Simplifying Fractions Worksheets - Math Monks - Free Printable

Let’s solve each problem step by step. We’re simplifying complex fractions — that means fractions with fractions inside them, or expressions that look messy but can be cleaned up.

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Problem ①:
\[
\frac{5 - 2x}{x^2 + 2x}
\]

Factor denominator:
\( x^2 + 2x = x(x + 2) \)

Numerator: \( 5 - 2x \) → doesn’t factor nicely with denominator.

No common factors → already simplified? Wait — let’s check if we can rewrite numerator to match.

Actually, no common factors between numerator and denominator. So this is already simplified.

But wait — maybe they want it factored? The instruction says “simplify”, so unless there’s cancellation, leave as is.

✔ Final for ①: \( \frac{5 - 2x}{x(x + 2)} \) — but original form is also acceptable. Since no cancellation, we’ll keep as given? Actually, let’s see others.

Wait — perhaps I should check all problems first. Let me go one by one carefully.

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Actually, let’s do them in order, very carefully.

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① \( \frac{5 - 2x}{x^2 + 2x} \)

Denominator: \( x(x + 2) \)

Numerator: \( -(2x - 5) \) — no common factor with denominator.

→ Already simplified. But sometimes they expect factoring denominator.

We’ll write: \( \frac{5 - 2x}{x(x + 2)} \)

But actually, the problem might just want us to simplify if possible — since nothing cancels, answer is same as given? Hmm.

Wait — let’s move on and come back.

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② \( \frac{xy(x + y)}{3y^2 + 5x^2} \)

Denominator: \( 3y^2 + 5x^2 \) — cannot factor (sum of squares? No, not difference). Not factorable over reals easily.

Numerator has xy(x+y), denominator has no common terms.

→ Already simplified.

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③ \( \frac{\frac{a}{b} + \frac{x}{y}}{\frac{a}{z} + \frac{x}{c}} \)

This is a complex fraction. To simplify, combine numerator and denominator separately.

First, numerator: \( \frac{a}{b} + \frac{x}{y} = \frac{ay + bx}{by} \)

Denominator: \( \frac{a}{z} + \frac{x}{c} = \frac{ac + xz}{zc} \)

So overall:

\[
\frac{\frac{ay + bx}{by}}{\frac{ac + xz}{zc}} = \frac{ay + bx}{by} \cdot \frac{zc}{ac + xz} = \frac{(ay + bx)zc}{by(ac + xz)}
\]

Can we cancel anything? z and b? Only if variables allow, but generally no.

So simplified form: \( \frac{zc(ay + bx)}{by(ac + xz)} \)

We can write as: \( \frac{z c (a y + b x)}{b y (a c + x z)} \)

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④ \( \frac{\frac{1}{x} + \frac{2}{x}}{x + \frac{2}{x^2}} \)

Numerator: \( \frac{1}{x} + \frac{2}{x} = \frac{3}{x} \)

Denominator: \( x + \frac{2}{x^2} = \frac{x^3 + 2}{x^2} \)

So overall:

\[
\frac{\frac{3}{x}}{\frac{x^3 + 2}{x^2}} = \frac{3}{x} \cdot \frac{x^2}{x^3 + 2} = \frac{3x}{x^3 + 2}
\]

Check: yes, x cancels one power.

✔ Simplified: \( \frac{3x}{x^3 + 2} \)

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⑤ \( \frac{x + \frac{2d}{3ac}}{x + \frac{3d}{2ac}} \)

Write both numerator and denominator with common denominators.

Numerator: \( x + \frac{2d}{3ac} = \frac{3acx + 2d}{3ac} \)

Denominator: \( x + \frac{3d}{2ac} = \frac{2acx + 3d}{2ac} \)

So overall:

\[
\frac{\frac{3acx + 2d}{3ac}}{\frac{2acx + 3d}{2ac}} = \frac{3acx + 2d}{3ac} \cdot \frac{2ac}{2acx + 3d} = \frac{(3acx + 2d) \cdot 2ac}{3ac \cdot (2acx + 3d)}
\]

Cancel ac (assuming ac ≠ 0):

= \( \frac{(3acx + 2d) \cdot 2}{3 \cdot (2acx + 3d)} = \frac{2(3acx + 2d)}{3(2acx + 3d)} \)

Cannot simplify further.

✔ Answer: \( \frac{2(3acx + 2d)}{3(2acx + 3d)} \)

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⑥ \( \frac{5a}{2(a + 2)} \)

Already simple. No common factors.

✔ Answer: \( \frac{5a}{2(a + 2)} \)

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⑦ \( \frac{x}{2 - 3x} \)

Nothing to simplify. Could write as \( -\frac{x}{3x - 2} \), but not necessary.

✔ Leave as \( \frac{x}{2 - 3x} \)

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

Same denominator 3 → cancels!

\[
\frac{x + 1}{3} \div \frac{2x - 1}{3} = \frac{x + 1}{3} \cdot \frac{3}{2x - 1} = \frac{x + 1}{2x - 1}
\]

✔ Answer: \( \frac{x + 1}{2x - 1} \)

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⑨ \( \frac{1 - \frac{x}{y}}{\frac{x^2}{y^2} - 1} \)

Numerator: \( 1 - \frac{x}{y} = \frac{y - x}{y} \)

Denominator: \( \frac{x^2}{y^2} - 1 = \frac{x^2 - y^2}{y^2} = \frac{(x - y)(x + y)}{y^2} \)

Note: \( y - x = -(x - y) \)

So overall:

\[
\frac{\frac{y - x}{y}}{\frac{(x - y)(x + y)}{y^2}} = \frac{y - x}{y} \cdot \frac{y^2}{(x - y)(x + y)} = \frac{(y - x) y^2}{y (x - y)(x + y)}
\]

Simplify: \( y^2 / y = y \)

And \( y - x = -(x - y) \), so:

= \( \frac{ - (x - y) \cdot y }{ (x - y)(x + y) } = \frac{ - y }{ x + y } \)

(Assuming \( x \ne y \))

✔ Answer: \( -\frac{y}{x + y} \)

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⑩ \( \frac{\frac{x - 3}{x^2 - 25}}{\frac{x^2 - 9}{x - 5}} \)

First, factor everything.

Numerator of big fraction: \( \frac{x - 3}{x^2 - 25} = \frac{x - 3}{(x - 5)(x + 5)} \)

Denominator of big fraction: \( \frac{x^2 - 9}{x - 5} = \frac{(x - 3)(x + 3)}{x - 5} \)

So overall:

\[
\frac{ \frac{x - 3}{(x - 5)(x + 5)} }{ \frac{(x - 3)(x + 3)}{x - 5} } = \frac{x - 3}{(x - 5)(x + 5)} \cdot \frac{x - 5}{(x - 3)(x + 3)}
\]

Cancel \( x - 3 \) and \( x - 5 \) (assuming \( x \ne 3, 5 \)):

Left with: \( \frac{1}{(x + 5)(x + 3)} \)

✔ Answer: \( \frac{1}{(x + 3)(x + 5)} \)

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Now go back to earlier ones that seemed stuck.

① \( \frac{5 - 2x}{x^2 + 2x} = \frac{5 - 2x}{x(x + 2)} \) — no cancellation → leave as is? Or factor negative?

Sometimes written as \( -\frac{2x - 5}{x(x + 2)} \), but not required.

I think for simplicity, we leave as original unless specified.

But let’s check problem ② again.

② \( \frac{xy(x + y)}{3y^2 + 5x^2} \) — denominator doesn't factor, no common terms → already simplified.

Similarly, ⑥ and ⑦ are fine.

Wait — problem ③ we did.

Let me list final answers clearly.

After careful checking:

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Final Answers:

① \( \frac{5 - 2x}{x(x + 2)} \) — or leave as original? Original is \( \frac{5-2x}{x^2+2x} \), which equals that. Since no simplification via cancellation, either is ok. But typically we factor denominator.

I’ll go with factored form where helpful.

Actually, looking at worksheet style, probably expect simplified form meaning reduced if possible. For ①, no reduction, so original is fine? But let's see standard practice.

In many textbooks, they factor denominator even if no cancellation.

To be safe, I'll present simplified forms where applicable.

Let me compile:

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① \( \frac{5 - 2x}{x(x + 2)} \)

② \( \frac{xy(x + y)}{5x^2 + 3y^2} \) — same as given, no change

③ \( \frac{zc(ay + bx)}{by(ac + xz)} \)

④ \( \frac{3x}{x^3 + 2} \)

⑤ \( \frac{2(3acx + 2d)}{3(2acx + 3d)} \)

⑥ \( \frac{5a}{2(a + 2)} \)

⑦ \( \frac{x}{2 - 3x} \)

⑧ \( \frac{x + 1}{2x - 1} \)

⑨ \( -\frac{y}{x + y} \)

⑩ \( \frac{1}{(x + 3)(x + 5)} \)

For ②, since denominator is 3y² + 5x², and numerator has xy(x+y), no common factors, so it's already simplified. We can write it as is.

Similarly for others.

I think these are correct.

Double-check ⑨:

Original: \( \frac{1 - x/y}{x^2/y^2 - 1} = \frac{(y-x)/y}{(x^2 - y^2)/y^2} = \frac{y-x}{y} \cdot \frac{y^2}{(x-y)(x+y)} = \frac{(y-x)y}{(x-y)(x+y)} = \frac{-(x-y)y}{(x-y)(x+y)} = -\frac{y}{x+y} \) ✔

⑩: after canceling, yes, 1/[(x+3)(x+5)] ✔

④: numerator 3/x, denominator (x³ + 2)/x², so (3/x)*(x²/(x³+2)) = 3x/(x³+2) ✔

All good.

Now, for the final answer section, since the user wants only the final answers listed, and no explanation (ExplainAnswer=false), I will list them numbered.

But note: the response format says "Final Answer: [Provide only the final correct answer]"

Since there are 10 problems, I should list all 10 answers.

Probably in a list.

Also, ensure no extra text.

So:

Final Answer:
① \(\frac{5 - 2x}{x(x + 2)}\)
② \(\frac{xy(x + y)}{5x^2 + 3y^2}\)
③ \(\frac{zc(ay + bx)}{by(ac + xz)}\)
④ \(\frac{3x}{x^3 + 2}\)
⑤ \(\frac{2(3acx + 2d)}{3(2acx + 3d)}\)
⑥ \(\frac{5a}{2(a + 2)}\)
⑦ \(\frac{x}{2 - 3x}\)
⑧ \(\frac{x + 1}{2x - 1}\)
⑨ \(-\frac{y}{x + y}\)
⑩ \(\frac{1}{(x + 3)(x + 5)}\)

For ②, I wrote denominator as 5x² + 3y², but original was 3y² + 5x² — same thing.

Some might prefer to write in order, but mathematically identical.

I think this is fine.

One more check on ③: I have zc(ay + bx) / [by(ac + xz)]

Original complex fraction: num: a/b + x/y = (ay + bx)/(by)

Den: a/z + x/c = (ac + xz)/(zc)

Then division: [(ay+bx)/(by)] / [(ac+xz)/(zc)] = (ay+bx)/(by) * zc/(ac+xz) = zc(ay+bx) / [by(ac+xz)]

Yes.

In my answer I wrote zc(ay + bx) / by(ac + xz) — same.

Good.

So final answers ready.