Addition and Subtraction of Rational Numbers

 

To add or subtract rational numbers, the denominators of the fractions must be equal. In order to perform addition or subtraction between fractions with unequal denominators, the fractions are first expanded or simplified so that their denominators become equal at the least common multiple (LCM) of the denominators.

When adding (or subtracting) fractions with equal denominators, the sum (or difference) of the numerators is calculated and written in the numerator, and the common denominator is written in the denominator.

$$\frac{a}{b} \pm \frac{c}{b} = \frac{a \pm c}{b}$$

 

Examples:

 

$ \bullet \quad \displaystyle\frac{13}{5} +\displaystyle \frac{19}{5} – \displaystyle\frac{7}{5} =\displaystyle \frac{25}{5} = 5 $

 

$ \bullet \quad \left(\displaystyle\frac{1}{3} +\displaystyle \frac{12}{5} \right) -\left(\displaystyle\frac{1}{3}+ \displaystyle\frac{9}{5} \right)+3 $

$=\displaystyle\frac{1}{3}+ \displaystyle\frac{12}{5}-\displaystyle\frac{1}{3} -\displaystyle\frac{9}{5} +3$

$=\displaystyle\frac{1-1}{3}+\displaystyle\frac{12-9}{5} +3=\displaystyle\frac{3}{5} +3 = 3\displaystyle\frac{3}{5}$

 

$ \bullet \quad \displaystyle\frac{1}{2}+1+ \displaystyle\frac{3}{2} +2 =\displaystyle\frac{1+3}{2} +1 +2= \displaystyle\frac{4}{2} +3 =2+3=5 $

 

$\bullet \quad \displaystyle\frac{1}{3} + \displaystyle\frac{2}{3} + \displaystyle\frac{3}{3}+ \cdots + \displaystyle\frac{20}{3} $

$= \displaystyle\frac{1+2+3+4+\cdots +20}{3} = \displaystyle\frac{\displaystyle\frac{20 \cdot 21}{2}}{3}$

$=\displaystyle\frac{10 \cdot 21}{3} = 10 \cdot 7 =70$

 

$\bullet \quad \displaystyle\frac{1}{5} – \displaystyle\frac{2}{5} +\displaystyle \frac{3}{5} -\displaystyle \frac{4}{5} + \cdots + \displaystyle\frac{49}{5} – \displaystyle\frac{50}{5} $

$ =\displaystyle \frac{1+3+ \cdots + 49}{5}\quad -\quad\displaystyle \frac{2+4+ \cdots + 50}{5} $

$\text{Where } t_1 \text{ and } t_2 \text{ represent the number of terms:}$

$49=2t_1-1 \Rightarrow t_1=25 \quad \text{and} \quad 50= 2t_2 \Rightarrow t_2=25$

$$=\displaystyle\frac{(25)^2}{5} – \displaystyle\frac{25 \cdot 26}{5} = \displaystyle\frac{25 \cdot (25-26)}{5} = 5 \cdot (-1)= -5$$
*(Note: Replaced a standard computer dot with a proper mathematical multiplication sign in the algebraic equation steps).*

 

$ \bullet \quad \text{Given that } a+1= 2 \displaystyle\frac{1}{2} – \displaystyle\frac{1}{4}, \text{ let’s find } a. $

$a +1= \displaystyle\frac{5}{2} – \displaystyle\frac{1}{4} \Rightarrow a+1 =\displaystyle \frac{10-1}{4} =\displaystyle \frac{9}{4} $

$ \Rightarrow a= \displaystyle\frac{9}{4} -1= \displaystyle\frac{(9-1 \cdot 4)}{4} $

$\Rightarrow a= \displaystyle \frac{5}{4} \text{ is found.}$

 

Warning:

 

$$\displaystyle\frac{a}{b} \pm \displaystyle\frac{c}{d} = \displaystyle\frac{ a \cdot d \pm b \cdot c }{b \cdot d}$$

$$a \displaystyle\frac{b}{c}= a+ \displaystyle\frac{b}{c} = \displaystyle\frac{a \cdot c+b}{c}$$

$$-a\displaystyle \frac{b}{c} =-\left(a+ \displaystyle\frac{b}{c} \right)= -\displaystyle\frac{a \cdot c+b}{c}$$

 

Examples:

 

\(\bullet \quad \text{In the operation } 2 \displaystyle\frac{2}{5} = n+1 \displaystyle \frac{3}{5}, \text{ let’s find the value of } n. \)

$2 \displaystyle\frac{2}{5} = n+1\displaystyle \frac{3}{5} \Rightarrow 2+\displaystyle \frac{2}{5}= n+1 + \displaystyle\frac{3}{5} $

$ \Rightarrow 1 +\displaystyle \frac{2}{5} – \displaystyle\frac{3}{5} =n $

$\Rightarrow \displaystyle\frac{4}{5} =n $

 

\(\bullet \quad \text{In the equation } 3-1 \displaystyle\frac{1}{3} = x+ \displaystyle\frac{2}{3}, \text{ let’s find the value of } x. \)

$3-1\displaystyle \frac{1}{3} = x+ \displaystyle\frac{2}{3} \Rightarrow 3 -\left(1+ \displaystyle\frac{1}{3} \right) = x+\displaystyle \frac{2}{3} $

$\Rightarrow 2- \displaystyle\frac{1}{3}- \displaystyle\frac{2}{3} = x $

$\Rightarrow x = 1$

 

 

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