1.2 Fractions (Review)

Terrance Berg

Chapter 1: Algebra Review

1.2 Fractions (Review)

Working with fractions is a very important foundational skill in algebra. This section will briefly review reducing, multiplying, dividing, adding, and subtracting fractions. As this is a review, concepts will not be explained in as much detail as they are in other lessons. Final answers of questions working with fractions tend to always be reduced. Reducing fractions is simply done by dividing both the numerator and denominator by the same number.

Example 1.2.1

Reduce $\dfrac{36}{84}.$

$\begin{array}{rl} \dfrac{36}{84}&\text{Both the numerator and the denominator are divisible by 4.} \\ \\ \dfrac{36\div 4}{84\div 4}=\dfrac{9}{21}&\text{Both the numerator and the denominator are divisible by 3.} \\ \\ \dfrac{9\div 3}{21 \div 3}=\dfrac{3}{7}&\text{Solution} \end{array}$

The previous example could have been done in one step by dividing both the numerator and the denominator by 12. Another solution could have been to divide by 2 twice and then by 3 once (in any order). It is not important which method is used as long as the fraction is reduced as much as possible.

The easiest operation to complete with fractions is multiplication. Fractions can be multiplied straight across, meaning all numerators and all denominators are multiplied together.

Example 1.2.2

Multiply $\dfrac{6}{7}\cdot \dfrac{3}{5}.$

$\begin{array}{rl} \dfrac{6}{7}\cdot \dfrac{3}{5} & \text{Multiply numerators and denominators, respectively.} \\ \\ \dfrac{18}{35} & \text{Solution} \end{array}$

Before multiplying, fractions can be reduced. It is possible to reduce vertically within a single fraction, or diagonally within several fractions, as long as one number from the numerator and one number from the denominator are used.

Example 1.2.3

Multiply $\dfrac{25}{24} \cdot \dfrac{32}{55}.$

$\begin{array}{rl} \dfrac{\cancel{25}\text{ }5}{\cancel{24}\text{ }3}\cdot \dfrac{\cancel{32}\text{ }4}{\cancel{55}\text{ }11} & \text{Reduce 25 and 55 by dividing by 5, and reduce 32 and 24 by dividing by 8.} \\ \\ \dfrac{5\cdot 4}{3\cdot 11}&\text{Multiply numerators and denominators across.} \\ \\ \dfrac{20}{33}&\text{Solution} \end{array}$

Dividing fractions is very similar to multiplying, with one extra step. Dividing fractions necessitates first taking the reciprocal of the second fraction. Once this is done, multiply the fractions together. This multiplication problem solves just like the previous problem.

Example 1.2.4

Divide $\dfrac{21}{16}\div \dfrac{28}{6}.$

$\begin{array}{rl} \dfrac{21}{16}\div \dfrac{28}{6}&\text{Take the reciprocal of the second fraction and multiply it by the first.} \\ \\ \dfrac{\cancel{21}\text{ }3}{\cancel{16}\text{ }8}\cdot \dfrac{\cancel{6}\text{ }3}{\cancel{28}\text{ }4} & \text{Reduce 21 and 28 by dividing by 7, and reduce 6 and 16 by dividing by 2.} \\ \\ \dfrac{3\cdot 3}{8\cdot 4} & \text{Multiply numerators and denominators across.} \\ \\ \dfrac{9}{32} & \text{Solution} \end{array}$

To add and subtract fractions, it is necessary to first find the least common denominator (LCD). There are several ways to find the LCD. One way is to break the denominators into primes, write out the primes that make up the first denominator, and only add primes that are needed to make the other denominators.

Example 1.2.5

Find the LCD of 8 and 12.

Break 8 and 12 into primes:

$\begin{array}{rrl} 8 &= &2 \times 2 \times 2 \\ 12 &= &2 \times 2 \times 3 \end{array}$

The LCD will contain all the primes needed to make each number above.

$\text{LCD}=\rlap{$\overbrace{2\times 2\times 2}^8$}2\times \underbrace{2\times 2\times 3}_{12}=4$

Adding and subtracting fractions is identical in process. If both fractions already have a common denominator, simply add or subtract the numerators and keep the denominator.

Example 1.2.6

Add $\dfrac{7}{8}+\dfrac{3}{8}.$

$\begin{array}{rl} \dfrac{7}{8}+\dfrac{3}{8}&\text{Same denominator, so add }7+3. \\ \\ \dfrac{10}{8}&\text{Reduce answer by dividing the numerator and denominator by 2.} \\ \\ \dfrac{5}{4} &\text{Solution} \end{array}$

While $\dfrac{5}{4}$ can be written as the mixed number $1 \dfrac{1}{4}$ , algebra almost never uses mixed numbers. For this reason, always use the improper fraction, not the mixed number.

Example 1.2.7

Subtract $\dfrac{13}{6}-\dfrac{9}{6}.$

$\begin{array}{rl} \dfrac{13}{6}-\dfrac{9}{6} & \text{Same denominator, so subtract }13-9. \\ \\ \dfrac{4}{6} & \text{Reduce answer by dividing by 2.} \\ \\ \dfrac{2}{3} & \text{Solution} \end{array}$

If the denominators do not match, it is necessary to first identify the LCD and build up each fraction by multiplying the numerator and denominator by the same number so each denominator is built up to the LCD.

Example 1.2.8

Add $\dfrac{5}{6}+\dfrac{4}{9}.$

$\begin{array}{rl} \rlap{$\overbrace{\phantom{2\times 3}}^6$}2\times \underbrace{3\times 3}_9& \text{LCD is }18. \\ \\ \dfrac{3\cdot 5}{3\cdot 6}+\dfrac{4\cdot 2}{9\cdot 2} & \text{Multiply the first fraction by 3 and the second by 2.} \\ \\ \dfrac{15}{18}+\dfrac{8}{18} & \text{Same denominator, so add }15 + 8. \\ \\ \dfrac{23}{18}&\text{Solution} \end{array}$

Example 1.2.9

Subtract $\dfrac{2}{3}-\dfrac{1}{6}.$

$\begin{array}{rl} \dfrac{2}{3}-\dfrac{1}{6} & \text{LCD is 6.} \\ \\ \dfrac{2\cdot 2}{2\cdot 3}-\dfrac{1}{6}& \text{Multiply the first fraction by 2.} \\ \\ \dfrac{4}{6}-\dfrac{1}{6} & \text{Same denominator, so subtract }4-1. \\ \\ \dfrac{3}{6}& \text{Reduce answer by dividing by 3.} \\ \\ \dfrac{1}{2} & \text{Solution} \end{array}$