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# Complex Fractions

## Solve fractions with fractional numerators and/or denominators.

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Complex Fractions

Gupta knows the area and width of a rectangle. He comes up with this equation for the length of the rectangle $\frac{\frac{2}{x^2-1}}{\frac{2x}{x+1}}$ . What is the length of the rectangle in simplified form?

### Guidance

A complex fraction is a fraction that has fractions in the numerator and/or denominator. To simplify a complex fraction, you will need to combine all that you have learned in the previous five concepts.

#### Example A

Simplify $\frac{\frac{9x}{x+2}}{\frac{3}{x^2-4}}$ .

Solution: This complex fraction is a fraction divided by another fraction. Rewrite the complex fraction as a division problem.

$\frac{\frac{9x}{x+2}}{\frac{3}{x^2-4}} = \frac{9x}{x+2} \div \frac{3}{x^2-4}$ .

Now, this is just like a problem from the Dividing Rational Expressions concept. Flip the second fraction, change the problem to multiplication and simplify.

$\frac{9x}{x+2} \div \frac{3}{x^2-4} = \frac{9x}{x+2} \cdot \frac{x^2-4}{3} = \frac{\overset{3}{\bcancel{9}x}}{\cancel{x+2}} \cdot \frac{\cancel{(x+2)}(x-2)}{\bcancel{3}} = 3x(x-2)$

#### Example B

Simplify $\frac{\frac{1}{x} + \frac{1}{x+1}}{4- \frac{1}{x}}$ .

Solution: To simplify this complex fraction, we first need to add the fractions in the numerator and subtract the two in the denominator. The LCD of the numerator is $x(x+1)$ and the denominator is just $x$ .

$\frac{\frac{1}{x} + \frac{1}{x+1}}{4- \frac{1}{x}} = \frac{{\color{red}\frac{x+1}{x+1}} \cdot \frac{1}{x} + \frac{1}{x+1} \cdot {\color{blue}\frac{x}{x}}}{{\color{blue}\frac{x}{x}} \cdot 4- \frac{1}{x}} = \frac{\frac{x+1}{x(x+1)} + \frac{x}{x(x+1)}}{\frac{4x}{x} - \frac{1}{x}} = \frac{\frac{2x+1}{x(x+1)}}{\frac{4x-1}{x}}$

This fraction is now just like Example A. Divide and simplify if possible.

$\frac{\frac{2x+1}{x(x+1)}}{\frac{4x-1}{x}} = \frac{2x+1}{x(x+1)} \div \frac{4x-1}{x} = \frac{2x+1}{\cancel{x}(x+1)} \cdot \frac{\cancel{x}}{4x-1} = \frac{2x+1}{(x+1)(4x-1)}$

#### Example C

Simplify $\frac{\frac{5-x}{x^2+6x+8} + \frac{x}{x+4}}{\frac{6}{x+2} - \frac{2x+3}{x^2-3x-10}}$ .

Solution: First, add the fractions in the numerator and subtract the ones in the denominator.

$\frac{\frac{5-x}{x^2+6x+8} + \frac{x}{x+4}}{\frac{6}{x+2} - \frac{2x+3}{x^2-3x-10}} = \frac{\frac{5-x}{(x+4){\color{red}(x+2)}} + \frac{x}{x+4} \cdot {\color{red}\frac{x+2}{x+2}}}{{\color{blue}\frac{x-5}{x-5}} \cdot \frac{6}{x+2} - \frac{2x+3}{(x+2){\color{blue}(x-5)}}} = \frac{\frac{5-x+x(x+2)}{(x+4)(x+2)}}{\frac{6(x-5)-(2x+3)}{(x+2)(x-5)}} = \frac{\frac{x^2+x+5}{(x+4)(x+2)}}{\frac{4x-36}{(x+2)(x-5)}}$

Now, rewrite as a division problem, flip, multiply, and simplify.

$\frac{\frac{x^2+x+5}{(x+4)(x+2)}}{\frac{4x-36}{(x+2)(x-5)}} = \frac{x^2+x+5}{(x+4)(x+2)} \div \frac{4x-36}{(x+2)(x-5)} = \frac{x^2+x+5}{(x+4)\cancel{(x+2)}} \cdot \frac{\cancel{(x+2)}(x-5)}{4(x-9)} = \frac{(x^2+x+5)(x-5)}{4(x+4)(x-9)}$

Intro Problem Revisit This complex fraction is a fraction divided by another fraction. Rewrite the complex fraction as a division problem.

$\frac{\frac{2}{x^2-1}}{\frac{2x}{x+1}} = \frac{2}{x^2-1} \div \frac{2x}{x+1}$ .

Flip the second fraction, change the problem to multiplication and simplify.

$\frac{2}{x^2-1} \div \frac{2x}{x+1} = \frac{2}{x^2-1} \cdot \frac{x+1}{2x} = \frac{\bcancel{2}}{\bcancel{(x+1)}(x-1)} \cdot \frac{\bcancel{(x+1)}}{\bcancel{2}x} = \frac {1}{x^2-x}$

Therefore, the length of the rectangle in simplified form is $\frac {1}{x^2-x}$ .

### Guided Practice

Simplify the complex fractions.

1. $\frac{\frac{5x-20}{x^2}}{\frac{x-4}{x}}$

2. $\frac{\frac{1-x}{x} - \frac{2}{x-1}}{1 + \frac{1}{x}}$

3. $\frac{\frac{3}{2x^2+12x+18} + \frac{x}{x^2-9}}{\frac{6x}{3x-9} - \frac{3}{x-3}}$

1. Rewrite the fraction as a division problem and simplify.

$\frac{\frac{5x-20}{x^2}}{\frac{x-4}{x}} = \frac{5x-20}{x^2} \div \frac{x-4}{x} = \frac{5 \cancel{(x-4)}}{x^{\cancel{2}}} \cdot \frac{\cancel{x}}{\cancel{x-4}} = \frac{5}{x}$

2. Add the fractions in the numerator and denominator together.

$\frac{\frac{1-x}{x} - \frac{2}{x-1}}{1+\frac{1}{x}} = \frac{\frac{x-1}{x-1} \cdot \frac{1-x}{x} - \frac{2}{x-1} \cdot \frac{x}{x}}{\frac{x}{x} \cdot 1+ \frac{1}{x}} = \frac{\frac{(x-1)(1-x)-2x}{x(x-1)}}{\frac{x+1}{x}} = \frac{\frac{-x^2+1}{x(x-1)}}{\frac{x+1}{x}}$

Now, rewrite the fraction as a division problem and simplify.

$\frac{-x^2+1}{x(x-1)} \div \frac{x+1}{x} &= \frac{-(x^2-1)}{x(x-1)} \cdot \frac{x}{x+1} \\&= \frac{-\cancel{(x-1)} \cancel{(x+1)}}{\cancel{x} \cancel{(x-1)}} \cdot \frac{\cancel{x}} {\cancel{x+1}} \\&= -1$

3. Add the numerator and subtract the denominator of this complex fraction.

$\frac{\frac{3}{2x^2+12x+18} + \frac{x}{x^2-9}}{\frac{6x}{3x-9} - \frac{3}{x-3}} &= \frac{\frac{x-3}{x-3} \cdot \frac{3}{2(x+3)(x+3)} + \frac{x}{(x-3)(x+3)} \cdot \frac{2(x+3)}{2(x+3)}}{\frac{6x}{3(x-3)} - \frac{3}{x-3} \cdot \frac{3}{3}} \\&= \frac{\frac{3(x-3)+2x(x+3)}{2(x+3)(x+3)(x-3)}}{\frac{6x-9}{3(x-3)}} \\&= \frac{\frac{2x^2+3x-9}{2(x+3)(x+3)(x-3)}}{\frac{\bcancel{3}(2x-3)}{\bcancel{3}(x-3)}}$

Now, flip and multiply.

$\frac{2x^2+3x-9}{2(x+3)(x+3)(x-3)} \div \frac{2x-3}{x-3} &= \frac{\cancel{(x+3)} \cancel{(2x-3)}}{2\cancel{(x+3)}(x+3)\cancel{(x-3)}} \cdot \frac{\cancel{x-3}}{\cancel{2x-3}} \\&= \frac{1}{2(x+3)}$

### Vocabulary

Complex Fraction
A fraction with rational expression(s) in the numerator and denominator.

### Practice

Simplify the complex fractions.

1. $\frac{\frac{2x}{5}}{\frac{8}{7}}$
2. $\frac{\frac{4}{x^2-9}}{\frac{6x}{x+3}}$
3. $\frac{\frac{7x^3}{x^2+5x+6}}{\frac{35x^2}{x+2}}$
4. $\frac{\frac{24x+3}{3x+1}}{\frac{16x+2}{6x^2-13x-5}}$
5. $\frac{\frac{4}{x-1} + \frac{1}{x}}{\frac{1}{x} -5}$
6. $\frac{\frac{3x}{x+4} - \frac{1}{x}}{\frac{3x-4}{x^2+6x+8}}$
7. $\frac{8- \frac{3x}{x+5}}{\frac{10}{x+5} + \frac{5}{x+1}}$
8. $\frac{\frac{x}{x+3} - \frac{4}{2x+1}}{\frac{3}{2x+1} + \frac{6}{x^2-9}}$
9. $\frac{\frac{x+3}{x} + \frac{2x}{5-x}}{\frac{3}{2x} - \frac{4x}{x-5}}$
10. $\frac{\frac{2x}{5x^2-13x-6} + \frac{1}{x-3}}{\frac{4}{5x+2} - \frac{5x}{5x^2-3x-2}}$
11. $\frac{\frac{3x}{x^2-4} + \frac{x+4}{x^2+3x+2}}{\frac{x+1}{x^2-x-2} - \frac{2x}{x^2+2x+1}}$

Use the following pattern to answer the next four questions.

$2+\frac{1}{1+\frac{1}{2}}, \ 2+\frac{1}{1+\frac{1}{2+\frac{2}{3}}}, \ 2 + \frac{1}{1+\frac{1}{2+\frac{2}{3+\frac{3}{4}}}}$

1. Find the next two terms in the pattern.
2. Using your graphing calculator, simplify each term in the pattern to a decimal.
3. Make a conjecture about this pattern and the number the terms appear to be approaching.
4. Find the sixth term in the pattern. Does it support your conjecture?

### Vocabulary Language: English

Complex Fraction

Complex Fraction

A fraction with rational expression(s) in the numerator and denominator (a fraction composed of other fractions) is known as a complex fraction.