Addition and Subtraction of Fractions.

The algebraic sum of two or more fractions which have the same denominator, is a fraction whose numerator is the algebraic sum of the numerators of the given fractions, and whose denominator is the common denominator of the given fractions. Hence,

add fractions,

Reduce the fractions to equivalent fractions having the same denominator; and write the algebraic sum of the numerators of these fractions over the common denominator.

133. When the denominators are simple expressions.

1. Simplify ${\displaystyle \frac{3a-4b}{4}}-{\displaystyle \frac{2a-b+c}{3}}+{\displaystyle \frac{a-4c}{12}}$.

The $\text{L.,C.,D.}=12$.

The multipliers, that is, the quotients obtained by dividing $12$ by $4$, $3$, and $12$, are $3$, $4$, and $1$.

Hence the sum of the fractions equals

$$\begin{gathered} \&\frac{9a-12b}{12}-\frac{8a-4b+4c}{12}+\frac{a-4c}{12} \\ \&=\frac{9a-12b-8a+4b-4c+a-4c}{12} \\ \&=\frac{2a-8b-8c}{12}=\frac{a-4b-4c}{6}. \end{gathered}$$

The preceding work may be arranged as follows:

The $\text{L.,C.,D.}=12$.

The multipliers are $3$, $4$, and $1$, respectively.

{l*{5}{cr}cl} 3(3a &-& 4b & & ) &=& 9a &-&12b & & &=& \text{1st numerator.} \ \llap{$-$} 4(2a &-& b &+& c) &=&-8a &+& 4b &-& 4c &=& \text{2d numerator.} \ 1( a &-& & &4c) &=& a & & &-& 4c &=& \text{3d numerator.} \ \cline{7-11} & & & & & & 2a &-& 8b &-& 8c \ & & & & &\rlap{or}& 2(a &-& 4b &-& 4c) &=& \text{the sum of the numerators.} \ \therefore\ \text{sum of fractions} = \frac{2(a - 4b - 4c)}{12} = \frac{a - 4b - 4c}{6}.

Exercise 46.

Find the sum of:

1. ${\displaystyle \frac{x+1}{2}}+{\displaystyle \frac{x-3}{5}}+{\displaystyle \frac{x+5}{10}}$.

2. ${\displaystyle \frac{2x-1}{3}}+{\displaystyle \frac{x+5}{4}}+{\displaystyle \frac{x-4}{6}}$.

3. ${\displaystyle \frac{7x-1}{6}}-{\displaystyle \frac{3x-2}{7}}+{\displaystyle \frac{x-5}{3}}$.