2. Check whether the value given in the brackets is a solution to the given equation  or not:

      (a) n + 5 = 19 (n = 1)

      (b) 7n + 5 = 19 (n = – 2)

      (c) 7n + 5 = 19 (n = 2)
      (d) 4p – 3 = 13 (p = 1)

      (e) 4p – 3 = 13 (p = – 4)

      (f) 4p – 3 = 13 (p = 0)

Answers (1)

(a) Put n = 1 in the equation, we have :

                           n  +  5  =  15

 or                      1   + 5   = 15 

 or                            6    \neq   15

Thus n = 1 is not a solution.

 

(b) Put  n  =  - 2, we have :

                          7n + 5 = 19

or                   7(-2)  + 5  =  - 14 + 5 = - 9  \neq 19.

So, n = - 2 is not a solution to given equation.  

 

(c) Put  n  =  2, we have :

                          7n + 5 = 19

or                   7(2)  + 5  = 14 + 5 = 19  = R.H.S

Thus n = 2 is the solution for the given equation.

 

(d) Put p = 1 , we have :

                          4p - 3  =  13

or                    4(1)  -  3   =  1   \neq   13 .

Thus p = 1 is not a solution.

 

(e)   Put p = - 4 , we get :

                       4p - 3  =  13

or                    4(1)  -  3   =  1   \neq   13 .

Thus p = 1 is not a solution.

 

 (f)   Put p = 0 , we get :

                       4p - 3  =  13

or                    4(0)  -  3   =  - 3   \neq   13 .

Thus p = 0 is not a solution.

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