Page 101 - Mathematics
P. 101

1
               Substitute  b =   into equation (ii)                    17
                             3                                7x −   4y =
               to obtain the value of a.                5.      5x −  4y =  11
                                                            
               From  a +  6b =  1.  It implies that,    6.       0.7x −   0.5y =  2.5

                     1                                      0.7x +  0.3y =  2.9               Tanzania Institute of Education
               a +  6    =  1.                                       0
                                                            
                     3                                7.       2x y+=
                       2
                       a += 1.                               x −  2y =  −  5
                             a = − 1                           4  +  1  =  3
                                                             
                        1          1                    8.        x  y
               But,  a =   and  b =  .  It follows            1  2
                        x          y                           +   =  −  1
                         1      1   1                         x  y
                                                             
               that,  1− =   and   =  .
                         x      3   y                        8 12  =  −  2
                                                                −
               Solving for the values of x and y        9.      x  y
                                                            
               gives,                                         5  +  3  =  1
                                                            
               x = − 1 and  y =  3.                           x  2y
                                                            
                                            3
               Therefore,  x = −  and  y =  is                0.2x +   0.3y =  0.1
                                1
               the solution of the simultaneous         10.     0.3x −  0.1y =  0.7
                                                             
               equation.
                                                              0.05x +   0.02y =  0.01
               Exercise 5.5                             11.     0.02x −  0.03y =  −  0.11
                                                             

               Solve the following simultaneous                1  x +  3y =  3
                                                             
                                                             
               equations by using elimination           12.      2
               method.                                        3  x −  2y =  −  2
                         y
                    2x +=   5                                 4
               1.     
                    4x y−=  7                                 u +  2v =  1
                                                        13.    
                    3pq+ =  6                                3u −  4v =  8
               2.     
                    5pq+ =  8                                 2pq+ =   4
                     5x −   2y =  16                   14.     −  3p +  5q =  −  19
                                                             
               3.     
                      x +  2y =  8                                1 n =  3                       Mathematics Form One
                    8x +   5y =  40                    15.      2m +   3
               4.                                             − 3mn+=  6
                     9x −  5y =  5                           




                                                  95




                                                                                        25/10/2024   09:51:32
   Mathematics form 1.indd   95                                                         25/10/2024   09:51:32
   Mathematics form 1.indd   95
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