Solve the following equation by trial and error method:
(i)3m – 14 = 4
Answers
LHS = 3m – 14
By substituting the value of m = 3
Then,
LHS = 3m – 14
= (3 × 3) – 14
= 9 – 14
= – 5
By comparing LHS and RHS
-5 ≠ 4
LHS ≠ RHS
Hence, the value of m = 3 is not a solution to the given equation.
Let, m = 4
LHS = 3m – 14
= (3 × 4) – 14
= 12 – 14
= – 2
By comparing LHS and RHS
-2 ≠ 4
LHS ≠ RHS
Hence, the value of m = 4 is not a solution to the given equation.
Let, m = 5
LHS = 3m – 14
= (3 × 5) – 14
= 15 – 14
= 1
By comparing LHS and RHS
1 ≠ 4
LHS ≠ RHS
Hence, the value of m = 5 is not a solution to the given equation.
Let, m = 6
LHS = 3m – 14
= (3 × 6) – 14
= 18 – 14
= 4
By comparing LHS and RHS
4 = 4
LHS = RHS
Hence, the value of m = 6 is a solution to the given equation.
Answer:
LHS = 3m – 14
By substituting the value of m = 3
Then,
LHS = 3m – 14
= (3 × 3) – 14
= 9 – 14
= – 5
By comparing LHS and RHS
-5 ≠ 4
LHS ≠ RHS
Hence, the value of m = 3 is not a solution to the given equation.
Let, m = 4
LHS = 3m – 14
= (3 × 4) – 14
= 12 – 14
= – 2
By comparing LHS and RHS
-2 ≠ 4
LHS ≠ RHS
Hence, the value of m = 4 is not a solution to the given equation.
Let, m = 5
LHS = 3m – 14
= (3 × 5) – 14
= 15 – 14
= 1
By comparing LHS and RHS
1 ≠ 4
LHS ≠ RHS
Hence, the value of m = 5 is not a solution to the given equation.
Let, m = 6
LHS = 3m – 14
= (3 × 6) – 14
= 18 – 14
= 4
By comparing LHS and RHS
4 = 4
LHS = RHS
Hence, the value of m = 6 is a solution to the given equation.