Question:

Complete the last column of the table
S. No.EquationValueSay, whether the Equation No. is Satisfied. (Yes/ No)
(i)x + 3 = 0x = 3 
(ii)x + 3 = 0x = 0 
(iii)x + 3 = 0x = – 3 
(iv)x – 7 = 1x = 7 
(v)x – 7 = 1x = 8 
(vi)5x = 25x = 0 
(vii)5x = 25x = 5 
(viii)5x = 25x = – 5 
(ix)\(\frac{m}{3}=2\)m = – 6 
(x)\(\frac{m}{3}=2\)m = 0 
(xi)\(\frac{m}{3}=2\)m = 6 

Updated On: Dec 11, 2023
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Solution and Explanation

(i) x + 3 = 0
L.H.S. = x + 3
By putting x = 3,
L.H.S. = 3 + 3 = 6 ≠ R.H.S.
∴ No, the equation is not satisfied.


(ii) x + 3 = 0
L.H.S. = x + 3
By putting x = 0,
L.H.S. = 0 + 3 = 3 ≠ R.H.S.
∴ No, the equation is not satisfied.


(iii) x + 3 = 0
L.H.S. = x + 3
By putting x = - 3,
L.H.S. = - 3 + 3 = 0 = R.H.S.
∴ Yes, the equation is satisfied.


(iv) x - 7 = 1
L.H.S. = x - 7
By putting x = 7,
L.H.S. = 7 - 7 = 0 ≠
∴ R.H.S. No, the equation is not satisfied.


(v) x - 7 = 1
L.H.S. = x - 7
By putting x = 8,
L.H.S. = 8 - 7 = 1 = R.H.S.
∴ Yes, the equation is satisfied.


(vi) 5x = 25
L.H.S. = 5x
By putting x = 0,
L.H.S. = 5 × 0 = 0 ≠ R.H.S.
∴ No, the equation is not satisfied


(vii) 5x = 25
L.H.S. = 5x
By putting x = 5,
L.H.S. = 5 × 5 = 25 = R.H.S.
∴ Yes, the equation is satisfied.


(viii) 5x = 25
L.H.S. = 5x
By putting x = - 5,
L.H.S. = 5 × ( - 5) = - 25 ≠ R.H.S.
∴ No, the equation is not satisfied.


(ix) \(\frac{m}{3}\)= 2
L.H.S. = \(\frac{m}{3}\)
By putting m = - 6,
L.H.S. = \(\frac{-6}{3}\)= - 2 ≠ R.H.S.
∴ No, the equation is not satisfied. ≠ R.H.S.


(x) \(\frac{m}{3}\) = 2
L.H.S. = \(\frac{m}{3}\)
By putting m = 0,
L.H.S. =\(\frac{0}{3}\) = 0
L.H.S. = R.H.S.
∴ Yes, the equation is satisfied


(xi) \(\frac{m}{3}\) = 2
L.H.S. =\(\frac{m}{3}\)
By putting m = 6,
L.H.S. = \(\frac{6}{3}\) = 2
L.H.S. = R.H.S.
∴ Yes, the equation is satisfied

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