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6x – 3x + y – y = 15 – 9   4x + 3x + 2y + (–2y) = 20 + 1  3x+  2y =  8
  3x = 6          4x + 3x + 2y – 2y = 20 + 1             
 3x  =  6         7x = 21                                 2x+ 3y =  12
 3  3             7x   21                                Solution
 Hence, x = 2.    7  =  7  .                             Label the equations:
 Therefore,  x = 2 and y = 3.  Hence, x = 3.                                                     Tanzania Institute of Education
                 Eliminate x to get y:                    3x+  2y =  8             (1)
 Example 5.31    The coefficent of  in (1) is 2x          2x+ 3y = 12              (2)
                                                         
 Solve the following simultaneous   The coefficent of  in (2) is 3x        Choose a variable to be eliminated.
 equations by elimination method:   Make the coefficient of x in (1) and   Eliminate x to get y:
  2x += 10      (2) to be the same.                     The coefficent of  in (1) is 3x
 y
  3x−  =        Multiply equation (1) by 3 and (2)      The coefficent of  in (2) is 2x
   2y  1
                 by 2:                                   Make the coefficients of x in (1) and
   Solution       3(2x +  ) y =  10 3×                  (2) to be the same.
 Label the equations:  
                       2 (3x −  2 ) 1 2×  y =           Multiply equation (1) by 2 and
  2x += 10  (1)                                         equation (2) by 3:
 y
  3x−  =  FOR ONLINE READING ONLY
                           30
                                           (5)
                      3y =
                 6x + 
   2y  1  (2)  −  6x− 4y =  2                           2(3x+   2 ) y =  8 2
                                                                        ×
                
 Choose the variable to be eliminated.     (6)            3(2x+  3 ) 12 3
                                                                  y =
                                                                         ×
                                                         
 Eliminate y to get x .  Then, subtract (6) from (5)
                   6x +   3y =  30                        6x +   4y =  16          (3)
 The coefficent of  in (1) is 1y  −                      6x +  9y =  36
                   6x−
                       4y =
                            2
                  
 The coefficent of  in (2) is 2y  6x – 6x + 3y – (–4y) = 30 – 2                     (4)
 Make the coefficients of y in (1)                       Then, subtract (3) from (4).
 and (2) to be the same.  6x – 6x + 3y + 4y = 30 – 2        6x +   9y =  36
 Multiply equation (1) by 2 and   7y = 28                −  6x +  4y =  16
                                                           
 equation (2) by 1.  7y  =  28                            6x −  6x +  9y−+  y =  −
                                                                        ( 4 ) 36 16
 ×
  2 (2x y+  ) =  10 2  7  7                              6x −  6x +  9y−  4y= 36 16
                                                                               −
                                                         
                 y = 4.
  1 (3x −  2 ) 1 1 ×  y =  Therefore, x = 3 and y = 4.  5y  = 20 .
 4x+   2y =  20  (3)  (3)                                5    5
           (4)                                          Hence, y = 4.
  3x−  2y = 1  (4)  Example 5.32
   Add equations (3) and (4):  Solve the following linear   Eliminate y to get x:                  Mathematics Form One
 4x+   2y =  20  simultaneous equations by              The coefficent of  in (1) is 2y
 +                                                      The coefficent of  in (2) is 3y
  3x−  2y = 1   elimination method:



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                                                                                        25/09/2025   15:01:12
   Mathematics form 1.indd   93
   Mathematics form 1.indd   93                                                         25/09/2025   15:01:12
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