Page 32 - ICSE Math 7
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Addition and subtraction

                    To add or subtract fractions, we have two cases:
                    (a)   If the fractions to be added or subtracted are like fractions, then just add or subtract their numerators
                        since their denominators are the same.
                    (b)   If the fractions to be added or subtracted are unlike fractions, then convert them into like fractions
                        and then perform addition or subtraction as explained above.
                        Alternatively,
                                      Numerator of the first fraction × (LCM of denominators ÷ denominator
                                        of the first fraction) + numerator of the second fraction × (LCM of
                                               denominators ÷ denominator of the second fraction)
                              Sum =
                                                              LCM of denominators
                         This rule can be used for any number of fractions. For subtraction of fractions, replace ‘+’ with
                        ‘–’ sign.
                    (c)   If the fractions to be added or subtracted are mixed fractions, then convert all the mixed fractions
                        into improper fractions. After performing addition or subtraction on these fractions reduce the
                        result to its simplest form and if the result is an improper fraction, convert it to a mixed fraction.

                    Example 9:  Simplify the following.
                                      20    9            1     3     7
                                  (a)     –      (b)  2   – 4   + 7
                                      21   14            3     5     10
                    Solution:     (a)  LCM of 21 and 14 = 42

                                      20  =  20 × 2  =  40                      ( 42 ÷ 21 = 2)
                                      21    21 × 2   42
                                       9   =   9 × 3   =  27                    ( 42 ÷ 14 = 3)
                                      14    14 × 3   42

                                      So,  20  –   9   =  40  –  27  =  13
                                          21   14    42   42   42
                                        1   2 × 3 + 1    7                                     Try These
                                  (b)  2   =           =
                                        3       3        3                                    1.   Find the value of x, if
                                        3
                                                                                                  3
                                                                                                      x
                                      4   =  4 × 5 + 3  =  23                                    1  + 2  = 3 17 .
                                                                                                          20
                                                                                                  5
                                                                                                      4
                                        5       5        5                                    2.   What should be added to
                                                                                                                      5
                                      7  7   =  7 × 10 + 7  =  77                                to get  ?            6
                                                                                                      7
                                        10       10        10                                         4
                                      LCM of 3, 5 and 10 = 30
                                      7  =  7 × 10  =   70
                                      3    3 × 10    30
                                      23    23 × 6   138
                                          =         =
                                       5    5 × 6     30
                                      77  =  77 × 3  =  231
                                      10    10 × 3    30

                                            1     3      7    7   23   77    70   138    231    163      13
                                      So, 2   – 4   + 7   =    –     +     =    –      +       =     = 5
                                            3     5     10    3    5   10    30    30     30     30      30


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