Page 21 - Maths Skills - 8
P. 21

Rational Numbers                                                                                        19


               Let’s Attempt


        Example 1:  What should be added to     - 7  to get  13  ?  Example 2:  Subtract  −   from  − 3  5  .
                                                 16         20                              5       6
        Solution:      Let the required number be x.                               5     3       5   3

                                                                   Solution:
                           - 7      13                                            6      5      6    5
                                 x
                               +=
                           16        20                                           LCM of 6 and 5 is 30.
                               13    −  7  13   7

                       or x =     −      =   +                                 (  55   )    (  36 )       25 18       7
                               20     16   20  16                                     30              30      30  .
                                  (13× 4) +(7× 5)
                                =                                  Example 3:  Multiply   7  and  3  .
                                         80                                               9      4

                                  52 +35    87     7                                                 7    3    71
                                =         =    =1                  Solution:     Required product  =    ×
                                     80     80     80                                                9  3  4     34

                               7                       7
                       Thus, 1    has to be added to -    to                                      =  7  .
                               80                     16                                            12
                       get  13  .
                             20
                                         -14       7                   - 6      5
        Example 4:  Divide the sum of         and    by the product of     and
                                           15                            11        12 .
                                                   9
                                   -14      7    −14    7   − 42  + 35  − 7
        Solution:      The sum of      and    =       +   =           =                                         ...(i)
                                            9
                                   15             15    9      45       45
                                      - 6      5    −  6    5    −  5
                       The product of     and     =     ×      =                                               ...(ii)
                                        11       12  11    12  2  22

                       Dividing (i) by (ii), we get   − 7  ÷  − 5  =  − 7  ×  − 22  =  154
                                                  45   22    45     5     225
        Example 5:  Give the additive inverse of  (i)  5          (ii)  - 3
                                                                        7
                                                      9
                                                   5    - 5        5   −  5
        Solution:      (i)  The additive inverse of  is     , since  +     =  0
                                                         9           9  9
                                                   9
                       (ii)   The additive inverse of  - 3  is  3  , since  − 3  +  3  = 0
                                                     7    7          7   7
                                                            2            -19                         7   2
        Example 6:  Give the multiplicative inverse of  (i)         (ii)            (iii)  5     (iv)    ×
                                                              3           17                         8   5
                                                        2     3         2   3
        Solution:      (i)  The multiplicative inverse of  is   ,  since    ×  =1
                                                          3   2         3   2
                                                        -        -19  17     −19    −17
                       (ii)  The multiplicative inverse of    is      , since     ×      =1
                                                         17       19          17     19
                                                            1
                      (iii)  The multiplicative inverse of 5 is  .
                                                            5
                      (iv)  Since   7     2     7  . The multiplicative inverse of   7  is  20  .
                                  8     5   20                               20    7
                                    4
   16   17   18   19   20   21   22   23   24   25   26