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SYSTEMATIC METHOD FOR SOLVING AN EQUATION  EXAMPIE:6   Solve the equation Bx = 24 and check the result.
        SOLUTION              8x = 24
 We can compare an equation with a balance.               8x      24  I dividing both sides by 81
                                       =
 If equal weights are put in the two pans. we find that the two pans   8  8
 remain in balance.                 x=3.
                                    x = 3 is the solution of the given equation.
 If we remove equal weights from the two pans, we find that the two  CHECK  Substituting x = 3 in the given equation, we get
 pans remain in balance.             LHS = 8 x 3 = 24 and RHS = 24.


 Thus, we can add (and, therefore, multiply) equal weights or amounts   LHS           RHS  when x = 3, we have: LHS = RHS.
                                            2
                                              x
 to both the pans to keep them in balance.  EXAMPIE:7 Solve the equation       = 18 and check the result.
                                            3
                                      2
                                           3
                                                  3
 Also, we can lessen (and. therefore, divide) equal amounts from both   SOLUTION   2    x 18 ⇒                         [multiplying both sides by   ]  2
                                        x×−
                                              18×
                               =
 pans to keep the pans in balance.   3  3  2      2                                     3
                                2 3
                                       x
                                                x
 Similarly, in the case of an equation, we have the following rules.  ⇒ × ×=  27 ⇒ = 27.
                                3 2
 Rule (i): We can add the same number to both the sides of an equation.  :. x= 27 is the solution of the given equation.
 Rule (ii): We can subtract the same number from both the sides of an equation.  CHECK  Substituting x = 27 in the given equation, we get
 Rule (iii):We can multiply both the sides of an equation by the same nonzero number.
 Rule (iv): We can divide both the sides of an equation by the same nonzero number.  LHS  =  2 ×  27 18 and RHS =
                                                       =
                                               3
 Using these rules, we can solve linear equations easily.
                                           when x = 27, we have: LHS = RHS.
        TRANSPOSITION You know that one can add or subtract a number from both sides of the
 EXAMPIE:4   Solve the equation x -5 = 7 and check the result.
 SOLUTION               x - 5 = 7  equation. So, for the equation x -4 = 5, we can write
                x - 5 + 5 = 7 + 5 [ adding 5 to both sides] => x = 12.      x - 4 + 4 = 5 + 4     x = 5 + 4.
 So, x = 12 is the solution of the given equation.  Similarly, for the equation x + 5 = 3, we can write
               x + 5 -5 = 3 -5      x = 3 -5.
 CHECK  Substituting x = 12 in the given equation, we get   In both these cases you will notice that after this operation, the number appears on the other side
          LHS = 12 - 5 = 7 and RHS = 7.    of the equation, but with the opposite sign. So, you can straightaway change the sign of a term
 :. when x = 12, we have: LHS = RHS.  and transfer it from one side of an equation to the other side. This is called
 EXAMPIE:5   Solve the equation 8 + x= 3 and check the result.  transposition.
 8 + x = 3
                 8 + x -8 = 3 -8 [subtracting 8 from both sidetimes]  EXAMPIE:8
 CHECK      Substituting x = -5 in the given equation, we get LHS = 8 -5 = 3 and RHS = 3.   Solve: 3x + 5 = 13 -x. Check the result.
 :. when x = -5, we have: LHS = RHS.  SOLUTION  3x +5 = 13 -x
                                     3x +x = 13 -5 [transposing-x to LHS and +5 to RHSI

                                     4x= 8

                            4x     8
                                 =     (dividing both sides by 41
                             4     4

                              x=2.

                         ∴    x = 2 is the solution of the given equation.
        CHECK                 Substituting x = 2 in the given equation, we get
                              LHS=3 × 2 + 5 = 11 and RHS = 13 - 2 = 11.

                         ∴    LHS =RHS, when x = 2.
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