The correct option is A 7x - 1 = 50
We have, x = 7
Put x = 7 in all the equations and equation which doesn't satisfy x = 7 can not be formed by taking x = 7.
(i) 2x+1=15
Put x = 7
2×7+1=15⟹15=15
So, x = 7 satisfies the equation 2x+1=15.
Hence equation 2x+1=15 can be formed by taking x = 7.
(ii) 7x−1=50
Put x = 7
7×7−1=50
⟹48=50, which is not true.
So, x = 7 does not satisfy equation 7x−1=50.
Hence equation 7x−1=50 can not be formed by taking x = 7.
(iii) x−3=4
Putting x = 7, we have
7−3=4, which is true.
So x = 7 satisfies the equation x−3=4.
Hence equation x−3=4 can be formed by taking x = 7.
(iv) x7−1=0
Putting x = 7, we have
77−1=0, which is true.
So x = 7 satisfies the equation x7−1=0.
Hence equation x7−1=0 can be formed by taking x = 7.