Complex fractions are fractions that have fractions inside the numerator, denominator, or both. To simplify them, it is often helpful to rewrite the numerator and denominator as single fractions, then divide by multiplying by the reciprocal. These examples include numerical complex fractions, algebraic complex fractions, negative exponents, and rational expression variations.
Simplify the following complex fractions
\(\textbf{1)}\) \(\displaystyle\frac{\frac{x}{5}+\frac{1}{3}}{\frac{1}{5}-\frac{1}{6}}\)
The answer is \(6x+10\)
\(\,\,\,\,\,\,\displaystyle\frac{\frac{x}{5}+\frac{1}{3}}{\frac{1}{5}-\frac{1}{6}}\)
\(\,\,\,\,\,\,\displaystyle\frac{\frac{3x}{15}+\frac{5}{15}}{\frac{6}{30}-\frac{5}{30}}\)
\(\,\,\,\,\,\,\displaystyle\frac{\frac{3x+5}{15}}{\frac{1}{30}}\)
\(\,\,\,\,\,\,\displaystyle\frac{3x+5}{15} \div \frac{1}{30}\)
\(\,\,\,\,\,\,\displaystyle\frac{3x+5}{15} \times \frac{30}{1}\)
\(\,\,\,\,\,\,\displaystyle\frac{(3x+5)\cdot 30}{15}\)
\(\,\,\,\,\,\,\displaystyle 2(3x+5)\)
\(\,\,\,\,\,\,\)The answer is \(6x+10\)
\(\textbf{2)}\) \(\displaystyle\frac{\frac{4x^3}{2y}}{\frac{x^2}{8y}}\)
The answer is \(16x\)
\(\,\,\,\,\,\displaystyle\frac{\frac{4x^3}{2y}}{\frac{x^2}{8y}}\)
\(\,\,\,\,\,\displaystyle\frac{4x^3}{2y}\div\frac{x^2}{8y}\)
\(\,\,\,\,\,\displaystyle\frac{4x^3}{2y}\cdot\frac{8y}{x^2}\)
\(\,\,\,\,\,\displaystyle\frac{32x^3y}{2x^2y}\)
\(\,\,\,\,\,\displaystyle 16x\)
\(\,\,\,\,\,\,\)The answer is \(16x\)
![Link to Youtube Video Solving Question Number 2](https://andymath.com/wp-content/uploads/2020/08/play-video.jpg")
\(\textbf{3)}\) \(\displaystyle\frac{1+\frac{1}{x}}{1-\frac{1}{x^2}}\)
The answer is \(\displaystyle\frac{x}{x-1}\)
\(\,\,\,\,\,\displaystyle\frac{1+\frac{1}{x}}{1-\frac{1}{x^2}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x+1}{x}}{\frac{x^2-1}{x^2}}\)
\(\,\,\,\,\,\displaystyle\frac{x+1}{x}\div\frac{x^2-1}{x^2}\)
\(\,\,\,\,\,\displaystyle\frac{x+1}{x}\cdot\frac{x^2}{x^2-1}\)
\(\,\,\,\,\,\displaystyle\frac{(x+1)x^2}{x(x-1)(x+1)}\)
\(\,\,\,\,\,\displaystyle\frac{x}{x-1}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{x}{x-1}\)
\(\textbf{4)}\) \(\displaystyle\frac{\frac{5}{x}-\frac{5}{4x}}{\frac{1}{x}-\frac{5}{8x}}\)
The answer is \(10\)
\(\,\,\,\,\,\displaystyle\frac{\frac{5}{x}-\frac{5}{4x}}{\frac{1}{x}-\frac{5}{8x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{20}{4x}-\frac{5}{4x}}{\frac{8}{8x}-\frac{5}{8x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{15}{4x}}{\frac{3}{8x}}\)
\(\,\,\,\,\,\displaystyle\frac{15}{4x}\div\frac{3}{8x}\)
\(\,\,\,\,\,\displaystyle\frac{15}{4x}\cdot\frac{8x}{3}\)
\(\,\,\,\,\,\displaystyle\frac{120x}{12x}\)
\(\,\,\,\,\,\displaystyle 10\)
\(\,\,\,\,\,\,\)The answer is \(10\)
\(\textbf{5)}\) \(\displaystyle\frac{\frac{8x^4}{3y^3}}{\frac{2x^2}{6y^3}}\)
The answer is \(8x^2\)
\(\,\,\,\,\,\displaystyle\frac{\frac{8x^4}{3y^3}}{\frac{2x^2}{6y^3}}\)
\(\,\,\,\,\,\displaystyle\frac{8x^4}{3y^3} \div \frac{2x^2}{6y^3}\)
\(\,\,\,\,\,\displaystyle\frac{8x^4}{3y^3} \cdot \frac{6y^3}{2x^2}\)
\(\,\,\,\,\,\displaystyle\frac{48x^4y^3}{6x^2y^3}\)
\(\,\,\,\,\,\displaystyle 8x^2\)
\(\,\,\,\,\,\,\)The answer is \(8x^2\)
\(\textbf{6)}\) \(\displaystyle\frac{\frac{x+3}{x-2}-1}{\frac{x+2}{x^2-4}}\)
The answer is \(5\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x+3}{x-2}-1}{\frac{x+2}{x^2-4}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x+3}{x-2}-\frac{x-2}{x-2}}{\frac{x+2}{x^2-4}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x+3-(x-2)}{x-2}}{\frac{x+2}{x^2-4}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{5}{x-2}}{\frac{x+2}{x^2-4}}\)
\(\,\,\,\,\,\displaystyle\frac{5}{x-2}\div\frac{x+2}{x^2-4}\)
\(\,\,\,\,\,\displaystyle\frac{5}{x-2}\cdot\frac{x^2-4}{x+2}\)
\(\,\,\,\,\,\displaystyle\frac{5}{x-2}\cdot\frac{(x-2)(x+2)}{x+2}\)
\(\,\,\,\,\,\displaystyle 5\)
\(\,\,\,\,\,\,\)The answer is \(5\)
\(\textbf{7)}\) \(\displaystyle\frac{\frac{1}{5x}+\frac{1}{6x}}{\frac{1}{5x}-\frac{1}{6x}}\)
The answer is \(11\)
\(\,\,\,\,\,\displaystyle\frac{\frac{1}{5x}+\frac{1}{6x}}{\frac{1}{5x}-\frac{1}{6x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{6}{30x}+\frac{5}{30x}}{\frac{6}{30x}-\frac{5}{30x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{11}{30x}}{\frac{1}{30x}}\)
\(\,\,\,\,\,\displaystyle\frac{11}{30x}\div\frac{1}{30x}\)
\(\,\,\,\,\,\displaystyle\frac{11}{30x}\cdot\frac{30x}{1}\)
\(\,\,\,\,\,\displaystyle 11\)
\(\,\,\,\,\,\,\)The answer is \(11\)
\(\textbf{8)}\) \(\displaystyle\frac{x^{-1}-y^{-1}}{x^{-2}-y^{-2}}\)
The answer is \(\displaystyle\frac{xy}{x+y}\)
\(\,\,\,\,\,\displaystyle\frac{x^{-1}-y^{-1}}{x^{-2}-y^{-2}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{1}{x}-\frac{1}{y}}{\frac{1}{x^2}-\frac{1}{y^2}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{y-x}{xy}}{\frac{y^2-x^2}{x^2y^2}}\)
\(\,\,\,\,\,\displaystyle\frac{y-x}{xy}\div\frac{y^2-x^2}{x^2y^2}\)
\(\,\,\,\,\,\displaystyle\frac{y-x}{xy}\cdot\frac{x^2y^2}{y^2-x^2}\)
\(\,\,\,\,\,\displaystyle\frac{xy(y-x)}{(y-x)(y+x)}\)
\(\,\,\,\,\,\displaystyle\frac{xy}{x+y}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{xy}{x+y}\)
\(\textbf{9)}\) \(\displaystyle\frac{\frac{2}{x}+\frac{3}{x}}{\frac{5}{x}}\)
The answer is \(1\)
\(\,\,\,\,\,\displaystyle\frac{\frac{2}{x}+\frac{3}{x}}{\frac{5}{x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{5}{x}}{\frac{5}{x}}\)
\(\,\,\,\,\,\displaystyle\frac{5}{x}\div\frac{5}{x}\)
\(\,\,\,\,\,\displaystyle\frac{5}{x}\cdot\frac{x}{5}\)
\(\,\,\,\,\,\displaystyle 1\)
\(\,\,\,\,\,\,\)The answer is \(1\)
\(\textbf{10)}\) \(\displaystyle\frac{\frac{3}{x}-\frac{1}{2x}}{\frac{5}{4x}}\)
The answer is \(2\)
\(\,\,\,\,\,\displaystyle\frac{\frac{3}{x}-\frac{1}{2x}}{\frac{5}{4x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{6}{2x}-\frac{1}{2x}}{\frac{5}{4x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{5}{2x}}{\frac{5}{4x}}\)
\(\,\,\,\,\,\displaystyle\frac{5}{2x}\div\frac{5}{4x}\)
\(\,\,\,\,\,\displaystyle\frac{5}{2x}\cdot\frac{4x}{5}\)
\(\,\,\,\,\,\displaystyle 2\)
\(\,\,\,\,\,\,\)The answer is \(2\)
\(\textbf{11)}\) \(\displaystyle\frac{\frac{x}{2}+\frac{x}{3}}{\frac{x}{6}}\)
The answer is \(5\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x}{2}+\frac{x}{3}}{\frac{x}{6}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{3x}{6}+\frac{2x}{6}}{\frac{x}{6}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{5x}{6}}{\frac{x}{6}}\)
\(\,\,\,\,\,\displaystyle\frac{5x}{6}\div\frac{x}{6}\)
\(\,\,\,\,\,\displaystyle\frac{5x}{6}\cdot\frac{6}{x}\)
\(\,\,\,\,\,\displaystyle 5\)
\(\,\,\,\,\,\,\)The answer is \(5\)
\(\textbf{12)}\) \(\displaystyle\frac{\frac{x+1}{x}-1}{\frac{1}{x}}\)
The answer is \(1\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x+1}{x}-1}{\frac{1}{x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x+1}{x}-\frac{x}{x}}{\frac{1}{x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x+1-x}{x}}{\frac{1}{x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{1}{x}}{\frac{1}{x}}\)
\(\,\,\,\,\,\displaystyle 1\)
\(\,\,\,\,\,\,\)The answer is \(1\)
\(\textbf{13)}\) \(\displaystyle\frac{\frac{1}{x}+\frac{1}{x+1}}{\frac{1}{x}}\)
The answer is \(\displaystyle\frac{2x+1}{x+1}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{1}{x}+\frac{1}{x+1}}{\frac{1}{x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x+1}{x(x+1)}+\frac{x}{x(x+1)}}{\frac{1}{x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{2x+1}{x(x+1)}}{\frac{1}{x}}\)
\(\,\,\,\,\,\displaystyle\frac{2x+1}{x(x+1)}\div\frac{1}{x}\)
\(\,\,\,\,\,\displaystyle\frac{2x+1}{x(x+1)}\cdot x\)
\(\,\,\,\,\,\displaystyle\frac{2x+1}{x+1}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{2x+1}{x+1}\)
\(\textbf{14)}\) \(\displaystyle\frac{\frac{2}{x-3}}{\frac{4}{x^2-9}}\)
The answer is \(\displaystyle\frac{x+3}{2}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{2}{x-3}}{\frac{4}{x^2-9}}\)
\(\,\,\,\,\,\displaystyle\frac{2}{x-3}\div\frac{4}{x^2-9}\)
\(\,\,\,\,\,\displaystyle\frac{2}{x-3}\cdot\frac{x^2-9}{4}\)
\(\,\,\,\,\,\displaystyle\frac{2}{x-3}\cdot\frac{(x-3)(x+3)}{4}\)
\(\,\,\,\,\,\displaystyle\frac{2(x+3)}{4}\)
\(\,\,\,\,\,\displaystyle\frac{x+3}{2}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{x+3}{2}\)
\(\textbf{15)}\) \(\displaystyle\frac{\frac{x}{x+2}+\frac{2}{x+2}}{\frac{x+2}{x}}\)
The answer is \(\displaystyle\frac{x}{x+2}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x}{x+2}+\frac{2}{x+2}}{\frac{x+2}{x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x+2}{x+2}}{\frac{x+2}{x}}\)
\(\,\,\,\,\,\displaystyle\frac{1}{\frac{x+2}{x}}\)
\(\,\,\,\,\,\displaystyle 1\div\frac{x+2}{x}\)
\(\,\,\,\,\,\displaystyle 1\cdot\frac{x}{x+2}\)
\(\,\,\,\,\,\displaystyle\frac{x}{x+2}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{x}{x+2}\)
\(\textbf{16)}\) \(\displaystyle\frac{\frac{a}{b}+\frac{b}{a}}{\frac{1}{ab}}\)
The answer is \(a^2+b^2\)
\(\,\,\,\,\,\displaystyle\frac{\frac{a}{b}+\frac{b}{a}}{\frac{1}{ab}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{a^2}{ab}+\frac{b^2}{ab}}{\frac{1}{ab}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{a^2+b^2}{ab}}{\frac{1}{ab}}\)
\(\,\,\,\,\,\displaystyle\frac{a^2+b^2}{ab}\div\frac{1}{ab}\)
\(\,\,\,\,\,\displaystyle\frac{a^2+b^2}{ab}\cdot ab\)
\(\,\,\,\,\,\displaystyle a^2+b^2\)
\(\,\,\,\,\,\,\)The answer is \(a^2+b^2\)
\(\textbf{17)}\) \(\displaystyle\frac{\frac{1}{x+2}-\frac{1}{x-2}}{\frac{4}{x^2-4}}\)
The answer is \(-1\)
\(\,\,\,\,\,\displaystyle\frac{\frac{1}{x+2}-\frac{1}{x-2}}{\frac{4}{x^2-4}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x-2}{(x+2)(x-2)}-\frac{x+2}{(x+2)(x-2)}}{\frac{4}{x^2-4}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x-2-(x+2)}{x^2-4}}{\frac{4}{x^2-4}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{-4}{x^2-4}}{\frac{4}{x^2-4}}\)
\(\,\,\,\,\,\displaystyle\frac{-4}{x^2-4}\div\frac{4}{x^2-4}\)
\(\,\,\,\,\,\displaystyle -1\)
\(\,\,\,\,\,\,\)The answer is \(-1\)
\(\textbf{18)}\) \(\displaystyle\frac{\frac{x^2-4}{x+2}}{\frac{x-2}{3}}\)
The answer is \(3\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x^2-4}{x+2}}{\frac{x-2}{3}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{(x-2)(x+2)}{x+2}}{\frac{x-2}{3}}\)
\(\,\,\,\,\,\displaystyle\frac{x-2}{\frac{x-2}{3}}\)
\(\,\,\,\,\,\displaystyle (x-2)\div\frac{x-2}{3}\)
\(\,\,\,\,\,\displaystyle (x-2)\cdot\frac{3}{x-2}\)
\(\,\,\,\,\,\displaystyle 3\)
\(\,\,\,\,\,\,\)The answer is \(3\)
\(\textbf{19)}\) \(\displaystyle\frac{\frac{2x}{x^2-1}}{\frac{4x}{x^2-2x+1}}\)
The answer is \(\displaystyle\frac{x-1}{2(x+1)}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{2x}{x^2-1}}{\frac{4x}{x^2-2x+1}}\)
\(\,\,\,\,\,\displaystyle\frac{2x}{x^2-1}\div\frac{4x}{x^2-2x+1}\)
\(\,\,\,\,\,\displaystyle\frac{2x}{(x-1)(x+1)}\cdot\frac{(x-1)^2}{4x}\)
\(\,\,\,\,\,\displaystyle\frac{2x(x-1)^2}{4x(x-1)(x+1)}\)
\(\,\,\,\,\,\displaystyle\frac{x-1}{2(x+1)}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{x-1}{2(x+1)}\)
\(\textbf{20)}\) \(\displaystyle\frac{\frac{1}{x}-\frac{1}{x+3}}{\frac{3}{x^2+3x}}\)
The answer is \(1\)
\(\,\,\,\,\,\displaystyle\frac{\frac{1}{x}-\frac{1}{x+3}}{\frac{3}{x^2+3x}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x+3}{x(x+3)}-\frac{x}{x(x+3)}}{\frac{3}{x(x+3)}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{x+3-x}{x(x+3)}}{\frac{3}{x(x+3)}}\)
\(\,\,\,\,\,\displaystyle\frac{\frac{3}{x(x+3)}}{\frac{3}{x(x+3)}}\)
\(\,\,\,\,\,\displaystyle 1\)
\(\,\,\,\,\,\,\)The answer is \(1\)
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