How to subtract complex fractions

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ACT Math › How to subtract complex fractions

Questions 1 - 4

Simplify,

$\frac{\frac{ab}{c} + \frac{c}{a}}{\frac{b}{c}} - a$

$\frac{c^2}{ab}$

$\frac{ba^2 + c}{ab}$

$\frac{a^2bc + bc^2}{a}$

$\frac{ab + c - a}{c^2}$

$\frac{b}{ac}$

Explanation

Convert the numerators and denominators into single fractions, then simplify.

Start by finding the lowest common denominator in both the numerator and denominator of the complex fraction.

$\frac{\frac{ab}{c} + \frac{c}{a}}{\frac{b}{c}} - a$

$\frac{\frac{ab}{c}\left ( \frac{a}{a} \right ) + \frac{c}{a}\left ( \frac{c}{c}\right )}{\frac{b}{c}} - a$

$\frac{\frac{a^2b}{ac} + \frac{c^2}{ac}}{\frac{b}{c}} - a$

Add fractions with like denominators.

$\frac{\frac{a^2b + c^2}{ac}}{\frac{b}{c}} - a$

Simplify. Divide complex fractions by multiplying the numerator by the reciprocal of the denominator.

$\frac{\frac{a^2b + c^2}{ac}}{\frac{b}{c}} - a = \frac{a^2b + c^2}{ac}\cdot \frac{c}{b} - a = \frac{a^2b + c^2}{ab} - a$

Solve.

$\frac{a^2b + c^2}{ab} - a\left ( \frac{ab}{ab} \right ) = \frac{a^2b + c^2}{ab} - \frac{a^2b}{ab} = \frac{c^2}{ab}$

Simplify:

$\frac{4}{\frac{4-12}{6}}-\frac{\frac{3}{4}}{5}$

$\frac{-63}{20}$

$\frac{40}{3}$

$\frac{-3}{7}$

$\frac{-15}{4}$

$\frac{3}{7}$

Explanation

Begin by simplifying the denominator of the first fraction:

$\frac{4}{\frac{4-12}{6}}-\frac{\frac{3}{4}}{5}=\frac{4}{\frac{-8}{6}}-\frac{\frac{3}{4}}{5}$

Now, remember that division of fractions is done by multiplying the numerator by the reciprocal of the denominator. Thus:

$\frac{4}{\frac{-8}{6}}-\frac{\frac{3}{4}}{5}=4*\frac{-6}{8}-\frac{3}{4}*\frac{1}{5}$

Simplify a bit:

$\frac{-3}{1}-\frac{3}{20}=\frac{-60}{20}-\frac{3}{20}=\frac{-63}{20}$

Subtract: $\frac{3}{2}-\frac{2}{3}-\frac{4}{5}$

$\frac{1}{30}$

$-\frac{4}{5}$

$-\frac{1}{2}$

$\frac{1}{6}$

Explanation

The least common multiple can be found by multiplying the denominators: 2, 3, and 5. The common denominator of these numbers is 30. Multiply the numerator with what was multiplied to the denominator of each term, and then solve.

$\frac{3}{2}-\frac{2}{3}-\frac{4}{5}$

$\frac{3(3*5)}{2(3*5)}-\frac{2(2*5)}{3(2*5)}-\frac{4(2*3)}{5(2*3)}$

$\frac{45}{30}-\frac{20}{30}-\frac{24}{30}=\frac{1}{30}$

What is $\frac{\frac{4}{9}}{2} - \frac{\frac{5}{12}}3{}$ ?

$\frac{1}{12}$

$\frac{3}{18}$

$\frac{5}{36}$

$\frac{1}{36}$

$\frac{1}{4}$

Explanation

First, simplify both sides. $\frac{\frac{4}{9}}{2}$ becomes $\frac{4}{18}$ and $\frac{\frac{5}{12}}{3}$ becomes $\frac{5}{36}$ . The LCF between $\frac{4}{18}$ and $\frac{5}{36}$ is 36. Thus, $\frac{8}{36} - \frac{5}{36} = \frac{3}{36}.$ This simplifies to $\frac{1}{12}$ .

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