Simplifying Complex Fractions Worksheet featuring eight problems to solve, ideal for math practice.

Let’s solve each problem one by one, step by step. We’ll simplify complex fractions by combining terms in the numerator and denominator first, then dividing (which means multiplying by the reciprocal).

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Problem ①:
\[
\frac{5}{x - 1} - \frac{3}{x + 1}
\]

This is not a complex fraction yet — it’s just subtraction of two rational expressions. To combine them, find a common denominator: \((x - 1)(x + 1)\)

\[
= \frac{5(x + 1) - 3(x - 1)}{(x - 1)(x + 1)}
\]

Expand numerator:

\[
5(x + 1) = 5x + 5 \\
3(x - 1) = 3x - 3 \\
\Rightarrow 5x + 5 - (3x - 3) = 5x + 5 - 3x + 3 = 2x + 8
\]

Denominator: \(x^2 - 1\) (difference of squares)

So,

\[
= \frac{2x + 8}{x^2 - 1}
\]

We can factor numerator: \(2(x + 4)\), but denominator doesn’t factor with that, so leave as is.

✔ Final Answer for ①: \(\boxed{\frac{2x + 8}{x^2 - 1}}\)

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

First, simplify numerator and denominator separately.

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

Common denominator: \(x(x - 1)\)

\[
= \frac{(x - 1) - x}{x(x - 1)} = \frac{x - 1 - x}{x(x - 1)} = \frac{-1}{x(x - 1)}
\]

Denominator: \(1 - \frac{1}{x} = \frac{x}{x} - \frac{1}{x} = \frac{x - 1}{x}\)

Now divide numerator by denominator:

\[
\frac{\frac{-1}{x(x - 1)}}{\frac{x - 1}{x}} = \frac{-1}{x(x - 1)} \cdot \frac{x}{x - 1}
\]

Cancel \(x\) top and bottom:

\[
= \frac{-1}{(x - 1)} \cdot \frac{1}{x - 1} = \frac{-1}{(x - 1)^2}
\]

✔ Final Answer for ②: \(\boxed{-\frac{1}{(x - 1)^2}}\)

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Problem ③:
\[
\frac{5 - \frac{2}{5}}{6 + \frac{1}{3}}
\]

Simplify numerator and denominator.

Numerator: \(5 - \frac{2}{5} = \frac{25}{5} - \frac{2}{5} = \frac{23}{5}\)

Denominator: \(6 + \frac{1}{3} = \frac{18}{3} + \frac{1}{3} = \frac{19}{3}\)

Now divide:

\[
\frac{23/5}{19/3} = \frac{23}{5} \cdot \frac{3}{19} = \frac{69}{95}
\]

Check if reducible: 69 = 3×23, 95 = 5×19 → no common factors.

✔ Final Answer for ③: \(\boxed{\frac{69}{95}}\)

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Problem ④:
\[
\frac{x + \frac{2d}{3ac}}{x + \frac{3d}{3ac}}
\]

Note: \(\frac{3d}{3ac} = \frac{d}{ac}\), so rewrite:

\[
\frac{x + \frac{2d}{3ac}}{x + \frac{d}{ac}}
\]

To simplify, get common denominators in numerator and denominator.

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

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

Now divide:

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

Cancel \(ac\):

\[
= \frac{3acx + 2d}{3} \cdot \frac{1}{acx + d} = \frac{3acx + 2d}{3(acx + d)}
\]

Cannot simplify further unless factoring helps — but 3acx + 2d ≠ 3(acx + d) because 3(acx + d) = 3acx + 3d.

So leave as is.

✔ Final Answer for ④: \(\boxed{\frac{3acx + 2d}{3(acx + d)}}\)

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Problem ⑤:
\[
\frac{x^2\left(\frac{2}{x^2} + \frac{1}{x}\right)}{x^2\left(\frac{4}{x^2} + \frac{1}{x}\right)}
\]

Notice \(x^2\) is multiplied in both numerator and denominator — we can cancel them out!

So simplifies to:

\[
\frac{\frac{2}{x^2} + \frac{1}{x}}{\frac{4}{x^2} + \frac{1}{x}}
\]

Now simplify numerator and denominator.

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

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

Now divide:

\[
\frac{(2 + x)/x^2}{(4 + x)/x^2} = \frac{2 + x}{x^2} \cdot \frac{x^2}{4 + x} = \frac{2 + x}{4 + x}
\]

Which is same as \(\frac{x + 2}{x + 4}\)

✔ Final Answer for ⑤: \(\boxed{\frac{x + 2}{x + 4}}\)

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Problem ⑥:
\[
\frac{\frac{1}{2} + \frac{1}{3} + \frac{1}{4}}{3 - \frac{4}{5}}
\]

Simplify numerator and denominator.

Numerator: Find LCD of 2,3,4 → 12

\[
\frac{1}{2} = \frac{6}{12}, \quad \frac{1}{3} = \frac{4}{12}, \quad \frac{1}{4} = \frac{3}{12} \\
\Rightarrow \frac{6 + 4 + 3}{12} = \frac{13}{12}
\]

Denominator: \(3 - \frac{4}{5} = \frac{15}{5} - \frac{4}{5} = \frac{11}{5}\)

Now divide:

\[
\frac{13/12}{11/5} = \frac{13}{12} \cdot \frac{5}{11} = \frac{65}{132}
\]

Check reducibility: 65 = 5×13, 132 = 12×11 = 2²×3×11 → no common factors.

✔ Final Answer for ⑥: \(\boxed{\frac{65}{132}}\)

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Problem ⑦:
\[
\frac{\frac{4}{5}}{\frac{1}{5} + \frac{2}{3}}
\]

Simplify denominator first.

LCD of 5 and 3 is 15.

\[
\frac{1}{5} = \frac{3}{15}, \quad \frac{2}{3} = \frac{10}{15} \\
\Rightarrow \frac{3 + 10}{15} = \frac{13}{15}
\]

Now divide:

\[
\frac{4/5}{13/15} = \frac{4}{5} \cdot \frac{15}{13} = \frac{4 \cdot 15}{5 \cdot 13} = \frac{60}{65}
\]

Simplify: divide numerator and denominator by 5 → \(\frac{12}{13}\)

✔ Final Answer for ⑦: \(\boxed{\frac{12}{13}}\)

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Problem ⑧:
\[
\frac{\frac{1}{4x} + \frac{2}{3x}}{\frac{1}{4x} + \frac{2}{3x}}
\]

Wait — numerator and denominator are IDENTICAL!

So this equals 1, as long as the expression is defined (i.e., x ≠ 0).

✔ Final Answer for ⑧: \(\boxed{1}\)

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Final Answer:
① \(\boxed{\frac{2x + 8}{x^2 - 1}}\)
② \(\boxed{-\frac{1}{(x - 1)^2}}\)
③ \(\boxed{\frac{69}{95}}\)
④ \(\boxed{\frac{3acx + 2d}{3(acx + d)}}\)
⑤ \(\boxed{\frac{x + 2}{x + 4}}\)
⑥ \(\boxed{\frac{65}{132}}\)
⑦ \(\boxed{\frac{12}{13}}\)
⑧ \(\boxed{1}\)