Explanation
Step -1: Considering option A.
We have given, x−5=1
Putting x=−1 we get,
−1−5=6
⇒−6=1 Which is not possible.
Here, L.H.S≠R.H.S
So, - 1 is not a solution of x−5=6.
Step -2: Considering option B.
We have given, 2x+5=7
2(−1)+5=7
⇒−2+5=7
⇒3=7 Which is not possible.
Here, L.H.S ≠ R.H.S
So, - 1 is not a solution of 2x+5=7.
Step -3: Considering option C.
We have given, 2(x−2)+6=0
2(−1−2)+6=0
⇒2(−3)+6=0
⇒−6+6=0
⇒0=0 Which is possible.
Here, L.H.S = R.H.S
So, - 1 is a solution of 2(x−2)+6=0.
Step -4: Considering option D.
We have given, 3x+5=4
3(−1)+5=4
⇒−3+5=4
⇒2=4 Which is not possible.
So, - 1 is not a solution of 3x+5=4.
Hence, option C. 2(x−2)+6=0 is correct answer.
We have given, x+1=0
−1+1=0
Here, L.H.S=R.H.S
So, - 1 is a solution of x+1=0.
We have given, 3x+4=1
3(−1)+4=1
⇒−3+4=1
⇒1=1 Which is possible.
So, - 1 is a solution of 3x+4=1.
We have given, 5x+7=2
5(−1)+7=2
⇒−5+7=2
⇒2=2 Which is possible.
So, - 1 is a solution of 5x+7=2.
We have given, x−1=2
−1−1=2
⇒−2=2 Which is not possible.
So, - 1 is not a solution of x−1=2.
Hence, option D. x−1=2 is correct answer.
Please disable the adBlock and continue. Thank you.