b )$, and no relation to $3 - b$, we stop here.

But that seems odd for a worksheet. Maybe I misread.

Looking back at image: Problem ⑨ is $\frac{3a - ab}{3a - ab^2}$

Ah — perhaps factor numerator: $a(3 - b)$
Denominator: $a(3 - b^2)$ — but note: $3 - b^2$ is not factorable nicely with integers.

Wait — unless it’s a trick? Or maybe it’s supposed to be $3a - ab$ over $3a - ab$? No.

Another thought: maybe write as $\frac{a(3 - b)}{a(3 - b^2)} = \frac{3 - b}{3 - b^2}$, and that’s the simplest.

But let’s check if $3 - b^2$ can be related to $3 - b$. For example, if b=1, then num=2, den=2 → 1. If b=0, num=3, den=3 → 1. If b=2, num=1, den=-1 → -1. So no constant simplification.

I think we have to leave it as $\frac{3 - b}{3 - b^2}$ after canceling ‘a’.

But wait — perhaps the denominator is $3a - ab^2 = a(3 - b^2)$, and we can write $3 - b^2 = (3 - b)(?)$ — no, because $3 - b^2$ is not divisible by $3 - b$ unless specific values.

For example, polynomial division: divide $3 - b^2$ by $3 - b$.

Let me do: $-b^2 + 0b + 3$ divided by $-b + 3$

Or better: let’s set u = b, divide $3 - u^2$ by $3 - u$.

$3 - u^2 = -(u^2 - 3)$, and $3 - u = -(u - 3)$

Divide $u^2 - 3$ by $u - 3$: quotient is u + 3, remainder 6? Because (u-3)(u+3) = u^2 - 9, so u^2 - 3 = (u-3)(u+3) + 6. Not clean.

So indeed, no simplification beyond canceling ‘a’.

But that seems unlikely for this level. Perhaps there’s a typo in my reading.

Wait — looking back at the image description: Problem ⑨ is $\frac{3a - ab}{3a - ab^2}$

Maybe it’s $\frac{3a - ab}{3a - ab}$? No, the image says ab^2.

Another idea: perhaps factor denominator as $a(3 - b^2)$ and notice that $3 - b^2 = ( \sqrt{3} )^2 - b^2$, but that introduces radicals — not appropriate.

I think for school level, they might expect us to factor and cancel what we can.

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

And that’s it. But let’s see if we can write it as $\frac{3 - b}{-(b^2 - 3)} = -\frac{3 - b}{b^2 - 3}$, but still not simpler.

Perhaps the problem is designed to have no further simplification, but that seems odd.

Wait — let’s look at problem ⑩: $\frac{x^2 + 3x}{x^2 - 9}$ — that will factor nicely.

Back to ⑨ — perhaps I should proceed and come back.

Actually, let me assume that it’s correct as is, and the answer is $\frac{3 - b}{3 - b^2}$ after canceling 'a'.

But to make it look nicer, we can write it as $\frac{3 - b}{3 - b^2}$, or factor negative: $\frac{ -(b - 3) }{ -(b^2 - 3) } = \frac{b - 3}{b^2 - 3}$, but still not helpful.

I recall that in some worksheets, they might have intended $3a - ab$ over $3a - ab$, but here it's ab^2.

Another thought: perhaps "ab^2" means a*b*2? No, in algebra, ab^2 means a times b squared.

I think I have to go with $\frac{3 - b}{3 - b^2}$

But let's calculate numerical value again: a=1,b=1: num=3-1=2, den=3-1=2, ratio=1. With formula: (3-1)/(3-1)=2/2=1 ok.

a=1,b=0: num=3, den=3, ratio=1. Formula: 3/3=1 ok.

a=1,b=2: num=3-2=1, den=3-4= -1, ratio=-1. Formula: (3-2)/(3-4)=1/-1=-1 ok.

So the expression $\frac{3 - b}{3 - b^2}$ is correct after canceling 'a'.

But perhaps we can leave it as $\frac{3 - b}{3 - b^2}$, or simplify by factoring denominator if possible, but it's not factorable over integers.

So I'll box that.

Wait — maybe the problem is $\frac{3a - ab}{3a - ab^2} = \frac{a(3 - b)}{a(3 - b^2)} = \frac{3 - b}{3 - b^2}$, and that's the answer.

But let's move on and come back if needed.

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Problem ⑩:
$\frac{x^2 + 3x}{x^2 - 9}$

Numerator: $x(x + 3)$
Denominator: $(x - 3)(x + 3)$
So: $\frac{x(x + 3)}{(x - 3)(x + 3)} = \frac{x}{x - 3}$

✔ Answer: $\frac{x}{x - 3}$

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Problem ⑪:
$\frac{1 + x}{1 - x^2}$

Note: $1 - x^2 = (1 - x)(1 + x)$
So: $\frac{1 + x}{(1 - x)(1 + x)} = \frac{1}{1 - x}$ (for $x ≠ -1$)

Also, $\frac{1}{1 - x} = -\frac{1}{x - 1}$, but usually we leave as $\frac{1}{1 - x}$

✔ Answer: $\frac{1}{1 - x}$

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Problem ⑫:
$\frac{4x - 12}{x^2 - 3x}$

Numerator: $4(x - 3)$
Denominator: $x(x - 3)$
So: $\frac{4(x - 3)}{x(x - 3)} = \frac{4}{x}$

✔ Answer: $\frac{4}{x}$

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Problem ⑬:
$\frac{7x}{y(x + y)}$

Already simplified. No common factors.

✔ Answer: $\frac{7x}{y(x + y)}$

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Problem ⑭:
$\frac{1}{x(x^2 + 3)}$

Already simplified. Denominator is factored, no common factors with numerator.

✔ Answer: $\frac{1}{x(x^2 + 3)}$

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Now back to Problem ⑨: $\frac{3a - ab}{3a - ab^2}$

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

We can write $3 - b^2 = - (b^2 - 3)$, and $3 - b = - (b - 3)$, so:

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

But $b^2 - 3$ doesn't factor nicely, so either form is acceptable. However, in many textbooks, they prefer to have leading positive in denominator, so perhaps $\frac{3 - b}{3 - b^2}$ is fine.

Notice that $3 - b^2 = ( \sqrt{3} )^2 - b^2$, but that's not helpful.

I think for this context, we'll leave it as $\frac{3 - b}{3 - b^2}$

But let me check online or standard practice — actually, upon second thought, perhaps the problem has a typo, and it's meant to be $3a - ab$ over $3a - ab$, but since it's written as ab^2, I'll go with the calculation.

Another idea: perhaps "ab^2" means a*b*2, i.e., 2ab? In some notations, but usually b^2 means b squared.

In the image, it's likely b squared.

Perhaps in the denominator, it's 3a - a b^2, and we can factor a, and that's it.

I think I have to accept that the simplified form is $\frac{3 - b}{3 - b^2}$

To make it look better, we can write it as $\frac{3 - b}{3 - b^2}$, or recognize that it's already simplified.

So for Problem ⑨, answer is $\frac{3 - b}{3 - b^2}$

But let's confirm with another value. Let a=2, b=1:

Numerator: 3*2 - 2*1 = 6 - 2 = 4
Denominator: 3*2 - 2*(1)^2 = 6 - 2 = 4 → ratio=1

Formula: (3-1)/(3-1) =2/2=1 ok.

a=2,b=0: num=6-0=6, den=6-0=6, ratio=1. Formula:3/3=1 ok.

a=2,b=3: num=6 - 2*3=6-6=0, den=6 - 2*9=6-18= -12, ratio=0

Formula: (3-3)/(3-9)=0/-6=0 ok.

So it's correct.

Perhaps we can leave it as is.

Some might write it as $\frac{3 - b}{3 - b^2} = \frac{ -(b - 3) }{ -(b^2 - 3) } = \frac{b - 3}{b^2 - 3}$, but I think the first form is fine.

I'll use $\frac{3 - b}{3 - b^2}$

But to match typical textbook style, perhaps factor out negatives, but I think it's okay.

Final decision for ⑨: $\frac{3 - b}{3 - b^2}$

Now, compiling all answers:

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 $\frac{3 - b}{3 - b^2}$
⑩ $\frac{x}{x - 3}$
⑪ $\frac{1}{1 - x}$
⑫ $\frac{4}{x}$
⑬ $\frac{7x}{y(x + y)}$
⑭ $\frac{1}{x(x^2 + 3)}$

For ⑨, I recall that in some contexts, they might expect to write it as $\frac{3 - b}{3 - b^2}$, but let me see if there's a different interpretation.

Another thought: perhaps the denominator is 3a - a b^2 = a(3 - b^2), and numerator a(3 - b), so the ratio is \frac{3 - b}{3 - b^2}, and since 3 - b^2 = (3^{1/2} - b)(3^{1/2} + b), but that's not rational, so we stop.

I think it's correct.

So final answers are as above.

Now, for the Final Answer section, since the user asked to solve the problems, and there are 14 problems, I need to list all answers.

The instruction says: "Provide only the final correct answer" — but since there are multiple, I should list them all.

Looking back at the response format: "Final Answer: [Provide only the final correct answer]"

But for multiple problems, I think we list them.

Perhaps box each or list numerically.

I'll list them in order.

So:

Final Answer:
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 $\frac{3 - b}{3 - b^2}$
⑩ $\frac{x}{x - 3}$
⑪ $\frac{1}{1 - x}$
⑫ $\frac{4}{x}$
⑬ $\frac{7x}{y(x + y)}$
⑭ $\frac{1}{x(x^2 + 3)}$

For ⑨, I feel uneasy. Let me search my memory — is there a way to simplify $\frac{3 - b}{3 - b^2}$?

Note that $3 - b^2 = 3 - b \cdot b$, not helpful.

Perhaps the problem is $\frac{3a - ab}{3a - ab^2} = \frac{a(3 - b)}{a(3 - b^2)} = \frac{3 - b}{3 - b^2}$, and in some books, they might write it as \frac{3-b}{3-b^2}, or perhaps leave it.

Another idea: perhaps factor denominator as - (b^2 - 3), and numerator as - (b - 3), so \frac{ -(b - 3) }{ -(b^2 - 3) } = \frac{b - 3}{b^2 - 3}, and b^2 - 3 is irreducible, so same thing.

I think it's fine.

Perhaps the intended answer is \frac{1}{b} or something, but from calculation, it's not.

Let me try b=3: num=3-3=0, den=3-9= -6, ratio=0. If it were 1/b, 1/3, not 0. So no.

I think we have to go with it.

So final answers as above.