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

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.