5p+2 =17 Putting p = -3 in L.H.S. 5(-3)+2 = -15+2 =-13 ∵ -13 ≠ 17 Therefore, p = -3 is not the solution. Putting p = -2 in L.H.S. 5(-2)+2 = -10+2 =-8 ∵ -8 ≠ 17 Therefore, p = -2 is not the solution. Putting p = -1 in L.H.S. 5(-1)+2 = -5+2 =-3 ∵ -3 ≠ 17 Therefore, p = -1 is not the solution. PuttingRead more
5p+2 =17
Putting p = -3 in L.H.S. 5(-3)+2 = -15+2 =-13
∵ -13 ≠ 17 Therefore, p = -3 is not the solution.
Putting p = -2 in L.H.S. 5(-2)+2 = -10+2 =-8
∵ -8 ≠ 17 Therefore, p = -2 is not the solution.
Putting p = -1 in L.H.S. 5(-1)+2 = -5+2 =-3
∵ -3 ≠ 17 Therefore, p = -1 is not the solution.
Putting p = 0 in L.H.S. 5(0)+2 = 0+2 = 2
∵ 2 ≠ 17 Therefore, p = 0 is not the solution.
Putting p = 1 in L.H.S. 5(1)+2 = 5+2 = 7
∵ 7 ≠ 17 Therefore, p = 1 is not the solution.
Putting p = 2 in L.H.S. 5(2)+2 = 10+2 = 12
∵ 12 ≠ 17 Therefore, p = 2 is not the solution.
Putting p = 3 in L.H.S. 5(3)+2 = 15+2 = 17
∵ 17 ≠ 17 Therefore, p = 3 is the solution.
n+5=19(n=1) Putting n = 1 in L.H.S., 1 + 5 = 6 ∵ L.H.S. ≠ R.H.S., ∴ n=1 is not the solution of given equation. Class 7 Maths Chapter 4 Exercise 4.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
n+5=19(n=1)
Putting n = 1 in L.H.S.,
1 + 5 = 6
∵ L.H.S. ≠ R.H.S.,
∴ n=1 is not the solution of given equation.
Solve the following equations by trial and error method: 5p+2 =17
5p+2 =17 Putting p = -3 in L.H.S. 5(-3)+2 = -15+2 =-13 ∵ -13 ≠ 17 Therefore, p = -3 is not the solution. Putting p = -2 in L.H.S. 5(-2)+2 = -10+2 =-8 ∵ -8 ≠ 17 Therefore, p = -2 is not the solution. Putting p = -1 in L.H.S. 5(-1)+2 = -5+2 =-3 ∵ -3 ≠ 17 Therefore, p = -1 is not the solution. PuttingRead more
5p+2 =17
Putting p = -3 in L.H.S. 5(-3)+2 = -15+2 =-13
∵ -13 ≠ 17 Therefore, p = -3 is not the solution.
Putting p = -2 in L.H.S. 5(-2)+2 = -10+2 =-8
∵ -8 ≠ 17 Therefore, p = -2 is not the solution.
Putting p = -1 in L.H.S. 5(-1)+2 = -5+2 =-3
∵ -3 ≠ 17 Therefore, p = -1 is not the solution.
Putting p = 0 in L.H.S. 5(0)+2 = 0+2 = 2
∵ 2 ≠ 17 Therefore, p = 0 is not the solution.
Putting p = 1 in L.H.S. 5(1)+2 = 5+2 = 7
∵ 7 ≠ 17 Therefore, p = 1 is not the solution.
Putting p = 2 in L.H.S. 5(2)+2 = 10+2 = 12
∵ 12 ≠ 17 Therefore, p = 2 is not the solution.
Putting p = 3 in L.H.S. 5(3)+2 = 15+2 = 17
∵ 17 ≠ 17 Therefore, p = 3 is the solution.
Class 7 Maths Chapter 4 Exercise 4.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
Check whether the value given in the brackets is a solution to the given equation or not: n+5=19(n=1)
n+5=19(n=1) Putting n = 1 in L.H.S., 1 + 5 = 6 ∵ L.H.S. ≠ R.H.S., ∴ n=1 is not the solution of given equation. Class 7 Maths Chapter 4 Exercise 4.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/
n+5=19(n=1)
Putting n = 1 in L.H.S.,
1 + 5 = 6
∵ L.H.S. ≠ R.H.S.,
∴ n=1 is not the solution of given equation.
Class 7 Maths Chapter 4 Exercise 4.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-4/