2. Check whether the value given in the brackets is a solution to the given equationor not:(a) n + 5 = 19 (n = 1)(b) 7n + 5 = 19 (n=-2) (c) In + 5 = 19 (n=2)(d) 4p – 3 = 13 (p = 1) (e) 4p – 3 = 13 (p = -4) (1) 4p – 3 = 13 (p = 0) About the author Ivy
Answer: see this is your answer Step-by-step explanation: (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. Reply
Answer:
see this is your answer
Step-by-step explanation:
(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.