Which of the following equations has $x = 2$ as a solution?
(A) $x + 2 = 5$
(B) $x - 2 = 0$
(C) $2x + 1 = 0$
(D) $x + 3 = 6$

Hint: An equation satisfies only when the left-hand side of the equation is equal to the right- hand side of the equation. So here we have to put the given value of x in that equation where the condition satisfies that the right hand side is equal to the left hand side.

Complete step by step answer:
In the above question, it is given that $x = 2$ and we have to find that equation from the above options in which the right hand side will be equal to the left hand side.
Now we will put the value of x in the above equations one by one and then find the correct option.
In option (A): We have $x + 2 = 5$
On putting $x = 2$ in the above equation, we get
$4 \ne 5$
On putting $x = 2$ in the above equation we get $4$ on the left hand side and $5$ on the right hand side.
Therefore (A) is the wrong option.

In option (B), $x - 2 = 0$
On putting $x = 2$ in the above equation, we get
$0 = 0$
On putting $x = 2$ in the above equation we get $0$ on the left hand side and $0$ on the right hand side.
Therefore (B) is the correct option.

In option (C): We have,
$2x + 1 = 0$
On putting $x = 2$ in the above equation, we get
$5 \ne 0$
On putting $x = 2$ in the above equation we get $5$ on the left hand side and $0$ on the right hand side.
Therefore (C) is the wrong option.

In option (D): We have,
$x + 3 = 6$
On putting $x = 2$ in the above equation, we get
$5 \ne 6$
On putting $x = 2$ in the above equation we get $5$ on the left hand side and $6$ on the right hand side.
Therefore (D) is the wrong option.

Therefore, option (B) is the only correct option.

Note:
The above equations are linear equations in one variable. If we show them graphically, then it represents a straight line either parallel to x-axis or parallel to y-axis passing through a particular point on the axis. It can be either positive or negative.
The line $x-2=0$ is a straight line parallel to y-axis as shown below passing through the point $(0,2)$.