Q.1 Complete the last column of the table.

(i) x + 3 = 0
LHS = x + 3
By substituting the value of x = 3
Then, LHS = 3 + 3 = 6
By comparing LHS and RHS
LHS $$\neq$$RHS
Therefore No,the equation is not satisfied
(ii) x + 3 = 0
LHS = x + 3
By substituting the value of x = 0
Then, LHS = 0 + 3 = 3
By comparing LHS and RHS
LHS $$\neq$$RHS
Therefore No, the equation is not satisfied.
(iii) x + 3 = 0
LHS = x + 3
By substituting the value of x = – 3 Then, LHS = – 3 + 3 = 0
By comparing LHS and RHS
LHS = RHS
Hence,
Yes, the equation is satisfied
(iv) x – 7 = 1
LHS = x – 7
By substituting the value of x = 7
Then, LHS = 7 – 7 = 0
By comparing LHS and RHS
LHS $$\neq$$ RHS
Hence
No, the equation is not satisfied
(v) x – 7 = 1
LHS = x – 7
By substituting the value of x = 8
Then, LHS = 8 – 7 = 1
By comparing LHS and RHS
LHS = RHS
Hence,
Yes, the equation is satisfied.
(vi) 5x = 25
LHS = 5x
By substituting the value of x = 0
Then,LHS = 5 × 0 = 0
By comparing LHS and RHS
LHS $$\neq$$RHS
Hence,
No, the equation is not satisfied.
(vii) 5x = 25
LHS = 5x
By substituting the value of x = 5
Then,LHS = 5 × 5 = 25
By comparing LHS and RHS
LHS = RHS
Hence,
Yes, the equation is satisfied.
(viii) 5x = 25
LHS = 5x
By substituting the value of x = -5
Then, LHS = 5 × (-5) = – 25
By comparing LHS and RHS
LHS $$\neq$$ RHS
Hence,
No, the equation is not satisfied.
(ix) m/3 = 2
LHS = m/3
By substituting the value of m = – 6
Then,LHS = -6/3 = – 2
By comparing LHS and RHS
LHS $$\neq$$RHS
Hence,
No, the equation is not satisfied.
(x) m/3 = 2
LHS = m/3
By substituting the value of m = 0
Then, LHS = 0/3 = 0
By comparing LHS and RHS
LHS $$\neq$$RHS
Hence,
No, the equation is not satisfied.
(xi) m/3 = 2
LHS = m/3
By substituting the value of m = 6
Then, LHS = 6/3 = 2
By comparing LHS and RHS
LHS = RHS
Hence,
Yes, the equation is satisfied.