Simplifying Complex Rational Expressions - Free Printable

You’ve uploaded an image titled “Complex Rational Expressions”, showing four complex rational expressions that need to be simplified. A *complex rational expression* is a fraction where the numerator, denominator, or both contain one or more fractions themselves.

The goal is to simplify each expression by eliminating the smaller fractions within the larger fraction — typically by multiplying numerator and denominator by the Least Common Denominator (LCD) of all the “inner” fractions.

Let’s solve each one step-by-step.

---

## 🔹 Expression 1:
\[
\frac{7 + \frac{1}{x}}{3 - \frac{1}{4}}
\]

Step 1: Simplify the denominator first.

\[
3 - \frac{1}{4} = \frac{12}{4} - \frac{1}{4} = \frac{11}{4}
\]

So now the expression becomes:
\[
\frac{7 + \frac{1}{x}}{\frac{11}{4}} = \left(7 + \frac{1}{x}\right) \cdot \frac{4}{11}
\]

Step 2: Distribute the multiplication:

\[
= \frac{4}{11} \cdot 7 + \frac{4}{11} \cdot \frac{1}{x} = \frac{28}{11} + \frac{4}{11x}
\]

Step 3: Combine into a single fraction (optional, but cleaner):

Common denominator is \(11x\):
\[
= \frac{28x}{11x} + \frac{4}{11x} = \frac{28x + 4}{11x}
\]

✔ Final Answer:
\[
\boxed{\frac{28x + 4}{11x}}
\]

---

## 🔹 Expression 2:
\[
\frac{\frac{1}{3} + \frac{2}{x}}{\frac{5}{9} + \frac{1}{4}}
\]

Step 1: Simplify denominator.

Find LCD of 9 and 4 → 36
\[
\frac{5}{9} = \frac{20}{36}, \quad \frac{1}{4} = \frac{9}{36} \Rightarrow \frac{20 + 9}{36} = \frac{29}{36}
\]

Step 2: Simplify numerator.

LCD of 3 and x → 3x
\[
\frac{1}{3} = \frac{x}{3x}, \quad \frac{2}{x} = \frac{6}{3x} \Rightarrow \frac{x + 6}{3x}
\]

Now the expression is:
\[
\frac{\frac{x + 6}{3x}}{\frac{29}{36}} = \frac{x + 6}{3x} \cdot \frac{36}{29}
\]

Step 3: Multiply:

\[
= \frac{(x + 6) \cdot 36}{3x \cdot 29} = \frac{36(x + 6)}{87x}
\]

Simplify 36/87 → divide numerator and denominator by 3:
\[
= \frac{12(x + 6)}{29x}
\]

✔ Final Answer:
\[
\boxed{\frac{12(x + 6)}{29x}}
\]

---

## 🔹 Expression 3:
\[
\frac{8 - \frac{1}{x^2}}{4 + \frac{1}{x}}
\]

This one is trickier — notice the numerator is a difference of squares if we think of it as:

\[
8 - \frac{1}{x^2} = \left(\sqrt{8}\right)^2 - \left(\frac{1}{x}\right)^2
\]

But √8 isn’t nice. Instead, let’s multiply numerator and denominator by the LCD of inner fractions: x²

Step 1: Multiply numerator and denominator by \(x^2\):

Numerator:
\[
x^2 \cdot \left(8 - \frac{1}{x^2}\right) = 8x^2 - 1
\]

Denominator:
\[
x^2 \cdot \left(4 + \frac{1}{x}\right) = 4x^2 + x
\]

So expression becomes:
\[
\frac{8x^2 - 1}{4x^2 + x}
\]

Step 2: Factor if possible.

Numerator: \(8x^2 - 1\) → difference of squares?
\(8x^2 - 1 = (\sqrt{8}x)^2 - 1^2\), not a perfect square with integer coefficients → leave as is.

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

No common factors between numerator and denominator.

✔ Final Answer:
\[
\boxed{\frac{8x^2 - 1}{x(4x + 1)}}
\]

---

## 🔹 Expression 4:
\[
\frac{7 + \frac{2}{x - 2}}{5 - \frac{1}{x + 2}}
\]

Step 1: Find LCD for numerator and denominator.

- Numerator: terms are 7 and \(\frac{2}{x - 2}\) → LCD = \(x - 2\)
- Denominator: terms are 5 and \(\frac{1}{x + 2}\) → LCD = \(x + 2\)

To eliminate *all* inner fractions, multiply numerator and denominator by the LCD of all denominators: \((x - 2)(x + 2)\)

Step 2: Multiply numerator and denominator by \((x - 2)(x + 2)\)

#### Numerator:
\[
\left(7 + \frac{2}{x - 2}\right) \cdot (x - 2)(x + 2)
= 7(x - 2)(x + 2) + 2(x + 2)
\]

Factor:
\[
= 7(x^2 - 4) + 2(x + 2) = 7x^2 - 28 + 2x + 4 = 7x^2 + 2x - 24
\]

#### Denominator:
\[
\left(5 - \frac{1}{x + 2}\right) \cdot (x - 2)(x + 2)
= 5(x - 2)(x + 2) - 1(x - 2)
\]

Factor:
\[
= 5(x^2 - 4) - (x - 2) = 5x^2 - 20 - x + 2 = 5x^2 - x - 18
\]

So expression becomes:
\[
\frac{7x^2 + 2x - 24}{5x^2 - x - 18}
\]

Step 3: Try factoring numerator and denominator.

Numerator: \(7x^2 + 2x - 24\)

Try factoring:
Looking for two numbers: multiply to \(7 \cdot (-24) = -168\), add to 2.

Factors of -168: 14 and -12 → 14 - 12 = 2 ✔

So:
\[
7x^2 + 14x - 12x - 24 = 7x(x + 2) - 12(x + 2) = (7x - 12)(x + 2)
\]

Denominator: \(5x^2 - x - 18\)

Multiply 5 × (-18) = -90. Need two numbers that multiply to -90, add to -1.

→ 9 and -10 → 9 - 10 = -1 ✔

\[
5x^2 - 10x + 9x - 18 = 5x(x - 2) + 9(x - 2) = (5x + 9)(x - 2)
\]

So expression becomes:
\[
\frac{(7x - 12)(x + 2)}{(5x + 9)(x - 2)}
\]

No common factors → cannot simplify further.

✔ Final Answer:
\[
\boxed{\frac{(7x - 12)(x + 2)}{(5x + 9)(x - 2)}}
\]

---

## 🎯 Summary of All Answers:

1. \(\displaystyle \frac{7 + \frac{1}{x}}{3 - \frac{1}{4}} = \boxed{\frac{28x + 4}{11x}}\)

2. \(\displaystyle \frac{\frac{1}{3} + \frac{2}{x}}{\frac{5}{9} + \frac{1}{4}} = \boxed{\frac{12(x + 6)}{29x}}\)

3. \(\displaystyle \frac{8 - \frac{1}{x^2}}{4 + \frac{1}{x}} = \boxed{\frac{8x^2 - 1}{x(4x + 1)}}\)

4. \(\displaystyle \frac{7 + \frac{2}{x - 2}}{5 - \frac{1}{x + 2}} = \boxed{\frac{(7x - 12)(x + 2)}{(5x + 9)(x - 2)}}\)

---

✔ Each was simplified by eliminating inner fractions using LCD multiplication, then simplifying/factoring where possible.

Let me know if you want to see any step expanded further!